Physics S3 SB

PHYSICS
For Rwanda Schools
Student’s Book
Senior 3
N. Kaboyo
C. Kariuki
P. Kimani
Published by
Longhorn Publishers (Rwanda) Ltd

Remera opposite COGE Bank
P.O. Box 5910
Kigali, Rwanda
Longhorn Publishers (Kenya) Ltd

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P.O. Box 18033-00500
Nairobi, Kenya
Longhorn Publishers (Uganda) Ltd

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Longhorn Publishers (T) Ltd

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Dar es Salaam, Tanzania
© N. Kaboyo, C. Kariuki, P. Kimani 2017
All rights reserved. No part of this publication may be reproduced, stored

in a retrieval system or transmitted in any form or by any means, electronic,
mechanical, photocopying, recording or otherwise without the prior written
permission of the Copyright owner.
First published 2017
ISBN 978 9997 77 112 4
Printed by Ramco Printing Works Ltd,

Unit 2, Ramco Industrial Complex,
Before Imara Daima Turn off, Mombasa Road,
P. O. Box 27750 - 00506,
Nairobi, Kenya.
Contents
1. Graph of linear motion ............................................................ 1
Key unit competence........................................................................................ 1

Unit Focus Activity .......................................................................................... 2
1.1 Uniform and non-uniform linear motion.................................................. 3
1.2 Plotting graphs of linear motion............................................................... 4
1.3 Analysing graphs of linear motion.......................................................... 14
Unit summary and new words................................................................ 18
Unit Test 1 .......................................................................................... 19
2. Friction Force and Newton’s Laws of Motion ............................. 24
Key unit competence ..................................................................................... 24

Unit Focus Activity ........................................................................................ 25
2.1 Newton’s First law of motion ............................................................... 26
2.2 Linear momentum and impulse............................................................. 28
2.3 Newton’s second law of motion ............................................................. 31
2.4 Newton’s third law of motion................................................................. 36
2.5 Conservation of linear momentum......................................................... 40
2.6 Coefficient of friction............................................................................. 47
Unit summary and new words ............................................................... 52
Unit Test 2 .......................................................................................... 53
3. Applications of Atmospheric Pressure....................................... 57
Key unit competence...................................................................................... 57

Unit Focus Activity......................................................................................... 58
3.1 Existence of atmospheric pressure.......................................................... 59
3.2 Factors influencing atmospheric pressure .............................................. 62
3.3 Instruments for measuring atmospheric pressure ................................... 64
3.4 Applications of atmospheric pressure...................................................... 67
Unit summary and new words ............................................................... 73
Unit Test 3 ........................................................................................... 74
iii
4. Renewable and non-renewable energy sources ........................... 76
Key unit competence...................................................................................... 76
Unit Focus Activity......................................................................................... 77
4.1 Energy sources...................................................................................... 78
4.2 Classification and characteristics of energy sources................................. 78
4.3 Energy transformations.......................................................................... 97
Unit summary and new words...............................................................103
Unit Test 4 ......................................................................................... 104
5. Heat Transfer and Quantity of heat..........................................107

Key unit competence ....................................................................................107
Unit Focus Activity........................................................................................108
5.1 Heat and temperature...........................................................................109
5.2 Modes of heat transfer..........................................................................110
5.3 Applications of heat transfer..................................................................118
5.4 Thermal expansion...............................................................................122
5.6 Quantity of heat...................................................................................138
Unit summary and new words ..............................................................162
Unit Test 5 ..........................................................................................163
6. Laws of thermodynamics .......................................................167

Key unit competence.....................................................................................167
Unit focus Activity.........................................................................................168
6.1 Introduction to themodynamics............................................................169
6.2 Internal energy of a system....................................................................170
6.3 First law of thermodynamics.................................................................171
6.4 Second law of thermodynamics.............................................................174
6.5 Heat exchange......................................................................................177
6.6 Change of state and kinetic theory of matter..........................................179
6.7 Applications of the principle of thermodynamics....................................183
6.8 Melting and solidification.....................................................................185
Unit summary and new words ..............................................................190
Unit Test 6 …..................................................................................... 191
iv
7. Introduction to Electromagnetic Induction …........................... 194
Unit focus activity..........................................................................................195
7.1 Demonstrations of electromagnetic induction........................................196
7.2 Factors affecting the magnitude of emf induced.....................................201
7.3 Laws of electromagnetic induction........................................................202
7.4 E.m.f induced in a straight conductor moving in a straight field..............207
7.5 E.m.f induced in a coil rotating in a uniform magnetic field....................210
7.6 Alternating current (a.c) generator........................................................215
7.7 Root-mean square (r.m.s) value............................................................217
7.8 Other applications of electromagnetic induction....................................220
Unit test 7 .......................................................................................... 224
8. Electrical Power Transmission ............................................... 228

Key unit competence.................................................................................... 228
Unit focus Activity........................................................................................ 229
8.1 Structure and working of a transformer.................................................230
8.2. Types of transformers ..........................................................................233
8.3 Transformers equation ........................................................................ 235
8.4 Power losses in transformers.................................................................240
8.5 Application of transformers .................................................................242
8.6 Electric power transmission...................................................................244
8.7 Environmental impact of power generation and transmission..................248
Unit test 8 …...................................................................................... 251
9. Electric Field Intensity ......................................................... 253

9.1 Electrostatic Force and coulombs Law..................................................255
9.2 Superposition of forces.........................................................................258
9.3 Electric field.........................................................................................261
9.4 Electric Field Intensity..........................................................................265
v
9.5 Electricfield patterns.............................................................................270
Unit test 9............................................................................................274
10. House Electric Installation .................................................... 277

Key unit competence ....................................................................................277
10.1 Standard symbols for electrical installation............................................279
10.2 Electrical lamps and fuses ....................................................................280
10.3 Types of electrical cables and their sizes.................................................286
10.4 Household wirings................................................................................290
10.5 Dangers of electricity............................................................................294
10.6 Electrical safety....................................................................................295
Unit test 10......................................................................................... 297
11. Basic alternating current circuits ........................................... 299

11.1 Standard symbols used in electric circuit and function...........................301
11.2 Differences between alternating current (a.c) and direct current (d.c).....302
11.3 The ciruit for analysing resistors, capacitors and inductors in a.c circuit..304
11.4 A single resistor connected in series to an a.c source..............................305
11.5 A single capacitor connected in series to an a.c source............................310
11.6 A single inductor connected in series to an a.c source.............................315
11.7 Resistor, inductor and capacitor (RLC) in series with an a.c source........321
Unit test 11 .........................................................................................328
12 Refraction of light .................................................................331

Key unit competence............................................................................331
Unit focus activity................................................................................332...
12.1 Phenomena of refraction of light.................................................334
12.2 Refraction of light through a prism.............................................355
12.3 Refraction of light through a thin lens ........................................367
vi
Unit summary and new words.....................................................397
Unit test 12 ...............................................................................399
13. Telecommunication Channels................................................ 406

Key unit competence ...........................................................................406
Unit Focus Activity ..............................................................................407
13.1 Definition of terms used in communication..................................407
13.2 Types of data Signals...................................................................409
13.3 Data transmission media.............................................................411
Unit test 13 ............................................................................... 429
14. Properties of physical processes affecting plant growth.............. 432

Key unit competence............................................................................432
14.1 Environmental factors and their impact on plant growth ..............434
14.2 Biotic factors..............................................................................452
Unit test 14 .............................................................................. 455
15. Environmental phenomena and related physical concept........... 459

Key unit competence ...........................................................................459
15.1 Environment and energy transfer ................................................461
15.2 Application of laws of thermodynamics in energy transfer in the
environment............................................................................... 461
15.3 Modes of heat transfer in the environment ..................................462
15.4 Noise Pollution...........................................................................466
15.5 Air pollution...............................................................................468
15.6 Structure and composition of the atmosphere ............................. 472
15.7 Climate change science...............................................................474
15.8 The hydrosphere and hydrologic Cycle........................................482
15.9 Clouds......................................................................................485
15.10 Cyclone and anticyclones...........................................................488
15.11 Global convectional currents and wind patterns...........................490
vii
15.12 Thermoregulation and the physics laws that govern it.................492
Unit test 15 ........................................................................................ 499
Appendix AP1.............................................................................501
Appendix AP2.............................................................................502
Grossary.....................................................................................505
References .................................................................................508
viii
UNIT 1 Graphs of linear motion
Key Unit Competence

By the end of this unit, the learner should be able to plot and analyse the graphs
of linear motion.
Learning objectives
Knowledge and understanding
• Describe graphs of uniform rectilinear motion.

• Plot distance - time graphs and velocity- time graphs.
• Identify uniform velocity and non-uniform velocity from displacement-time
graphs.
• Identify uniform acceleration and non-uniform acceleration from
velocity–time graphs.
• Explain why the slope of a distance-time graph gives speed.
• Explain why the slope of a distance-time gives speed.
Skills
• Appreciate use of suitable scale in plotting graphs.
• Recognise that the area under velocity-time graph represents distance covered by the
body.
• Determine the speed of a body from a distance time graph.
• Interpret velocity-time graphs.
Attitude and value

• Appreciate that the slope of distance -time graph gives velocity while the slope of
velocity-time graph gives acceleration.
• Appreciate that linear motion can be represented using graphs.
1
Introduction
Unit Focus Activity

Materials
• A small ball
• A dish
Steps
1. Place the dish on a level ground.
2. Place the ball directly above the dish at a reasonable height.
3. Allow the ball to fall freely onto the dish as shown in Fig. 1.1. Observe the
ball as it hits the dish and make consecutive rebouncies.
Fig. 1.1: A bouncing ball
4. Compare the consecutive rebounces. What do you notice? Account for the
differences in rebounce heights (if any).
2
Graphs of linear motion
5. Starting from the time the ball hits the dish, sketch for the first three
rebounces its:
(a) Displacement – time graph
(b) Speed – time graph
(c) Velocity – time graph
In our daily lives, we come across or interact with objects in motion. For example,
people, animals and objects are from time to time involved in motion in various
directions. The motion in a straight line is called linear motion, also referred to
as rectilinear motion.
Graphs of linear motion help us to visualise and analyse various aspects of the
motion including distance and displacement covered, speed and velocity, direction
of motion and acceleration.
In this unit, we will draw motion graphs for objects moving in linear motion,
analyse and interpret the motions as represented in the graphs.
1.1 Uniform and non-uniform linear motion
Activity 1.1 To identify uniform and non-uniform linear motion
(a) Distinguish between uniform and non-uniform motion.

(b) Identify uniform linear motions and non-uniform linear motions from the
following graphs. Give a reason in each case to support your decision.
Displacement (m)
Displacement (m)
Δs
Δt
Time (s) Time (s)
(a) (b)
Displacement (m)
Displacement (m)
Time (s) Time (s)

(c) (d)
Fig. 1.2 (a) to (d): Graphs of uniform and non-uniform linear motion
3
Speed (m/s)
Speed (m/s)
Time (s) Time (s)
(e) (f)
Velocity (m/s)
Speed (m/s)
2t
0
t Time (s)
0 -1
t 2t
Time (s)
(g) (h)
Fig. 1.2 (e) to (f): Graphs of uniform and non-uniform linear motion
Uniform linear motion is a type of motion in which the body moves with constant
velocity. In other words, it moves with zero acceleration or along a straight path
with constant speed. An example of this kind of motion is a car moving along a
straight section of road at a constant speed. Non uniform linear motion is the
kind of motion in which a body moves with a varying velocity. An example of a
non-linear motion is a bouncing ball.
1.2 Plotting graphs of linear motion
1.2.1 Distance-time graphs
(a) Distance-time graph of an object moving with uniform speed
Activity 1.2 To plot a distance-time graph for an object moving

with uniform speed
Materials
• Graph paper
• Geometrical set
4
Steps
Table 1.1 shows the distance travelled with time by an object.
Table 1.1:Distance travelled with time by an object
Distance (m) 0 8 16 24 32 40
Time (s) 0 2 4 6 8 10
1. Plot a graph of distance against time from the data given in Table 1.1.
2. Describe the motion of the object based on the graph obtained.
Note: When drawing graphs determine a convenient scale that will

give a graph that fits at least 34 of a normal graph paper.
The change in the position of an object with time can be represented on the
distance-time graph. Such a graph helps us to visualise and analyse various aspects
of the motion, e.g. the average speed or velocity of the body. Let us consider a
moving body whose distance changes with time as shown in the table 1.2.
Distance (km) 0 6 18 27 35 42 48
Time (s) 0 5 15 22.5 29 35 40
The distance-time graph of the body is as shown in Fig 1.3.
50
s2
Distance (m)
40
s1
30 A C
20
10
10 20 t1 30 t2 40
Time (s)
Fig. 1.3: Distance-time graph
5
From the graph in Fig. 1.3, the object travels equal distance in equal time intervals.
It moves with uniform speed.Therefore, for a body moving with uniform speed, it's
distance-time graph is a straight line. In other words, the distance is increasing at a
uniform rate. The term velocity is used instead of speed if the direction is specified.
Determination of Speed/Velocity from distance-time graph

In the graph in Fig 1.3, the distance travelled from A to B is obtained as follows:
Distance = s2 – s1 = 45 m – 30 m = 15 m
(where s1 is the initial distance and s2 is the final distance.)
The time taken = t2 – t1 = 37.5 s – 25.0 s = 12.5 s
(where t1 is time at s1 and t2 is time at S2.)
Distance travelled 15 m
Speed = =
time taken 12.5 s
= 1.2 m/s
∴ We see that,
speed = slope of the graph (see triangle of ABC in Fig. 1.3).
Since the speed is not changing, this is a case of non-accelerating motion.
(b) Distance-time graph for an object moving with non-uniform speed
Activity 1.3 To plot a distance-time graph of an object with

non-uniform speed
Material
• Graph paper • Geometrical set
Steps
1. Consider a case of an object moving as shown in table 1.3
Table 1.3 Distance travelled with time by an object
Time (s) 0 2 4 6 8 10 12
Distance (m) 0 1 4 9 16 25 36
2. Plot a graph of distance against time from data given in Table 1.3.
3. Describe this type of motion based on your graph.
4. Find the gradient of the graph at point (6.2, 10). What does it represent?
The motion of a body moving with non-uniform speed can be presented on a

graph. Such a graph helps us to identify time intervals when the body was either
at rest, accelerating or moving at a constant speed. It also helps us to determine
its velocity at any instance or average velocity in any given time interval.
6
Let us consider a case similar to the one in Activity 1.3. The distance covered
with time by a body was recorded as shown in table 1.4.
Distance (m) 0 5 12.5 25 45

Time (s) 0 2 4 6 8
The distance-time graph of the body is as shown in Fig. 1.4
60
55
50
40
30
Distance (m)
20
10 s
2
s1
1 2 3 4 5 6 7 8 9 10
t1 t2 t3 t4
Time t(s)
Fig. 1.4: Distance-time graph for non-uniform motion
The graph shows variation in the rate of distance covered in time. This represent
motion with non-uniform speed.
7
Determination of speed/velocity from the graph

Change in distance ∆s
The slope (gradient) of the graph i.e Change in time or gradient = ∆t where
∆s is change in distance and ∆t is change in time, represents speed/velocity of the
object at any given point. Therefore, we can find the speed of the object at any
instance by finding the slope of the graph at that point.
To do this, we draw a tangent to the curve at that point i.e a line touching the
graph at only that point, then find the gradient of that line.
s –s 10 – 0 10
At point A, instantaneous speed vA = 2 1 = 3.5 – 0.5 = 3 = 3.3 m/s
t 2 – t1
s3 – s2 55 – 10 45
At point B, instantaneous speed vB= = 10 – 4 = 6 = 7.5 m/s
t 4 – t3
We can see that the speed of the object is higher at point B than at point A hence
the body is accelerating.
Exercise 1.1
1. With the help of a sketch graph, explain how a body undergoes uniform
linear motion.
2. What does the gradient from a distance-time graph represent? Explain.
3. Uwimana carried out an experiment with an object moving at different
distances in different times and recorded the findings as shown on the table 1.5.
Time(s) 20 30 40 50 60
Distance (m) 12 16 20 24 28
(a) Plot a graph of distance (y-axis) against time.
(b) Find the slope of the graph.
(c) Describe the motion of the object.
(d) Give two reasons why drawing graphs of linear motion is important.
4. By giving an example, differentiate between uniform and non-uniform motion.
1.2.2 Velocity-time graphs

(a) Velocity-time graph for an object moving at a constant velocity
Activity 1.4 To show the variation of velocity with time using a

ticker-timer
1. The table 1.6 shows velocity and time for a car moving in straight line.
Table 1.6:Velocities of a car at given times
Time (h) 0 15 25 30
velocity (ms )-1
40 40 40 40
2. Plot a graph of velocity against time.
3. Describe the motion based on the graph obtained.
8
The velocity-time graph for a body moving at constant velocity is a straight

line parallel to the x-axis. (Fig 1.5).
A B
Velocity(m/s)
Time (s)

Fig 1.5: Velocity time graph
Since velocity is constant, then
displacement covered
Velocity = time
∴ Velocity × Time = Displacement

Looking at the graph in Fig. 1.5, we see that the distance covered from A to
B is (given by velocity × time covered) is equal to the area of the rectangle
bounded by AB and the time axis. Therefore, distance covered by a body
moving at constant velocity or speed from a velocity time graph is equal to
the area under the graph in the section under consideration.
Example 1.1
Fig. 1.6 shows the graph of velocity against time for body.
50
B C
40
30
Velocity (km/h
20
10
A D
t1 1 2 3 t2 4
Time (hour)
Fig. 1.6: Velocity-time graph for a body with constant velocity
9
Find the distance covered by the object between time t1 and t2.
Solution
To determine the distance moved by the object between time t1 and t2 from
the graph, we draw perpendicular lines t1 and t2 to the graph lines to get
rectangle A B C D.
The distance moved = Velocity × Time
= Area of rectangle ABCD
= AB × AD
= 40 m s-1 × (3.5 – 0.5)s
= (40 × 3) m
= 120 m
The velocity does not change with time. This is a case of non-accelerated motion.
Facts
1. The area under velocity-time graph gives the distance covered by
the body.
2. The gradient/slope of the velocity-time graph represents acceleration
and in this case the slope is zero hence acceleration is zero.
(b) Velocity – time graph for an accelerating/decelerating object.
Activity 1.5 To determine the distance moved by an accelerating

object from a velocity-time graph
Materials
• A long tape • wooden block • a runway
• ticker-timer • trolley • carbon disc
• cellotape
Steps
1. Pass the long tape under the carbon disc of the ticker-tape timer. Use
cellotape to attach it to the trolley.
2. Now, increase the angle of inclination of the runway using the wooden block
until the trolley is seen to be moving with increasing speed down the runway
(See Fig 1.7).
ticker-timer
cellotape
carbon disc
trolley
tape runway
Support
Fig. 1.7: To determine uniform velocity using a ticker-tape timer
10
3. Release the trolley and start the ticker-timer. What do you notice about the
separation of adjacent dots on the tape? Explain.
4. Cut the tape through the dot produced just before the trolley was released.
5. Count five dot-to-dot spaces and cut the tape again. (If the dots are too close
together to distinguish them, then you will have to estimate the ten spaces.)
6. Starting from your last cut, count ten more spaces, and cut again.
7. Repeat this, until you have a collection of consecutive tapes, each one longer
than the one before it. Number your tapes, in the order 1 onwards.
8. Draw a horizontal line on a sheet of paper. Make a 'bar chart' by sticking
the tapes in order vertically side by side, so that their bottoms just touch the
horizontal line. The first and shortest tape should be at the left hand end of
the line as shown in Fig.1.8.
B
Velocity (cm/s)
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 Time (s)
Fig. 1.8: Tape chart for constant acceleration

9. What does the horizontal axis represent in this case?. Mark your horizontal
axis in intervals of 0.2s. What are the units for the horizontal axis?
10. Draw a vertical line through the zero of the time axis. This vertical line is the
velocity axis. When time was zero for the trolley journey, velocity was zero.
(The trolley started from standstill, or rest.)
11. Mark the scale on the vertical axis, in centimetres per second.
12. Draw a smooth line through the top-centre of each tape on your velocity-
time graph. Describe in words what the graph says about the trolley motion.
When the velocity of a body is not constant, the body is either accelerating or
decelerating. Consider an object whose motion is shown in the velocity-time
graph in Fig 1.9.
11
Velocity (m/s)
∆v
∆t
Time (s)
Fig. 1.9
Change in velocity ∆v
The gradient = = ∆t = acceleration
Time taken
Thus, the gradient of a velocity-time graph gives the acceleration of the object.
When the graph has a uniform slope (gradient), the body has uniform acceleration.
Consider a motion of a body whose velocity at regular time is as shown in table 1.7.
Table 1.7
Time (s) 0 5 10 15 20 25 30
velocity (ms-1) 0 9 18 27 36 45 54
The velocity-time graph of the body is as shown in Fig. 1.10.
50
40 v2 B
18 m
Velovity (ms-1)
30
40 ms-1
v1 C
20
10
A
10 t1 20 t2 30 40
Time (sec)
Fig. 1.10: Graph of velocity against time
12
Change in velocity between t1 and t2 = v2 – v1

= 40 m/s – 22 m/s
= 18 m/s
Change in time = t2 – t1
= 22.5 s – 12.5 s
= 10 s
v2 – v1
Slope = t – t = 1810m/s
s
= 1.8 m/s2
2 1
The velocity changes at the same rate in unit time. This is a case of uniformly
accelerated motion.
When a velocity-time graph is a curve, then the body is moving with either increasing
or decreasing acceleration as shown in the graphs in Fig. 1.11 and Fig, 1.12.
Velocity
Velocity
Time Time
Fig. 1.11: Graph of body moving Fig. 1.12: Graph of body moving
with increasing acceleration with decreasing acceleration

The instantaneous acceleration of the object at any instant of a body movig with
non-uniform speed is given by the gradient of the graph at that point, in the same
way we determined the instantaneous speed from the graph of a body moving with
non-uniform speed earlier in this unit.
Note:
The slope/gradient of the velocity-time graph represents the
acceleration of the body.
Exercise 1.2
1. Differentiate between speed and velocity.
2. In a velocity-time graph, the slope obtained stand for?
3. The velocity of a body increases from 60 km/h to 90 km/h in 20 seconds.
Calculate its acceleration.
4. An applied force changes the velocity of an object from 20 m/s to 36 m/s in
0.01 seconds. What is the acceleration produced?
13
1.3 Analysing graphs of linear motion

Example 1.2
Table 1.8 shows the data for the motion of a vehicle over a period of 7s.
Table 1.8:Displacement of a car with time
Time (s) 0 1 2 3 4 5 6 7
Displacement (m) 0 20 40 60 80 95 105 110
(a) Plot a graph of displacement against time.
(b) Describe the motion of the vehicle for the first 4 s.
(c) (i) Determine the velocities at 5.0 s and 6.5 s. Hence or otherwise determine
the average acceleration of the vehicle over this time interval.
(ii) Comment on your answer.
Solution
(a) Fig. 1.13 shows the expected graph
140 Y
B
120
20 m
A
100
C
Displacement (m)
80
60 X Z
40
t = 2s
20
1 2 3 4 5 6 7 8
Time (sec)
Fig. 1.13: Displacement-time graph
14
(b) The graph shows that vehicle moves with constant velocity (uniform
velocity) in the first 4 s i.e. since the displacement is directly proportional
to time i.e. linear relationship. The acceleration is zero.
(c) Instantaneous velocity t = 5.0 s = slope at point t = 5.0 s.
∆YZ (140 – 60) m/s
` = ∆XZ =
(8 – 2.5) s
80 m/s
= 5.5 s
= 14.5 m/s
Instantaneous velocity at t = 6.5 s = slope at point t = 6.5 s
∆BC (121 – 100) m/s
= ∆AC = (8.5 – 4.5) s
21 m/s
= 4.0 s
= 5.25 m/s
v–u (5.25 – 14.5) m/s
Average acceleration = ∆v
t
= t = (5.5 – 40) s
-9.25 m/s
= 1.5 s
≈ -6.17 m/s2
(ii) The negative sign means that the vehicle is decelerating.
Example 1.3
Figure 1.14 shows a velocity-time graph of an object with uniform motion.
Velocity (m/s)
Time (s)
Fig 1.14: Velocity-time graph
(a) Describe the motion of the object.

(b) Sketch the displacement-time graph of the motion (take motion upward as
positive).
15
Solution
(a) The object is decelerated uniformly and finally comes to rest.
(b) Fig. 1.15 shows the displacement time graph for the motion.
Displacement (m) Time (s)

Example 1.4
Two cars X and Y are driven on the same 120 km trip. Car X travels at
80 km/h all the time. Car Y starts at the same time as, driving at 96 km/h, but
the driver stops for fifteen minutes after he has travelled for half an hour then
continues at the same speed.
(a) Plot a graph of speed against time for two cars.
(b) Which car is the first to arrive at the destination.
Solution
(a) Fig 1.16 shows the motion and the expected graph of the two cars.
100 Car Y
Car X
80
Speed (Km/hr)
60
40
20
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6

Time (hr)
Fig. 1.16: Speed-time graph of two cars
16
(b) Car X travels 120 km at 80 km/h

120 km
The time it takes = Distance covered = 80 km/h = 1.5 h
Speed
Car Y travels the full distance of 120 km at 96 km/h
120 km
Its total time in motion = Distance covered = 96 km/h = 1.25 h
Speed
Car Y rests for 15 minutes = 15 h = 0.25 h
60
∴ total time by car Y = (1.25 + 0.25)h
= 1.5 h
∴ The two cars arrived at the same time.
Exercise 1.3
1. What do you understand by:

(a) Accelerated motion.
(b) Non-accelerated motion.
(c) Give 2 examples for the motions in (a) and (b).
2. Show that:
(a) The slope of a displacement-time graph gives the velocity.
(b) The slope of a velocity-time graph gives acceleration.
(c) The area under the graph of velocity-time gives the displacement.
3. The data for a journey made by a car is given in table 1.9
Table 1.9:Data for a car's journey
Time interval Duration of interval (h) Speed (km/h)

1 0.10 32
2 0.40 96
3 0.20 32
(a) How far does the car go in the first interval?
(b) Determine the total distance of the journey.
(c) Plot a graph of speed against time for this journey.
(d) Indicate the area on your graph which corresponds to the first
22.5 km/h of travel.
17
Unit summary and new words

• Uniform linear motion is the motion of a body moving with constant velocity
in a straight line.
• A displacement-time graph shows the variation of the displacement of an
object with time. For example;
d d
t t
(a) Body moving with constant (b) Body at rest
velocity away from origin original
d d
t t
(c) Body moving with increasing (d) Body moving with decreasing
acceleration acceleration

• The gradient (slope) of a displacement-time graph at any instance gives a
velocity at that instance.
• A velocity-time graphs shows the variation of the velocity of a body with time.
For example,
v v v
t t t
Body moving constant or Body moving zero Body moving constant or
uniform acceleration acceleration uniform deceleration
Fig. 1.18: Velocity-time graph
18
• Instantaneous speed is the speed of a body at a specific moment in time.

• The gradient of a velocity-time graph at any instance gives the acceleration
of the object at that point.
• The area under a velocity-time graph gives you the displacement of the
object in the time interval under consideration.
• The following are acceleration – time graph for bodies moving with constant
acceleration and uniformly increasing acceleration respectively.
a a
t t
(a) Body moving (b) Body moving with uniformly
constant acceleration increasing acceleration
Fig. 1.19: Acceleration-time graph
• When choosing the scale for plotting a graph, it must be simple and uniform.
Unit Test 1
For questions 1 – 7, select the correct answer from the choices given.
1. The slopes of a velocity-time graph is the
A. speed of body B. velocity of body
C. acceleration of body D. distance the body travelled
2. A body has a constant velocity when
i) acceleration is increasing.
ii) it is moving in a straight line.
iii) the net force on the body is zero.
A. (iii) only B. (i) and (ii) only
C. (i) & (iii) only D. (ii) and (iii) only
3. The velocity-time graph in Fig. 1.20 shows the motion of an object moving
with
A. decreasing acceleration B. constant acceleration
C. an increasing acceleration D. constant velocity
19
Velocity (m/s)
Time (s)
Fig. 1.20: Motion graph for an object
4. A body moves with uniform acceleration if

A. Its momentum remains constant.
B. It covers equal distances in equal time.
C. The velocity change by equal amount in equal times.
D. The net force on the body is zero.
5. A cyclist travelling at a constant acceleration of 2 m s-2 passes through two
points A and B in a straight line. If the speed of A is 10 ms-1 and the points
are 75 m apart, find the speed at B.
A. 15.8 m s-1 B. 17.3 m s-1
C. 20.0 m s-1 D. 400.0 m s-1
6. A lift accelerates from rest for 3 s. It then moves at uniform velocity for 15 s then
decelerates uniformly for 2 s before coming to rest. Which of the following
velocity - time graph in Fig. 1.21 represents the motion of the lift.
A. B.
V
V
3 15 20 t(s)
3 40 20 t(s)
C. D.
V
V
3 40 t(s)
3 18 20 t(s)
Fig. 1.21: velocity - time graph for a lift
20
7. Which of the following displacement time graph in Fig. 1.22 shows a car
moving away from traffic lights at a steady speed towards the starting point?
Displacement
Displacement
Displacement
Displacement
Time(s)
Time(s) Time(s) Time(s)

A B C D
Fig. 1.22:Displacement time graphs
12
8. The graph in Fig. 1.23 represents variation of 10 B
Velocity(m/s)
velocity with time of two athletes A & B. 8
(a) Describe the motion of A and B 6
(b) What distance was covered by B in 50 s? 4 A

2
0
10 20 30 40 50 t(s)
9. (a) What is meant by uniform velocity? Fig. 1.23: Velocity-time graph

(b) A car travelling with a uniform velocity of 25 ms-1 for 5 s travels and
then comes to rest under a uniform deceleration in 8 s.
(i) Sketch a velocity-time graph of the motion.
(ii) Find the total distance travelled.
10. Define the following:
(a) Instantaneous velocity (b) Average velocity
11. Explain how you can determine:
(a) Velocity from a displacement-time graph.
(b) Displacement from a velocity-time graph.
(c) Acceleration from velocity-time graph.
12. Sketch a velocity-time graph to show the motion of a body moving with:
(a) uniform acceleration. (b) zero acceleration.
13. Table 1.10 shows the instantaneous speed of a vehicle at intervals of
1 second.
Table 1.10:Instantaneous speeds of a car with time
Time (s) 0.0 1.0 2.0 3.0 4.0 5.0 6.0

Speed (m/s) 10.0 12.4 14.8 17.2 19.6 22.0 24.4
21
(a) Plot a graph of speed against time .

(b) Use your graph to determine:
(i) How fast the vehicle is travelling at 2.6 s and 4.8 s.
(ii) How far the vehicle travelled between the two instants in part (i)
(iii) Slope of the graph.
14. The distance-time graph of a motorbike travelling along a road is as shown
in Fig. 1.24. Plot a graph of its speed against time.
50
45
40
35
30
Distance (m)
25
20
15
10
0____
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
Time (sec)
Fig. 1.24: Distance-time graph of a motorbike
Be Safe!!
Do you know that motorcycles contribute most of the deaths caused by road
accidents in Rwanda? Always follow traffic rules. Wear helmets whenever
you are on a motorbike, and never be more than two people on a motorbike.
22
15. A train moves according to the speed-time graph shown in Fig. 1.25.
90
80
70
Speed (km/hr)
60
50
40
30
20
10
0
0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18
Time (hr)
Fig. 1.25: Speed-time graph of a train
How far does it travel in the first six minutes?
23
Friction Force and Newton’s Law of Motion
Friction Force and Newton’s Laws of

UNIT 2 Motion
Key Unit Competence

By the end of this unit, the learner should be able to perform experiments involving
Newton’s laws of motion and friction force.
Learning objectives

• Explain inertia.
• Outline and explain factors affecting inertia.
• State and explain Newton’s laws of motion.
• Describe why objects resist changes.
• Describe linear momentum and its conservation.
• Describe motion of objects on a horizontal plane with or without friction.
• Determine coefficient of friction.
Skills
• Design experiments to illustrate Newton’s laws of motion.
• Investigate effects of friction force on motion.
• Apply Newton’s laws principles in solving motion problems.
Attitude and value

• Appreciate the effect of friction force on moving body.
• Realise the effect of air resistance on the speed of moving objects.
• Appreciate the importance of Newton‘s laws of motion in life.
• Protection of persons by safety belts in a moving car.
24
Introduction
Unit Focus Activity

Materials:
• Two balls • Clear flat ground
Steps
1. Place one ball A, at a point on the flat ground.
2. Place the second ball B, on the ground at small distance from the first ball.
Kick ball A carefully so that it moves to strike ball B (Fig. 2.1).
A B

Fig. 2.1: One ball moving to hit a stationary one
3. Describe and explain what happens to the velocity of each ball after collision.
4. If instead a smaller ball C was used in place of ball A but kicked with the same
force, compare the velocities of ball B in both cases. Explain the difference if
any.
25
5. What quantity possessed by balls A and C is passed on to ball B? What two

factors determine the magnitude of this quantity?
6. State Newton's second law of motion and explain how it governs the events
during and after the collisions of the balls.
7. Why does ball B start slowing down and eventually stop after some time and
distance on the ground?
8. Share your observations and facts to the rest of the class in a class discussion.
Everyday, we interact with forces. The forces can cause different effects such as
change in motion, pressure as well as turning moments on an object. The effects
of force on motion of a body are summarised by Newton’s three laws of motion.
In this unit, we will investigate each of these laws in details.
2.1 Newton’s First law of motion
Activity 2.1 To demonstrate inertia using a coin and cardboard

Materials
• Table or bench • a smooth cardboard
• coin
Steps
1. Place the cardboard on the table with a small section of it extending beyond
the edge of the table. Place the coin on the cardboard (Fig. 2.2).
Card Coin
Pull card slowly
Pull card quickly
Table
Fig. 2.2: A coin on top of a cardboard

2. Pull the cardboard gradually and observe what happens to the cardboard
and the coin.
3. Return the cardboard and the coin to their initial position. Now pull the
cardboard abruptly and observe what happens to the cardboard and the
coin.
4. Explain your observations in steps 2 and 3.
5. Discuss in your groups and make a presentation on the role of safety belts in
a car.
26
The effects of a force on a body either at rest or in uniform motion summed up

under Newton’s first law of motion which states that a body remains in its state of
rest or uniform motion in a straight line unless acted upon by an external force.
Newton’s first law of motion suggest that matter has an in-built reluctance to
change its state of motion or rest. For instance, when a moving bus comes to an
abrupt stop, the passenger lurch forward, i.e. tend to keep on moving. Likewise,
when a bus surge forward, the passengers are jerked backwards, i.e. tend to resist
motion.
The property of matter to resist change to its state of motion i.e. either to remain at
rest or to continue moving in straight line is known as inertia (latin word meaning
laziness). This explains why cars have seat belts. The seat belts (see Fig. 2.3),
hold passengers onto the seat incase the vehicle comes to a stop or decelerates
suddenly, preventing them from lurching forward. This reduces any chances of
serious injury incase of an accident.
Hey! Never forget to put

on your safety belt when
in a moving vehicle
Fig. 2.3: A person wearing safety belt
Note: Newton’s first law of motion is also known as the law of inertia
Factors affecting inertia of a body

(a) Mass of a body
The mass of a body is a measure of its inertia. A large mass require a large force
to produce a given acceleration or deceleration than a smaller mass. A larger mass
therefore has a greater inertia.
(b) Acceleration of a body
As the acceleration of a body increases so does its tendency to continue at a
constant velocity.
27
(c) Force applied on a body

When the force applied on a body is increased, its tendency to remain at rest is
reduced. This would result to movement of the body from its resting state.
(d) Friction acting on a body
The law of inertia states that an object / body will keep moving at a constant
velocity unless a force is applied in it. An example of such a force is friction. It
is a force that makes a body to slow down. Without it, the body would continue
moving at the same velocity without slowing down.
Exercise 2.1
1. Define the term ‘inertia.’

2. State the law of inertia.
3. Briefly explain why wearing safety belts in moving vehicles is very important.
2.2 Linear momentum and impulse

2.2.1 Linear momentum
Activity 2.2 To illustrate linear momentum

Materials
• Two hammers (light and heavy) • Four identical nails • Wooden block
Steps
1. Take two nails and drive them into two pieces of wood using a light hammer.
2. Hit the first one gently and the second nail very hard. What happens in each case?
3. Repeat the activity using a heavy hammer (Fig. 2.4). What do you notice?.
light hammer heavy hammer
A gentle hit A hard hit A gentle hit A hard hit
Fig. 2.4: Driving a nail into wood.

4. Highlight two factors on which the penetration distance of the nail depends.
5. Discuss with your classmate what the term 'inear momentum' means.
28
To drive a nail into wood, a certain rate of motion (velocity) and mass of the
hammer is required. The quantity involving both motion and mass of a body is
called linear momentum. It is denoted by the letter p and is called momentum in short.
Linear momentum of an object is defined as the product of the mass and the velocity of
the object. i.e.
momentum, p = mass × velocity
In symbols p=m×v
The SI unit of momentum is kg m/s. Momentum is a vector quantity. The direction

of momentum is the same as that of the velocity.
Example 2.1
A car of mass 600 kg moves with a velocity of 40 m/s. Calculate the momentum
of the car.
Solution
Momentum = mass × velocity
= 600 kg × 40 m/s
= 24 000 kg m/s
Example 2.2
A body A of a mass 4 kg moves to the left with a velocity of 7 m/s. Another body
B of mass 7 kg moves to the right with a velocity of 6 m/s. (Fig. 2.5).
4 kg 7 kg
–7 m/s B +6 m/s
A
Fig. 2.5
Calculate (a) the momentum of A, (b) the momentum of B, (c) the total
momentum of A and B.
Solution
Let us assign positive sign to indicate movement to the right and a negative sign
to indicate movement to the left.
(a) Momentum of A = 4 kg × (–7) m/s = –28 kg m/s
(b) Momentum of B = 7 kg × (+6) m/s = +42 kg m/s
29
(c) Total momentum = (momentum of A) + (momentum of B)

= –28 kg m/s + 42 kg m/s
= +14 kg m/s
= 14 kg m/s to the right
2.2.2 Impulse
Activity 2.3 To demonstrate impulse using a ball.
Material
• ball • pin • wall
Steps
1. Take an inflated ball provided to you and remove some air from it using the
pin. (Be careful not to destroy the inner tube).
2. Press the ball using your finger. What do you observe? Withdraw the finger.
What do you observe? Explain.
3. Give the ball a strong kick towards a wall and observe the point of contact
between your foot and the ball during the time of contact and between the
wall and the ball respectively. What do you observe? What happens to the
point of contact on the ball after bouncing back? Explain.
4. Discuss with your classmates what impulse is.
5. Suggest examples in daily lives that demonstrate impulse.
6. Discuss the difference between impulse and linear momentum.
7. Make short notes on your findings and report to the whole class.
When a force F acts on an object for a very short time t, it produces an impact,
usually referred to as impulse on the object.
Impulse is defined as the product of force and time i.e.
Hence Impulse = Force × time
In symbols: I = Ft
The SI unit of impulse is newton - second (N s).
When an impulsive force acts on an object, it produces a change in the momentum
of that object. The velocity of that object changes from an initial value u to a final
value v. Its mass m remains constant.
30
Experiments have shown that the impulse acting on the object is equal to the
change in momentum it produces on the object.
Impulse = Change in momentum
Ft = mv – mu
Example 2.3
A hammer strikes a metal rod with a force of 20 N. If the impact lasts 0.4 s,
calculate the impulse due to this force.
Solution
Impulse = Force × time
= 20 N × 0.4 s
= 8 Ns
2.2.3 Distinction between impulse and linear momentum

Impulse and linear momentum are two quantities that confuses many. Table 2.1
will help us to differentiate the two.
Table 2.1:Differences between impulse and momentum
Impulse Linear momentum

1. It is the product of velocity and the
1. Impulse is product of the force and
mass of an object
the time the force acts on an object
Momentum = Mass × Velocity
Impulse = force (F) × time (t)
2. Its SI unit is kilogram-metre per

2. Its SI units is Newton-second (N s)
second (kg m/s)
The knowledge of impulse will help us to understand the law of conservation of

momentum which we shall discuss later in this unit.
2.3 Newton’s second law of motion
Activity 2.4 To demonstrate and define Newton’s second law of motion

Materials
• Two trolleys (heavier and light one)
• Spiral spring or rubber bands
31
Steps
1. Place the two trolleys on a smooth flat surface (floor or a table surface).
2. Connect the heavier trolley to the lighter one using a spiral spring or rubber
bands provided.
3. Move the trolleys away from each other till they are about 1m apart.
4. Release them at the same time. Observe the difference in their velocities and
acceleration. Which trolley accelerates faster?
5. Based on your observations in this activity, suggest relationship between the
applied force, mass of an object and the acceleration produced by the force
on the body.
6. Explain your observation using Newton’s Second law of motion in term of
momentum?
7. Compare your finding with those of other classmates.
Note:
It is a good idea to visit youtube on their website link: https://www.
youtube.com/watch?v=AFwbcWIUwLQ for more demonstration on
the Newton’s laws of motion.
As we have already learnt, one of the effects of a force is that it changes the state
of motion of an object. i.e. it causes a body at rest to move and a moving body to
accelerate or come to rest. Any change in the velocity of a body causes a change
in its momentum.
Newton summarised this effect on a body in his second law of motion which
states that the rate of change of momentum is directly proportional to the resultant force
in a body and it takes place in the direction in which the force acts.
Mathematically, the law is represented as follows:
change in momentum
Force (F) =
time taken
final momentum – initial momentum

Force(F) = time taken
If m is the mass of the body and taking u and v to represent initial and final velocities
respectively, while t representing time,
32
Initial momentum = mass × initial velocity (mu)

Final momentum = mass × final velocity (mv)
Change in momentum = final momentum – initial momentum
= mv – mu
Rate of change of momentum = mv – mu
t
v–u
=m t
but = v – u = a
t
Therefore rate of change of momentum = ma
But F α ma
Thus, F = kma where k is a constant of proportionality.
Experiments show that k = 1, Therefore,
F = ma.
This is the mathematical representation of Newton’s second law. The relationship
F = ma shows that the greater the force applied on an object the more acceleration
it causes on the object.
If mass is 1 kg and acceleration is 1 m/s2, then the force is 1 N. This is the definition
of 1 newton i.e. 1 newton is the force which when it acts on a mass of 1 kg, it gives an
acceleration of 1m/s2.
Example 2.4
A truck of mass 2.5 tonnes accelerate at 7.5 m/s2. Calculate the force generated
by the truck’s engine to attain this acceleration.
Solution
F = ma
= (2.5 × 1 000) kg × 7.5 m/s2
= 18 750 N
Example 2.5
An object of mass 4 kg accelerates to 5 m/s2. Calculate the resultant force.
Solution
F = ma
= 4 kg × 5 m/s2
= 20 N
33
Example 2.6
Calculate the acceleration produced by a force of 20 N on an object of mass 300 kg.
Solution
F 20 N
a= m =
300 kg
= 0.066 7 m/s2
Example 2.7
Table 2.2 shows the values of force, F, and the acceleration, a, for the motion of
a trolley on a friction compensated runway.
Table 2.2
Force F (N) 0.2 0.4 0.6 0.8 1.2

Acceleration, a (m/s2) 0.90 1.8 2.7 3.6 5.4
(a) Plot a graph of force, F against acceleration, a.

(b) Use your graph to determine the force when the acceleration is 4.0 m/s2
(c) Calculate the mass of the trolley, in grams, from your answer in (b).
Solution
(a) The graph is shown in Fig. 2.6.
(b) As shown in the graph, force, F = 0.9 N, when acceleration, a = 4.0 m/s2.
F
(c) Since F = ma, m = a
1.2
×
0.9 N
=
4.0 m/s2 1.0
= 0.225 kg
0.8
×
Force (N)
The mass of the trolley = 225 g

0.6
×
Note: 0.4
×
The slope of the graph of

0.2
Force (F) against acceleration
×
(a) gives the mass of trolley.

0 1 2 3 4 5 6
Acceleration m/s2)
Fig. 2.6: A graph of acceleration against force
34
Example 2.8
A car of mass 1 500 kg is brought to rest from a velocity of 25 m/s by a constant
force of 3 000 N. Determine the change in momentum produced by the force
and the time it takes the car to come to rest.
Solution
Data: m = 1 500 kg u = 25 m/s
v = 0 m/s2 F = 3 000 N
Change in momentum = mv – mu
= (1 500 × 0) – (1 500 × 25)
= –37 500 N s
(negative sign show the direction)
Impulse (Ft) = Change in momentum
Ft = 37 500
37 500 N s
t= = 125 s or 2 min 5 seconds
3 000 N
Exercise 2.2
1. (a) State Newton’s second law of motion.

(b) Use Newton’s second law of motion to derive the equation F = ma.
(c) Define the unit of force, ‘the newton’ using F = ma.
2. (a) What is meant by ‘impulse of a force’?
(b) What is the relationship between impulse and the change in momentum?
(c) When a soccer goalkeeper is being trained on how to catch hard balls,
he/she is taught how to pull back the hands while catching the ball.
Explain how the technique works.
3. A 0.06 kg tennis ball drops freely from the left hand of a player. When in the
air, it hits a racquet and leaves with a horizontal speed of 58 m/s.
(a) Calculate the impulse produced on the ball.
(b) If the time of contact with the racquet is 0.025 seconds, calculate the
average force exerted on the ball.
4. A bullet of mass 7 g travels with a velocity of 150 m/s. It hits a target and
penetrates into it. It is brought to rest in 0.04 s. Find the:
(a) distance the bullet travels in the target.
(b) average retarding force exerted on the bullet.
5. A ball of mass 45 g travelling horizontally at 30 m/s strikes a wall at right angles
and rebounds with a speed of 20 m/s. Find the impulse exerted on the ball.
35
6. Fig. 2.7, shows a graph of the force on a tennis ball when served during a
game. Find the mass of the ball if it leaves the racket with a velocity of 40m/s.
(Assume the ball is stationary before it is struck.)
900
800
700
600
Force (N)
500
400
300
200
100
0 1 2 3 4 5 6 7 8
Time (× 10–3 s)
Fig. 2.7: A graph of force against time

7. A truck of mass 4 000 kg starts from rest on a horizontal rails. Find the speed
4 seconds after starting if the impulsive force by the engine is 1 500 N.
2.4 Newton’s third law of motion
Activity 2.5 To demonstrate action and reaction force
Materials
• Carton • Cello tape • 4 pins
• A large balloon • A straw
Steps
1. Cut out one rectangular and 4 equal circular pieces from the carton to act as
the body and wheels of a trolley.
2. Pass the pins through the centre of the wheels to act as the shafts and fix the
pins onto the body using cellotape. Ensure the wheels are able to rotate freely
about the shafts.
36
3. Fix the straw into the mouth of the balloon using the cello tape and seal the
mouth airtight. Fix the straw firmly onto the body of the trolley using cello
tape as shown in Fig 2.8.
4. Inflate the balloon through the straw and then seal the mouth of the straw
with the finger to prevent air from coming out. Place the trolley on a smooth
horizontal surface
5. Remove the finger suddenly from the mouth of the straw so that air from the
balloon comes out at once (Fig 2.8). Observe what happens to the trolley.
Fig. 2.8: A demonstration of action and reaction
6. In what direction does the air move as it leaves the mouth of the straw? In
which direction does the trolley move? Explain the behaviour of the trolley.
State the law that governs the behaviour of the trolley and other objects under
similar conditions.
It is easier to study effects of forces on an object by considering one force at a

time. However, in reality, a single force cannot exist by itself. Two forces always
occur when two objects push or pull each other. These forces are called action
and reaction force.
Newton’s third law of motion state that if a body X exerts a force on another body
Y, Y exert an equal and opposite force on X, i.e. to every action force there is an
equal and opposite reaction force.
37
Fig. 2.9 shows a real life example where action and reaction force is experienced.
Action
Reaction
Boat
Fig 2.9: A person jumping from a boat
When one jumps off (action force) shore from a boat his/her forward force, exerts
a backward force (reaction)on the boat. The boat moves backwards dragging with
it his/her legs and then person tends to fall into the water, (see Fig. 2.9).
Example 2.9
Fig. 2.10 shows a block of mass 4.5 kg resting on a
smooth horizontal surface. It is attached to another
block of mass 1 kg using a light inextensible string
passing over a frictionless pulley.
Determine
(a) The acceleration of the system. Fig. 2.10: Mass system over
a pulley
(b) The tension in the string.
Solution
(a) The weight of the 1.5 kg block (1.5 × 10 = 15 N) acting downwards makes the
two blocks to move together as one block of mass of 6 kg (i.e 1.5 + 4.5 = 6 kg).
There is no frictional force hence this weight is the resultant force
Resultant force = mass × acceleration
15 N = (1.5 + 4.5) kg × a
Acceleration, a = 15 N = 2.5 m/s2
6 kg

(b) Let the tension in the string be T. Since the block is accelerating downwards,
weight of the 1.5 kg block is greater than tension T.
38
The resultant force acting downwards is (15 – T) N.

Resultant force = mass × acceleration
15 N – T = 1.5 × 2.5
Tension T = 15 N – 3.75 N = 11.25 N
Exercise 2.3
1. Give three real life situations where Newton's third law of motion is
experienced.
2. In groups, discuss the working principle of the following:-
(a) Rockets and jet propulsion
(b) The garden sprinklers
nozzle closed water moves
3. A garden sprinkler (Fig. 2.11) out (action)
has three of the four jets blocked.
60 cm3 of water jets out of the nozzle closed
fourth jet every second (density of
water is 1 g/cm3). If the area of the
mouth of the jet is 15 mm2, find:
nozzle closed
(a) the velocity of the water
Fig. 2.11: sprinkler moves back (reaction)
coming out.
(b) the minimum force required to prevent rotation.
(c) Explain how you would, measure the force in (b).
4. A rocket pushes out exhaust gas at a rate of 150 kg/s. The velocity of the gas is
250 m/s. Calculate the forward thrust on the rocket.
5. A block of mass 8 kg rests on a smooth horizontal surface. It is being dragged
by another block of mass 2 kg attached to it by a light in extensible string
passing over a frictionless pulley.
(a) Draw the setup described above.
(b) Calculate the acceleration of the system.
(c) Calculate the tension in the string.
6. Stunt men in films often fall safely from tall buildings.
(a) Two main forces act on such a man when falling. Name these forces.
(b) To reduce chances of injury, such a man usually lands on deep soft,
inflatable mattresses placed at the landing point. Explain how the
mattresses achieve this purpose.
39
2.5 Conservation of linear momentum

2.5.1 Law of conservation of momentum
Activity 2.6 To demonstrate the conservation of linear momentum
Materials
• A table • Marbles
Steps
1. Place the marbles a distance from each other on table in a straight line.
(See Fig. 2.12)
Fig. 2.12: Marbles arranged in a straight line

2. Strike the one near you towards the next. What did you observe about their
movement after collision? Explain.
Suppose two objects A and B of masses mA and mB move in the same direction
with different velocities (See fig. 2.13). And there are no other external unbalanced
force acting on them.

A B
mA uA mB uB
Fig. 2.13: Collisions
40
On collision, A pushes B with a force FA and B reacts by pushing A with an equal

and opposite force FB. Such that FA = –FB.
Since the time spent in colliding is the same, A experiences an impulse –FBt from
B while B also experiences an impulse FAt from A.
Thus –FBt = FAt …… (i)
Impulse = change in momentum = mv – mu
Letting the final velocities of A and B be vA and vB respectively
Then, –FBt = mAvA – mAuA ……(ii)
FAt = mBvB – mBuB……(iii)
Equating (ii) and (iii)
–(mAvA – mAuA) = mBvB – mBuB
– mAvA + mAuA = mBvB – mBuB

mAuA + mBuB = mAvA + mBvB
But mAuA + mBuB = total momentum before collision
and mAvA + mBvB = total momentum after collision
∴ Total momentum before collision = total momentum after collision
The momentum has been conserved. This conclusion is summarised as the law
of conservation of momentum.
The law of conservation of momentum states that when two or more bodies
collide, their total momentum remains constant provided no external forces act
on them.
2.5.2 Collisions
There are two types of collisions, namely elastic and inelastic collision.
2.5.2.1 Elastic collisions

When the total kinetic energy is conserved after collision, the collision is said to
be elastic. This is only possible between atoms or molecules. However, the collision
between two smooth marble balls is approximately elastic.
41
Activity 2.7 To demonstrate an elastic collision
Materials
• Two trolleys A and B • A ticker timer
• A tape • Weighing balance • Two carbon papers
• Different masses (200 g and 300 g) • A runway
Steps
1. Determine the mass of trolleys A and B. Compensate the runway for friction.
2. Attach a tape to each of the trolleys. Pass the two tapes through a ticker
timer using two carbon papers (Fig. 2.14).
ticker timer
two
carbons trolley B
buffer rod
trolley A (at rest)
runway
Fig. 2.14: Elastic collision
3. Position trolley A halfway down the runway. Make sure it is stationary.

4. Start the ticker timer and push trolley B aiming its buffer rod towards the
trolley A.
5. Allow the two trolleys to continue moving after the collision.
6. Calculate from the tapes, the velocity of trolley B before collision uB and the
velocity of trolley A and B after collision (vA and vB).
7. Repeat the experiment with trolleys A and B loaded with different masses.
Record the results in a table (Table 2.3).
8. Determine the total momentum and the total kinetic energy before and after
collision. Compare their values before and after collision.
Table 2.3
Before collision After collision
Trolley A Trolley B Total Total Trolley A Trolley B Total Total
momentum K.E momentum K.E
mA uA mAuA mB uB mBuB mAuA + mBuB m
–1 AuA + m
2 1– u 2 m v mAvA mB vB mBvB mAvA + mBvB m
B B A A
1– AvA2 + m
–1 BvB
2
2 2 2 2
42
The results of a similar activity shows the tapes shown in Fig. 2.15.
After collision Collision Before collision
Tape B For trolley B
Velocity vB Velocity uB
After collision Before collision
Tape A For trolley A
Velocity vA Velocity vA = 0
Fig. 2.15: Ticker timer tapes
If no other forces are acting, the total momentum and total energy before and
after collision is found to be the same. These types of collisions are not common
because in every collision, some energy is always converted to other forms. (Tell
your classmates which other forms?)
Example 2.10
A cannon of mass 800 kg fired a cannon ball of mass 3 kg at a velocity of
120 m/s. Find the recoil velocity of the cannon.
Solution
m1 = 3 kg, u = 0 m/s, v1 = 120 m/s, m2 = 800 kg, u2 = 0, v2 = ?
Total initial momentum = total momentum
m1u1 + m2u2 = m1v1 + m2v2
3 × 0 + 800 × 0 = 3 × 120 + 800 × v2
0 = 360 + 800v2
-800v2 = 360
360
v2 = - = -0.45 m/s
800
The cannon recoiled backwards at a velocity of 0.45 m/s. The negative value in
the velocity shows that the cannon moved (recoiled) in the opposite direction.
43
2.5 2 .2. Inelastic collision
Activity 2.8 To demonstrate inelastic collision

Materials
• Two trolleys A and B • A cork
• A pin or a sharp buffer rod • Masses 200 g, 300 g
• Runway • A weighing balance
Steps
1. Determine the mass of two trolleys A and B. Raise the runway to compensate
for friction.
2. Attach a cork at the back of trolley A. Position trolley A halfway down the
runway. Make sure it is stationary.
3. Attach a pin or a sharp buffer rod at the front of trolley B and a tape to pass
through the ticker timer.
4. Start the ticker timer. Push the trolley B aiming the pin towards the cork on
trolley A (Fig. 2.16).
ticker timer
trolley B
cork
buffer rod trolley A (at rest)
runway
Fig. 2.16: Setup to demonstrate inelastic collision
5. Allow the trolleys to move together after collision.

6. Repeat the experiment with trolleys A and B loaded with different masses.
Record the results in a table (Table 2.4).
7. Determine the total momentum and the total kinetic energy before and after
collision. Compare these quantities before and after collision.
8. Record your results in a tabular form as shown in table 2.4.
Table 2.4
Trolley B Total Total Total
momentum K.E momentum K.E
mB uB mBuB mBuB m
1– u 2 mAB vAB mABvAB m
B B
1– v 2
AB AB
2 2
44
9. From the tapes, measure the velocity (uB) of trolley B before collision, and
then velocity (vAB) of trolley A and B after collision. Comment on your
answer.
A similar activity that was conducted by some students produced the tapes
shown in Fig. 2.17.
Collision
Both trolleys after collision Trolley B before collision
Tape B
vAB vB
Fig. 2.17: Ticker timer tape
From the tape, we observed that during the collision only momentum is
conserved, but there is a loss of kinetic energy.
We have seen that when the two bodies collide, their total momentum is conserved.
The total kinetic energy is however not usually conserved. Some kinetic energy
is converted into sound and heat. Collisions where the total kinetic energy is not
conserved are called inelastic collisions. A completely inelastic collision is one which
two bodies stick together after collision.
Example 2.11
A mass of 3 kg moving with a velocity of 4 m/s collides with another mass of
2 kg which is stationary. After collision the two masses stick together. Calculate
the common velocity for the two masses after collision.
Solution
u1 = 4 m/s u2 = 0 m/s
3 kg (2 + 3) kg
2 kg

Momentum before collision = m1u1 + m2u2 = (2 × 0) + (3 × 4)

= 12 kg m/s
Momentum after collision = (m1 + m2)v = (3 + 2)v = 5v
Momentum before collision = momentum after collision
12 kg m/s = 5 v
∴ Common velocity, v = 2.4 m/s
45
Example 2.12
A 5 kg mass moving with a velocity of 10 m/s collides with a 10 kg mass
moving with a velocity of 7.0 m/s along the same line. If the two masses
join together on impact, find their common velocity if they were moving:
(a) in opposite direction (b) in the same direction
Solution
(a) u1 = 10 m/s u2 = 7 m/s

5 kg (5 + 10) kg
10 kg
Velocity of 5 kg mass = +10 m/s ; velocity of 10 kg mass = –7 m/s

(5 × 10) + (10 × –7) = (5 + 10)v
50 + (–70) = 15v
v = –1.33 m/s
The minus sign means that the joined mass moves in the initial direction of the 10 kg mass.
(b) u1 = 10 m/s u2 = 7 m/s v?
5 kg 10 kg (5 + 10) kg

(5 × 10) + (10 × 7) = 15 × v
50 + 70 = 15 v
v = +8.0 m/s
The plus sign means that the joined mass move in the initial direction of the two
individual masses.
Exercise 2.4
1. A mass of 8 kg travelling to the right at 2.5 m/s collides with a 2 kg mass

travelling to the left at 3.0 m/s. Find
(a) the momentum of each mass.
(b) their total momentum.
2. An object of mass 20 kg collides with a stationary object of mass 10 kg.
46
The two objects join together and move at a velocity of 5 m/s. Find the initial
velocity of the moving object.
3. A car of mass 600 kg travels at 20 m/s towards a stationary pick-up of
1 200 kg. After colliding, the two stick and move together. Find their common
velocity.
4. A truck of mass 3 000 kg moving at 3 m/s collides head on with a car of mass
600 kg. The two stop dead on collision. At what velocity was the car travelling before
collision?
5. If a 2 kg ball travelling north at 6 m/s collides with 4 kg ball travelling in the same
direction at 4 m/s, the velocity of the 4 kg ball is increased to 5.5 m/s to the north.
What happens to the 2 kg ball?
6. A small car of mass 500 kg is involved in a head-on collision with a heavy car of
mass 4 000 kg travelling at 20 m/s. The small car is thrown onto the bonnet of the
heavy car which continues after impact at 4 m/s in the original direction. How fast
was the small car moving?
7. A bullet of mass 10 g is shot from a gun of mass 20 kg with a nuzzle velocity
of 100 m/s. If the barrel is 20 cm long, determine:
(a) the acceleration of the bullet.
(b) the recoil velocity of the gun.
2.6 Coefficient of friction

In Senior 1, we were introduced to frictional force. We defined friction as the force
that oppose the relative motion of two surfaces that are in contact. In Senior 2,
we also learnt that the two factors that affect the friction between two surfaces
are the nature of the surfaces and the normal reaction (R).
In this section, we will learn how to determine the coefficient of friction between
two surfaces.
2.6.1 Determination of coefficient of friction
Activity 2.9 To describe motion of object on a horizontal plane with or

without friction
Materials
• A solid block • Smooth and rough horizontal surfaces
• Spring balance
Steps
1. Attach the spring balance to the solid block.
2. Pull the solid block across the rough surface through a particular distance.
Record the force applied.
47
3. Repeat step 2 but this time on a smooth surface. Compare the two forces
recorded. What do you observe? Which surfaces was easier to move the
block? Explain.
Look at Fig. 2.18. A solid block is being pulled over a horizontal surface by an
applied force, F. Frictional force (Fr) acts in the opposite direction to oppose the
movement of the block.
R
Applied force F
Frictional force (Fr)
W
Fig. 2.18: Forces acting on a solid block on a horizontal surface
When the block is just about to move, solid friction between the block and the
surface is called static friction. Static friction is the force opposing motion between
surfaces when the surfaces are just about to move. When the block is moving,
friction force is reduced and is called dynamic friction. Dynamic friction is the
opposing force motion when there is relative motion between surfaces.
Suppose some weights are placed on top of the block, would the frictional force
remain the same? In other words, what is the relationship between the frictional
force and the normal reaction (R)?
The following experiment will help us establish the relationship between frictional
force and normal reaction.
Activity 2.10 To establish the relationship between frictional force and

normal reaction
Materials
• A 50 g wooden block with rough surface.
• Four 50 g mass
• A rough bench
• A spring balance
48
Steps
1. Set up the apparatus as shown in Fig. 2.19.
Wooden block Normal reaction R
Spring balance
F Pull
Friction
F
w = mg
Fig. 2.19: Determining coefficient of friction

2. Pull the string until the block is just about to move. Record the reading on
the spring balance.
Note that the spring balance reads the value of frictional force which is
acting in the opposite direction in this experiments. (F is the frictional force
between the surfaces.
Frictional force = Applied force F).
3. Place one 50 g mass on the block and pull the string again. Record the
reading on the spring balance.
4. Repeat steps 2 and 3 for two, three and four 50 g masses on the block masses
and the record the reading of force, F on the spring balance in Table 2.5.
Table 2.5
Mass (g) 50 100 150 200 250
Reaction R (N)
Spring reading (F) (N)
5. Draw a graph of F(N) against R(N).
The results from a similar activity showed that the readings on the spring balance
F
increase with the increase of mass. When a column of R was added and completed
F
in the table, it was noted that the value of R was constant.
When a graph of F against R was plotted, a straight line that is passes through
the origin was obtained (Fig. 2.20).
49
Frictional force FR (N)

F
Normal reaction R (N)
Fig. 2.20: A graph of F against N
The gradient of the line gives:

change in F ∆F
Gradient = change in R = ∆ R
From the graph (Fig. 2.19), we conclude that, frictional force opposing the
movement of a solid is directly proportional to the weight of the solid i.e. F α W.
From Newton’s third law of motion:
Weight (w) of an object placed on the bench is equal and opposite to the normal
reaction (R) between the surface of the bench and the block in contact.
Frictional force, F α w
Therefore, F α Normal reaction (R) (i.e. F ∝ R)
Therefore, F = µR, where µ is a constant called the coefficient of friction.

As we have already learnt, the solid friction force is either static or kinetic fiction
depending on the state of motion of the solid.
(a) The coefficient of static friction, µs, is the ratio of static frictional force to the
normal reaction R.
F
µs = R
Coefficient of static friction, µS, has no unit since it is a ratio of forces.
One has to apply a force larger than the limiting static frictional force (Fs) for
a body to move.
(b) When the body is sliding along the bench with a constant velocity, the
frictional force is now called kinetic or dynamic friction (Fk). It is calculated as
Fk = µk R, where R is the normal force and µk is the coefficient of kinetic
friction.
50
Example 2.13
A force of 25 N just limits the motion of a block of mass 50 kg which is being
dragged on the horizontal ground. Calculate the coefficient of static friction force.
Solution
Fs = µsR where µs is the coefficient of static friction
Weight of the block = Normal reaction = 50 kg × 10 N/kg = 500 N
Fs = µsR
∴ 25 N = µs × 500 N
25 N
µs = 500 N = 0.05
µs = 0.05
Example 2.14
Fig. 2.21 shows a block of mass 200 kg being dragged at constant velocity with
a force 40 N at angle 60º to the horizontal.
Block
N
40
F=
200 kg 60°
Fr
Rough surface
Fig. 2.21: A block being dragged at a constant velocity horizontally
Determine the coefficient of kinetic friction (µk).

Solution
F r = µkR
Since the body is not accelerating,
Friction = horizontal component of F.
Fr =Fh = Fcosθ = 40 × cos 60 = 40 × 0.5 = 20 N
R = Weight - vertical component of 40 N
= 2000 N - 40sin60°
= 1965.36 N
Substituting for Fr and R in, Fr = µkR, we get:
51
20 N = µk × 1965.36 N
20 N
µk = 1965.36 N = 0.01
The coefficient of kinetic friction is 0.01.

• Newtons first law of motion states that a body continues in its state of rest or
uniform motion in a straight line unless compelled to act otherwise by some
external force.
• Momentum is the product of mass and velocity of a body.
(p = m × v)
• Impulse = force (N) × time(s) = Ft
• Impulse = change in momentum (Ft = mv – mu)
• Newton’s second law of motion states that the rate of change of momentum
is directly proportional to the force on a body and takes place in the direction
in which the force acts.
• F = ma is the mathematical statement of Newton’s second law of motion.
• The law of conservation of momentum states that when one or more bodies
collide, their total momentum remain constant provided no external forces
are acting.
mAuA – mBuB = mA vA – mB vB
• In an elastic collision, the bodies separate after colliding. Both momentum
and kinetic energy are conserved.
• In an inelastic collision, the bodies fuse and move together after colliding.
Momentum is conserved but kinetic energy is not.
• Newton’s third law states that for every action there is an equal and
opposite reaction.
• Frictional force by solid is directly proportional to the normal reaction (R)
of the solid, that is,
F α R
Therefore, F = µR where µ is called coefficient of friction
• There are two kind of friction force; static and kinetic.
52
The coefficient of static friction is given by

F
µs = R
While the coefficient of kinetic dynamic friction is given by
F
µk = R
k
Unit Test 2
For questions 1 - 5, select the most appropriate answer.
1. The following are factors affecting inertia of a body, which one is not?
C Acceleration of body B. Linear momentum
C. Mass of a body D. Friction acting on a body
2. Which one of the following is the correct definition of momentum?
C Sum of two forces acting on a body.
B. Product of mass and density of a body.
C. Product of mass and velocity of a body.
D. Product of impact and velocity of a body.
3. Which one of the following statements is true about inelastic colision?
A. Momentum and kinetic energy are both conserved.
B. Momentum is conserved but kinetic energy is not.
C. Momentum is not conserved but kinetic energy is conserved.
D. Both momentum and kinetic energy are not conserved.
4. Frictional force by a solid is directly proportional to the area of surfaces in
contact.
A. True B. False
5. There are more than two states of friction force between solids.
A. True B. False
6. (a) Explain why Newton’s first law is also called the law of inertia.
(b) Describe an experiment to illustrate the Newton’s first law of motion.
7. Table 2.6 shows the values of the resultant force, F, and time, t, for a bullet
travelling inside the gun barrel after the trigger is pulled.
Table 2.6
Force (N) 360 340 300 240 170 110
Time (10 s) 3 4 8 12 17 22
–3
53
(a) Plot a graph of force, F, against time, t.

(b) Determine from the graph.
(i) the time required for the bullet to travel the length of the barrel.
(Assume that the force becomes zero just at the end of the barrel).
(ii) the impulse of the force.
(c) Given that the bullet emerges from the muzzle of the gun at a velocity of
240 m/s, calculate the mass of the bullet.
8. A block of mass of 3 kg is pulled by a spring balance that records a force of
10 N. The ticker tape (Fig. 2.22) attached at the back of the mass shows a
group of 10 spaces between them having the lengths as shown in Fig 2.22.
The ticker timer makes 50 dots per second.
10
10 spaces 10 spaces 10 spaces
spaces
10 cm 20 cm 30 cm 40 cm
Fig. 2.22
(a) Draw and show how the dots are spaced out on the tape.
(b) Find (i) the acceleration of the block
(ii) the balanced force acting on the block.
9. The graph in (Fig. 2.23) shows how the force applied on a 20 kg mass varies
with time.
F (N)
30
20
10
0.1 0.2 0.3 0.4 0.5 0.6 0.7

t (s)
10
20
Fig. 2.23: A graph of F against t
54
(a) Find the total impulse after 0.7 s.

(b) If the body starts from rest, draw the:
(i) acceleration-time graph.
(ii) velocity-time graph.
(iii) momentum-time graph.
10. A ball of mass of 2.0 kg, travelling at 1.5 m/s collides with another ball of
mass 3.0 kg travelling at 0.8 m/s in the same direction. If they stick and move
together, calculate.
(a) the common velocity after collision.
(b) the change in momentum for each ball.
11. A rugby player of mass 75 kg, running east at 8.0 m/s, collides with another
player of mass 90 kg and who is running directly towards him at 5.0 m/s. If the
two players cling together after the tackle, what will be their common velocity?
12. Fig. 2.24 is a graph of acceleration against applied force.
24
20
acceleration (m/s2)
16
12
0
10 20 30 40 50 60 70
Force (N)
Fig. 2.24 A graph of acceleration against force
Find:
(a) the force that results in an acceleration of 14 m/s2.
(b) the gradient and state its unit.
55
13. A bullet of mass 15 g travelling at 400 m/s becomes embedded onto a block
of wood of mass 300 g which is at rest. Calculate the initial speed of the
block immediately after collision.
14. A bullet of mass 20 g travelling horizontally at a speed of 200 m/s embeds
itself in a block of wood of mass 850 g suspended from a light inextensible
string so that it can swing freely. Find the:
(a) velocity of the bullet and block immediately after collision.
(b) height through which the block rises.
15. Two ice hockey players travelling in opposite directions get entangled and
move while locked together. Player A has a mass of 120 kg and is travelling
at 5 m/s. Player B has a mass of 90 kg and is travelling at 4 m/s. Calculate.
(a) the initial momentum of player A.
(b) the initial momentum of player B.
(c) the speed with which they both move with after being locked together.
(d) In which direction did they move after collision?
16. A trolley of mass 2 kg travelling from left to right at 4 m/s collides elasticity with
another trolley of mass 4 kg travelling from right to left at 1 m/s. If the speed of
the 2 kg trolley after collision is 0.8 m/s, what is the speed of the 4 kg trolley?
17. A block weighing 400 N is pushed along a surface. If it takes 110N to get the
block moving and 70 N to keep the block at a constant velocity, what are the
coefficient of friction µs and µk?
18. State:
(a) three advantages of frictional force
(b) three disadvantages of frictional force
19. A force F pushes towards the left on a box. A friction force, f, between the
floor and the box resists the movement of the box. These are the only forces
acting in the horizontal direction. For the following three cases, state which
is bigger (or the same size), F or f and why.
(a) The box does not move.
(b) The box moves to the left with constant velocity.
(c) The box moves to the left and accelerates.
(d) The box moves to the left and decelerates
56
Application of Atmospheric Pressure
Applications of Atmospheric
UNIT 3 Pressure
Key Unit Competence

By the end of the unit the learner should be able to explain the existence of
pressure in gas and the application of atmospheric pressure.
Learning objectives

• Explain the existence of force exerted by air on a surface.
• Explain the relationship between atmospheric pressure and altitude.
• Relate atmospheric pressure, density of air and altitude.
• Outline the applications of atmospheric pressure
Skills
• Explain the force atmospheric pressure exerts on earth surface.
• Discuss factors affecting atmospheric pressure.
• Explain applications of atmospheric pressure in real life.
• Evaluate factors influencing atmospheric pressure.
Attitude and value

• Appreciate existence of atmospheric pressure.
• Predict change in weather by observing changes in atmospheric pressure.
• Be aware of effects of changing of atmospheric pressure on human body as the altitude
increases (climbing mountains).
57
Introduction
Unit Focus Activity

Materials
• A glass • Two beakers • A straw
• Delivery tube • Water
Steps
1. Put some clean water in a glass. Insert the straw and start drinking through
the straw (Fig. 3.1 (a)).
2. Using two beakers, delivery tube and water, setup a working siphon as shown
in Fig. 3.1 (b).
(a) (b)
Fig. 3.1 Some applications of atmospheric pressure
58
3. Define atmospheric pressure.

4. Explain the application of atmospheric pressure in the working of:
(a) drinking straw
(b) siphon
5. Identify two more applications of atmospheric pressure.
Atmospheric pressure is the pressure resulting from the weight of the air column
acting on the earth’s surface. In this unit, we will start by carrying out activities to
demonstrates the existence of atmospheric pressure, discuss factors influencing the
pressure, instruments used to measure the pressure then discuss the application
of atmospheric pressure.
3.1 Existence of atmospheric pressure
Activity 3.1 To illustrate the existence of atmospheric pressure

using an inverted glass tumbler
Materials
• A glass tumbler with water • A cardboard
Steps
1. Fill a glass tumbler with water and cover the top with a cardboard (Fig. 3.2(a)).
2. Invert the tumbler while holding the cardboard (Fig. 3.2(b)). Suddenly
remove the hand holding the cardboard. What do you observe?
3. Now, turn the tumbler side ways (Fig. 3.2(c). What do you observe?
4. Suddenly remove the cardboard covering the tumbler. What happens to the
water inside the tumbler? Explain.

(a) (b) (c)
Fig. 3.2: Demonstrating atmospheric pressure using a glass tumbler
59
You should have observed that the cardboard does not fall when the glass is
inverted vertically upwards or sideways. This take place because the pressure due
to the column of air in the atmosphere (i.e atmospheric pressure) is greater than
the combined pressure due to the column of water and air inside the glass. Hence
atmospheric pressure keeps the cardboard intact. The water does not flow out.
Activity 3.2 To illustrate the existence of atmospheric pressure

using a can
Materials
• A thin-walled can • A cork
• Water `` • Source of heat
Steps
1. Pour some water in a large thin-walled can.
2. Boil off the water in the can and immediately cork the can.
3. Allow it to cool and observe what happens.
4. Explain your observations to other group members.
Before the can in this activity is heated and corked, the air inside and outside
the can exerts pressure equally on the walls of the container that balances with
atmospheric pressure (Fig. 3.2(a)).
When the can is heated (Fig. 3.3 (b)), the steam that is formed expels the air
inside the can. After cooling the can, the steam condensed and a partial vacuum
is formed inside (Fig. 3.3 (c)). The air pressure inside the can decreases. The
atmospheric pressure acting on the surface of the can from outside is greater than
the air pressure inside. It therefore makes the can to crust or collapse inward as
shown in Fig. 3.3 (c).
(a) (b) (c)
Atmosphere
pressure
Fig. 3.3: Demonstration of atmospheric pressure
60
Activity 3.3 To demonstrate existence of atmospheric pressure

atmospheric pressure using a wet coin.
Materials
• A coin • bench • Water
Steps:
1. Place a coin on top of a bench. Lift it and note the ease with which you do that.
2. Now, wet the bench surface and the coin also. Place the coin on the wet
surface.
(a) Lifting a dry coin from table top (b) Lifting a wet coin from table top
Fig. 3.4: Demonstrating atmospheric pressure by lifting a coin.
3. Lift the coin up, note the ease with which you do that this time round?
Explain clearly your observation.
4. Compare your observations on steps 1 and 3. Explain the difference in the
two observations.
The water between the coin and the bench expels air, reducing the air pressure
under the coin. The atmospheric pressure above the coin presses it to the bench,
making it more difficult to lift the coin.
Activities 3.1 to 3.3 shows the existence of atmospheric pressure. This atmospheric
pressure play an important role in our daily lives as we shall learn later in this unit.
Exercise 3.1
In groups of three, conduct a research from the internet and reference books on
how you can explain the existence of atmospheric pressure using Magdeburg
hemisphere.
61
3.2 Factors influencing atmospheric pressure
Activity 3.4 To find out factors that influence atmospheric pressure
Materials
• Internet enabled computers • Reference books
Step
1. Suggest the factors that influence atmospheric pressure.
2. Explain to your classmates how the factors you have suggested in step 1
influence atmospheric pressure.
3. Now conduct a research from the internet or reference books on factors that
influence atmospheric pressure. Write them down.
4. In your research, find out why athletes prefer training at high altitudes.
5. Give a summarised presentation on your findings to the whole class through
your secretary.
(a) Altitude
The main factor that affects atmospheric pressure at a given location is the altitude
(or height above sea level) of the location. The maximum air density is at the earth’s
surface. The air density decreases with the height away from the surface of the
earth (See Fig. 3.5).
This is because the pull of the ear th's g ravity on the air is less.
density = 0.4 kg/cm3
layers 11 000 m
of air density = 1kg/cm3
2 000 m
density = 1.3 kg/cm3
Fig. 3.5: Atmospheric pressure varies with altitude
The fewer number of gas molecule at higher altitude means fewer molecular
collisions and a decrease in atmospheric pressure.
The following are some of the effects of increase in altitude increases to the
human body.
62
1. At high altitude (about 1500 m and above), there is still approximately

21% of oxygen in the air, but since the atmospheric pressure at this level is
low, the pressure exerted by oxygen is reduced. Due to pressure gradient,
more energy is required by the lungs to take in oxygen. Hence, to adapt
to the difficulty of obtaining oxygen, the body will increase the production
of red blood cells and hemoglobin in the blood, as well as reduce muscles
metabolism to enable a more efficient use of oxygen. This effect lasts for
upto two weeks after returning to the lower altitude giving athletes who is
training at high altitude a competitive advantage.
2. When altitude increases, the atmospheric pressure drops, tissue in the
human body expand. The expansion of tissues in and surrounding the joints
aggravates nerves causing pain.
3. A drop in atmospheric pressure such as in an ascending airplane also causes
headaches, particularly sinus headache. This is because gases in the sinuses
and ears are at higher pressure than those of the surrounding air.The pressure
tries to equalise, causing pain in the face and ears.
4. Sometimes when we are at high altitude, our noses may start bleeding. There
are many reasons for nose bleeding, but bleeding at high altitude is associated
with the surrounding weather conditions. As one climbs up the mountain
(change in altitude), the atmospheric pressure is reduced, temperature
decrease and the water vapour content becomes less. The presence of water
vapour in the air acts as a lubricant for the movement of air through the nose.
Its reduction increases friction and irritation of nasal membrane resulting to
bleeding.
(b) Temperature
When atmospheric air is heated (such as by radiation from the sun), the air
molecules become active. The space between the neighbouring air molecules
increases and reduces air density. Lowering the air density decrease the amount
of pressure exerted by the air i.e. atmospheric pressure. Therefore, given equal
volume of air, warm air is less dense than cold air and exert less pressure.
(c) Water vapour concentration (humidity)

When water evaporates into the atmosphere its molecules take the place of gas
molecules in the air. The water vapour is less denser than dry air. Hence, wet air
being less dense exerts less pressure than dry air. Therefore, atmospheric pressure
decreases with increase of humidity (water vapour in the atmosphere).
63
(d) Wind Pattern

Atmospheric pressure is also influenced by wind patterns. Winds causes
convergence (moving together) and divergence (moving apart) of air at the earth’s
surface. When the wind converges, air molecule increases exerting more pressure
on the surface whereas it exerts less pressure when the wind diverges since air
molecules decrease in number.
The combination of these factors make atmospheric pressure an important
parameters in predicting weather.
In general, weather becomes stormy when atmospheric pressure falls due to air
becoming warmer or humid. The convergence of air masses at the surface of the
earth causing convection and rising of air. It becomes fair when atmospheric
pressure rises due to drier, colder air and the divergence of air masses. Therefore,
if a weather person on television says that the barometer is falling, you might want
to get your raincoat out!
Exercise 3.2
1. List four factors that influences atmospheric pressure.

2. Explain why it is difficult to cook food while on top of a mountain.
3. Briefly explain why athletes in Rwanda normally enter residential training
camp in Musanze District for training and not Rusizi District near Rusizi river.
4. Discuss how the temperature and wind patterns affects atmospheric pressure.
5. A student in senior three started nose bleeding while they were in a trip at
the top of Mt. Karisimbi.
(a) Explain the possible reason for her nose bleeding
(b) Discuss how you can help her to stop nose bleeding.
3.3 Instruments for measuring atmospheric pressure
Activity 3.5 To measure atmospheric pressure using mercury

barometer
Material:
• Mercury barometer
Caution
Mercury barometer is heavy, fragile and explosive. Care must be taken when
handling it.
64
Steps
1. Take the mercury barometer provided and observe its calibrations. What is
the height of the column of mercury in it? Record it down.
2. Tell your classmate why mercury is used as a barometric liquid and not water.
3. Predict what will happen to the level of the mercury as you climb up a high.
Explain.
4. Discuss with your group members other instruments used to measure
atmospheric pressure apart from the mercury barometer.
5. Give a summarised report on your findings to the whole class in a class
discussion.
(a) Mercury Barometer

A mercury barometer consist of a thick-walled glass tube, which is closed at one
end. The tube is completely filled with mercury and inverted repeatedly to remove
air bubbles. The tube is then completely filled again with mercury and inverted
into a trough containing mercury.
The mercury column drops until it reaches
a height of 76 cm above the lower level of
the mercury meniscus (see Fig. 3.6) if the
barometer is at sea level.
When the barometer is placed in a region
with lower atmospheric pressure e.g. high
on the mountain, the height of the mercury
column in the tube drops to a level showing
the atmospheric pressure at that place..
This barometer can be used in the laboratory
or weather station, but it is not easy to move
it from one place to another.
(b) Fortin Barometer

The simple mercury barometer discussed Fig. 3.6: Mercury barometer
above cannot be used for accurate
measurement of atmospheric pressure. An
improved version called the Fortin barometer is used where high level of accuracy
is required. Fig. 3.7 shows a Fortin barometer.
65
(a) Drawing of Fortin barometer (b) Photgraph of Fortin barometer

Fig. 3.7: Fortin Barometer
Before taking the reading, the level of mercury surface in the reservoir is adjusted
by turning the adjusting screw until the surface of the mercury just touches the
tip of the ivory pointer. The mirror-like mercury surface produces an image of
the tip which helps to make the adjustment very accurate. The height of mercury
is then read from the main scale and the vernier scale. Any change in air pressure
makes the surface to move up and down hence this adjustment is necessary before
the barometer is read.
(c) Aneroid Barometer
The aneroid barometer is another example of a portable barometer. It consist
of a sealed metal chamber in the form of a flat cylinder with flexible walls. The
chamber is partially evacuated and the spring helps in preventing it from collapsing.
Fig. 3.8 shows an aneroid barometer.
(a) drawing of aneroid barometer (b) Photograph of aneroid barometer

Fig. 3.8: Aneroid Barometer
66
The chamber expands and contracts in response to changes in atmospheric

pressure. The movement of the chamber walls is transmitted by a mechanical
lever system which moves a pointer over a calibrated scale.
Normally, the pointer indicates a particular value of atmospheric pressure of the
surrounding. Any changes in pressure is noticeable by the movement of the pointer
to either side of this atmospheric value on the scale.
The aneroid barometer has no liquid and is wieldy used as an altimeter by
mountaineers or pilots to determine altitude. The scale can be calibrated to give
readings of altitude equivalent to a range of values of atmospheric pressure.
An aneroid barometer is also used as a weather glass to forecast the weather. Rainy
clouds form in areas of low pressure air. This is shown by the fall in the barometer
reading which often means that bad weather is coming.
Exercise 3.3
1. Name a liquid that is used for constructing instruments that measures

atmospheric pressure. Give a reason why the liquid is preferred over others.
2. Describe how you can measure atmospheric pressure at the top of a mountain.
3. Describe the working of:
(a) Mercury barometer.
(b) Aneroid barometer.
4. Explain how you can test for the vacuum in a barometer.
5. The atmospheric pressure in a particular day in Kigali was measured as
740 mmHg. Express this Nm–2. (Assume density of mercury is 13600 kgm–3
and g = 10 N/kg
6. A senior three student plans to make a barometer using sea-water of density
1025 kgm–3. If the atmospheric pressure is 104 000 Nm–2, what is the
minimum length of the tube that the student will require?
3.4 Applications of atmospheric pressure
Activity 3.6 To demonstrate some applications of atmospheric

pressure
Materials
• Drinking straw • Rubber sucker
• Glass • Empty beakers
• Drinking water • Syringe
67
Steps
1. Take a drinking straw provided to you and dip it in the glass with clean
drinking water.
2. Sip the water using the straw. What do you observe? Explain.
3. Dip the nozzle syringe in the water. What do you observe? Explain your
observation.
4. Take two empty beakers and fill one with water.
5. Now, discuss with your classmates how a rubber sucker, syringe and lift
pump work.
6. Note down the main points from your discussion.
7. Give a summarized report on your findings to the whole class through a
discussion.
(a) Drinking straw

Sucking through a straw, reduces the air pressure inside the straw.The atmospheric
pressure forces the water into your mouth through the straw (Fig 3.9).
Fig. 3.9: Drinking straw
(b) Syringe
A syringe consists of a tight-fitting piston in a barrel (Fig. 3.10(a)). It is used by
doctors to give injections.
Consider the case which the piston is not dipped in a liquid. When the piston is
pulled with the nozzle open, space is created in the barrel lowering the pressure
inside. Air from outside is pushed in by atmospheric pressure. Since the barrel is
also open to the outside, both the top and the bottom of the piston are under the
same force but in different directions. Hence the piston moves freely. The same
68
happens when the piston is pushed only, that the pressure increases inside the
tube and is balanced by the atmospheric pressure once the air is pushed out of
the barrel by the piston.
When the nozzle is closed and the piston is pushed, the pressure inside increases
and the movement of the piston is restricted.
Consider the case when the nozzle of the syringe inside a liquid (Fig. 3.10(b)).
piston rod
barrel
(a) Outside a liquid (b) Inside a liquid

Fig. 3.10: A syringe
When the piston is pulled (upstroke) the pressure inside reduces and the
atmospheric pressure on the surface of the liquid pushes the liquid into the barrel.
During a downstroke, the pressure inside increases and the liquid is expelled from
the barrel.
(c) Lift pump
A lift pump is used to raise liquids from a low level to a high level e.g raising water
from a well, drawing paraffin from a drum etc. The pump consists of a cylindrical
metal barrel with a delivery tube (Fig. 3.11). Inside the barrel, there is a piston
and two valves. Before starting to operate the pump, some of the liquid to be
drawn is poured on top of the piston in order to have a good air tight seal round
the piston and valve 2.
piston (plunger)
lever delivery tube

metal barrel
2 valve 2
1
valve 1
Fig. 3.11: A lift pump
69
The pump is operated by means of a lever which moves the plunger up and down
the barrel.
(i) Upstroke
During upstroke, the air between valve 1 and 2 expands and its pressure reduces
below atmospheric pressure. The atmospheric pressure on the water surface forces
water up past valve 1 into the space between valves 1 and 2. At the same time
valve 2 closes due to its weight as shown in Fig. 3.12(a).
outlet
up
tube
lever down
2
lever
2
1 1
water
(a) upstroke (b) downstroke
Fig. 3.12: Working of a lift pump
(ii) Downstroke
During downstroke, valve 1 closes due to its weight and the weight of the water
in the space between valves 1 and 2. As the plunger moves down, it forces valve
2 to open. Water escapes into the space above valve 2. As the process is repeated,
the plunger lifts the water out through the delivery tube.
Since atmospheric pressure can only support a water column of about 10 m, this
pump cannot raise water above a height of 10 m. The situation becomes even
worse when the pump is used in areas well above sea level where atmospheric
pressure is low.
(d) Siphon
A flexible tube may be used to empty fixed containers e.g. petrol tanks in cars,
which are otherwise not easy to empty directly. When used in this manner, the
flexible tube is called a siphon.
To empty the liquid in the container, the siphon is first filled with the liquid. One
end is pushed into the liquid and the other one left hanging as shown in Fig. 3.13.
The liquid comes out of the end C.
70
A
B
C
air in
Fig. 3.13 : A siphon
The pressure at B is atmospheric since B is on the same horizontal level as the

surface A. The pressure at C in the liquid is greater than the atmospheric pressure
by the pressure due to the height BC. This difference in pressure causes the water
to flow out of the container.
(e) Automatic flushing unit
Fig. 3.14 shows a flushing unit system.
tap
inverted U-tube
water
.
Fig. 3.14: Automatic flushing unit
It is normally used in urinals and flush toilets. When the water in the tank fills
above the top of the inverted U-tube, a pressure different between the two arms
is created. This causes the water to flow out of the tank. The tap can be adjusted
to enable the flushing unit to flush at pre-determined intervals.
(f) Rubber Sucker

A rubber sucker (Fig. 3.15) can be used for lifting heavy objects with flat smooth
surface and for hanging things on walls and windows.
71
Atmospheric
Partial vacuum
pressure
(Lower air pressure)
(Greater pressure)
Fig. 3.15: A Rubber Sucker
When a rubber sucker is pressed against the surface, usually a glass or tiled surface,
the air in the rubber sucker is forced out. This causes the space between surfaces
and sucker to have low pressure (partial vacuum). The external atmospheric
pressure, which is much higher acts on the rubber suckers, pressing it securely
against the wall.
(g) Vacuum cleaner
A vacuum cleaner (Fig. 3.16) applies the principle of atmospheric pressure to
remove dust particles.
Tas
debu
Exhaust
Motor
listrik Filter
Sikat
berputar
Fan
Intake
Fig. 3 16: A vacuum Cleaner
When the vacuum cleaner is switched on, the fan sucks out the air from the space
inside creating a partial vacuum. The atmospheric pressure outside, which is
greater, then forces air and dust particles into the filter bag. This traps the dust
particles but allows the air to flow out through an exit at the back.
72
Exercise 3.4
1. Explain the action of a drinking straw.

2. Draw and explain the features of a siphon.
3. Explain how a rubber sucker and vacuum cleaner works.
4. Draw and explain the function of a syringe.
5. What is the function of the compressed air in the force pump?

• Atmospheric pressure is the pressure due to the column of air above the
surface of the earth.
• Factors that influence atmospheric pressure include
(i) Altitude
(ii) Temperature
(iii) Water vapour concentration
(iv) Wind pattern
• Atmospheric pressure decreases with the increase in altitude.
• Atmospheric pressure decreases with increase in air temperature.
• Atmospheric pressure increases with convergence of winds and decreases
with divergence of winds.
• Instruments used to measure atmospheric pressure include:-
(i) Mercury barometer
(ii) Fortin barometer
(iii) Aneroid barometer
• Some of the situation in which atmospheric pressure is applied include:
(i) When using a drinking straw
(ii) When using a Syringe
(iii) When siphoning liquids
(iv) When cleaning using a vacuum cleaner
(v) When hanging clothes using a rubber sucker on a smooth wall
(vi) In an automatic flushing unit
73
Unit Test 3
1. The following are factors influencing atmospheric pressure. Which one is
not?
A. Temperature C. Cloud cover
B. Water vapour concentration D. Altitude
2. Which one of the following instruments is used for measuring atmospheric
pressure?
A. Temperature C. Thermometer
B. Barometer D. Lactometer
3. Which one of the following statements is correct?
A. When altitude increases atmospheric pressure also increases.
B. When altitude increases atmospheric pressure remains constant.
C. When altitude increases atmospheric pressure also decreases.
D. There is no relationship between increase in altitude and atmospheric
pressure.
4. Use the words given to fill in the spaces.
atmosphere, barometer,atmospheric,density
Earth surface is surrounded by a thick layer of air called _____. The _____
of air varies earth's surface to the outer place. The pressure exerted by air is
called _____ pressure and its measured using an instrument called _____.
5. State and briefly explain three factors that influence atmospheric pressure.
6. The Fig. 3.17 shows a rubber sucker. Explain why the sucker sticks on a flat
surface.
Fig. 3.17: Rubber sucker
74
7. The air pressure at the base of a mountain is 76.0 cm of mercury while at the top
it is 43.0 cm of mercury. Given that the average density of air is 1.25 kg m–3 and
the density of mercury is 13600 kg m–3, calculate the height of the mountain.
8. Describe briefly how an automatic flushing unit operates.
9. Explain how you can use mercury barometer to measure atmosphere pressure
in your school.
10. Outline and briefly discuss four applications of atmospheric pressure in our
daily lives.
11. Explain how altitude affects atmospheric pressure.
12. Hibimana and Hakizimana, Senior three students put hot drinking water
in a plastic container and then placed it on the ice in a basin as shown
in Fig. 3.18.
Fig. 3.18: Plastic container with hot water on ice
Explain why some of the ice cubes stick onto the bottle.
Hey! Drink safe water!

Always drink boiled water to avoid waterborne diseases such as typhoid.
75
Renewable and Non-renewable energy sources
Renewable and non-renewable

UNIT 4 energy sources
Key Unit Competence

By the end of the unit, the learner should be able to differentiate between renewable
and non-renewable energy sources and give examples.
Learning objectives
• Outline renewable and non-renewable energy sources.

• Describe basic features of renewable and non renewable energy sources.
Skills
• Compare energy sources in Rwanda and the rest of the world.
• Classify energy sources as renewable and non-renewable energy sources.
• Estimate the rate of fuel consumption in power stations.
• Analyse transformation of energy into different forms.
Attitude and value

• Recognize sources associated with carbon dioxide emission.
• Appreciate that in most instances, the sun is the prime energy source for world
energy.
• Appreciate higher rate of energy consumption in developed countries is from large
deposits of fossil fuels.
• Be aware of the moral and ethical issues associated with nuclear weapons.
• Adapt scientific method of thinking.
• Recognize the need of acquiring knowledge for analysing and modeling physical
processes.
76
Introduction
Unit Focus Activity

1. On a piece of paper provided by your teacher, write down:
(a) The meaning of renewable and non-renewable sources of energy
(b) At least eight sources of energy.
(c) Classify the sources you have listed as either renewable or non-renewable.
Organise your classification in a table.
(d) Three ways of conserving non-renewable sources of energy.
(e) Describe the energy transformations in
(i) a swinging pendulum
(ii) a loud speaker
2. Present your work to the rest of the class during the class discussion.
Energy is useful in our daily lives. We need energy for any kind of work we do.
Energy comes from different sources. Some are renewable while others are
non-renewable.
In this unit, we are going to learn in details about these sources of energy.
77
4.1 Energy sources
Activity 4.1
To identify and define energy sources
Material
• A chart showing pictures of different sources of energy
Steps
1. Look at the pictures in Fig. 4.1.
Wind Coal Biomas Propane
Uranium Water Natural gas Sun Petroleum

Fig 4.1:
2. What name collectively describes the objects in the pictures?
3. Discuss with your classmates the meaning of the terms ‘source’ and ‘energy
source’.
4. With the help of your teacher, compile your findings and note them in your books.
The word ‘source’ means the beginning of something. An energy source is a system
which produces energy in a certain way. Examples of energy sources are energy
from water, wind, the sun, geothermal sources, biomass sources such as energy
crops, fuels such as coal, oil, and natural gas.These sources show that energy
exists freely in nature.
4.2. Classification and characteristics of energy sources

Activity 4.2
To classify energy sources
Steps
1. Distinguish between renewable and non-renewable sources.
2. Revisit activity 4.1 and categorize the energy sources shown in the picture as
either renewable or non-renewable sources.
78
Energy sources are classified as renewable or non-renewable sources.
4.2.1 Renewable sources
Activity 4.3 To find out what renewable resources are

Steps
1. Discuss with your classmate the meaning of renewable sources of energy.
2. Identify at least three (3) characteristics of renewable sources of energy.
3. Describe three examples of renewable sources of energy in Rwanda and the
World.
Renewable energy sources are energy sources that are continually replenished.
They exist infinitely (they never run out).
4.2.1.1 Characteristics of renewable resources
• These resources are capable of regeneration.
• They are renewed along with exploitation and hence are always available for use.
• The regeneration of these sources involves some ecological processes on a
time scale.
• The renewable sources become non-renewable if used at a greater rate than
the environment’s capacity to replenish them.
4.2.1.2 Examples of renewable sources of energy

Renewable energy includes biomass, wind, hydro-power, geothermal and solar
sources. Renewable energy can be converted to electricity, which is stored and
transported to our homes and industries for use.
(a) Wind energy
Activity 4.4
To conduct a research on wind energy
Materials:
• Reference books
• Internet
Steps
1. Why is wind energy regarded as a renewable source of energy?
2. Outline advantages and disadvantages of using wind energy.
79
3. Compare and discuss your findings with other groups in your class.
4. With the help of the teacher, note down your findings in your note books.
The term "wind energy" or "wind power" refers to the energy produced by wind.
It is used to rotate turbines that convert kinetic energy into other forms that can
be used to do specific tasks such as grinding grain, pumping water or generate
electricity to power homes, businesses, schools, and industries.
Wind is produced as a result of giant
convection currents in the Earth's
atmosphere which are driven by heat
energy from the sun. This means
that the kinetic energy in wind is a
renewable energy resource; as long as
the sun exists, the winds will always be.
Figure 4.2 shows a wind turbine.
Wind turbines have huge blades
mounted on a tall tower. The blades
are connected to nacelle or housing that
contain gears linked to a generator. As
the wind blows, it transfers some of its
kinetic energy to the blades which turn Figure 4.2: A wind turbine
and drive the generator. Several wind
turbines may be grouped together in windy locations to form wind farms.
As at the writing of this book, the Rwanda government had put a number of
initiatives on exploring ways of implementing wind power generation at suitable
locations in the country. However, small scale exploitation of wind power like
mills for pumping water or generating homestead electricity is in common use
in some regions of Rwanda.
Advantages of wind energy

• Exploitation and utilisation of wind energy has no associated fuel costs.
• No harmful polluting gases are produced by wind energy.
Disadvantages of wind energy

• Wind farms are noisy and may cause noise pollution for people living near
them.
• The amount of electricity generated depends on the strength of the wind. If
there is no wind, there is no electricity.
80
Project 4.1. To make a simple wind turbine
Materials
• Manilla paper • A pair of sciscors
• Pencil • A nail • Stapler
Steps
1. Cut a square piece from the manila paper.
2. Use a ruler to draw diagonal lines from corner to corner. Make a small mark
along each diagonal line about 2 cm from the center of the square piece.
3. Cut along the diagonal lines toward the center until you reach the 2 cm
mark.
4. Fold alternating corners onto the center and staple the layers together, but
make sure to leave space between staples in the very center.
5. When all four 'blades' are folded in, push a straight nail through all the
layers at the center. Remove the nail and push the pencil through the hole
to act as the ‘shaft’. The turbine is now complete (Fig 4.3). Make sure the
turbine is free to rotate on the pencil
Fig. 4.3: A simple wind turbine
6. Hold the turbine in the direction of the wind. The wind currents blow the
curved part of the blades, causing them to spin.
81
(b) Water energy
Activity 4.5 To research on the exploitation, advantages and

disadvantages of water energy
Materials:
• Reference books • Internet enabled computers
Steps
1. Conduct a research from reference books or the internet on how electricity
is generated using waves, tides and flowing water (HEP).
2. Discuss the disadvantages associated with each one of them.
3. Compare and discuss your findings with the rest of your classmates.
4. With the help of your teacher, note down your findings in your notebooks.
Moving water mainly produces energy in the form of wave power, tidal barrage,
and hydroelectric power.
Wave energy
The water in the sea rises and falls because of waves on the surface.Wave machines
use the kinetic energy in this movement to drive electricity generators.
Tidal barrage
Huge amounts of water move in and out of river mouths each day because of
the tides. A tidal barrage is a barrier built over a river estuary to make use of the
kinetic energy in the moving water. The barrage contains electricity generators,
which are driven by the water rushing through tubes in the barrage.
Hydroelectric power (HEP)

Hydroelectric power stations use the kinetic energy in moving water. The water
comes from vast reservoirs behind a dam built across a river valley. The water high
up in the dam posses gravitational potential energy. This is transformed to kinetic
energy as the water rushes down through tubes inside the dam. The moving water
drives electrical generators which are built at the base of the dam.
Examples of HEP stations in Rwanda include Nyabarongo Power Station with a
capacity of 28 megawatts, located in Mushishiro, Muhanga District and Rukarara
II, which has a capacity of 2.2 megawatts in Nyamagabe district.
82
Fig 4.4 Shows Nyabarongo dam.
Fig. 4.4: Nyabarongo dam in Mushishiro in Muhanga district
Hydro dams can generate large amounts of electricity. However, dry periods can
drain the reservoirs resulting to less power production. The flooding of reservoirs
behind dams and slowing of the flow of the river below the dam can have a serious
impact on the ecology around the dam. The number of sites suitable for new
dams is limited.
Fig 4.5 and 4.6 shows distribution and transmission lines in Rwanda respectively.
Fig 4.5 An electricity distribution line
83
Fig. 4.6: Higher capacity transmission lines being set up
For your safety

1. Never let yourself or anything else (a tool, scaffolding, construction
materials or a tree branch) come within three metres of a medium-voltage
wire when pruning or felling a tree.
2. These high-voltage lines are not insulated hence nothing should come near
them. Carrying out work near these lines or planting anything in the right-
of-way without proper authorization is prohibited.
3. Don't climb fences near hydropower facilities. They are there for your
safety!
Advantages of water energy

• Water power in its various forms is a renewable energy resource and has no
associated fuel costs.
• No harmful polluting gases are produced.
• Tidal barrages and hydroelectric power stations are very reliable and can be
turned on quickly.
Disadvantages of water energy

• It has been difficult to scale up the designs for wave machines to produce
large amounts of electricity.
• Tidal barrages destroy the habitat of estuary species, including wading birds.
• Hydroelectricity dams may flood farmlands and push people away from
their homes.
• The rotting vegetation underwater releases methane, which is a greenhouse
gas that contributes to global warming and ozone layer depletion.
84
(c) Solar energy
Activity 4.6 To demonstrate solar energy using a convex lens
Materials:
• Convex lens
• Thin piece of paper
Steps
1. Choose a clear bright day with a lot of sunshine.
2. Place a convex lens in the path of sunlight.
3. Place a thin piece of paper under the lens. Ensure that light is focused on the
paper.
4. Wait for some minutes. What happens to the paper? Explain.
"Solar" is a Latin word for "sun". Solar power is energy from the sun. It is a powerful
source of energy. Without it, there would be no life. It has been considered Earth's
main source of energy for many years because of the vast amounts of energy that
it makes freely available, if harnessed by modern technology.
Unfortunately, the sun is not available in the night, and on some days, clouds and
rains and other natural conditions prevent the sun's powerful rays from reaching
us. This means that it is not always available. This is why we cannot rely on solar
energy alone.
An example of a solar power station in Rwanda is Ngoma solar power station
located in Agahozo village (ASYU) in Rubuna Sector, Rwamagana District;
which produces 8.5 megawatts. The largest solar power station in the world as
at the time of writing this book is Cestas Solar Plant in France whose capacity is
300 megawatts.
Transducers that tap solar energy and convert it to other forms of energy include
solar cells (photovoltaic cells) and solar thermal heaters.
Solar cells( Solar panels)

Solar cells also called photovoltaic cells or PV devices are devices that convert
light energy directly into electrical energy. In these cells, there are semiconductors
(silicon alloys and other materials). You may have seen small solar cells on
calculators or some mobile phones. Larger arrays of solar cells are used to power
road signs in remote areas, and even larger arrays are used to power satellites in
orbit around Earth.
85
(a) Placing solar panel on (b) Cleaning solar panel

the roof top
Fig. 4.7: Solar panel
Solar thermal heaters
Project work 4.2

To make a solar balloon
Materials:
• A small trash bag
• 13 µm thick black bag
• Scissors
• A sewing thread
• Bottle cap
Steps
1. Choose your day: clear sky, little or no wind, early in the morning before
thermal activity.
2. Get the trash bag and tie its neck with a sewing thread.
3. Cut other four pieces of the sewing thread and tie them on the neck of the trash
bag around it and use them to attach a small weight like a plastic bottle cap.
4. Inflate the flash bag (Fig. 4.8)
5. Lay the trash bag on the ground clearly exposed to the sun.
6. Observe and explain what will happen when the bag is gradually heated by
the sun.
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Fig 4.8: A simple solar balloon.
Solar hot water systems (also known as Solar thermal) harness heat from the sun by
capturing energy which is radiated by the sun within solar panels or collectors.
Solar thermal heaters should not be confused with Solar Photovoltaic (PV), devices
which are designed to generate electricity.
The heat energy is then moved down pipes to the hot water cylinder within your
home, reducing the need to use gas, oil or electricity to heat the required water.
A pump pushes cold water from a storage tank through pipes in the solar panel.
The water is heated by heat energy from the sun and returns to the tank. In some
systems, a conventional boiler may be used to increase the temperature of the
water. They are often located on the roofs of buildings where they can receive
the most sunlight.
Fig 4.9 shows the outline of how they work.
Figure 4.9 showing a solar thermal heater
Solar thermal power plant. Here, a concentration of the sun's energy by many
panels is used to heat up water into steam, which is then used to turn turbines
to produce electricity.
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Guess what!
Solar power stations usually require a lot of space to capture a lot of the sun's
energy!
The parabolic trough system uses this kind of system. Here, troughs are designed
to direct the sun's energy to absorber tubes as long as the sun is up.
Many of these parabolic troughs are installed to collect massive amounts of energy
for the rods to heat water to turn turbines.
Advantages of solar energy

• Solar energy is a renewable energy resource and it has no associated fuel costs.
• No harmful polluting gases are produced.
Disadvantages of solar energy

• Solar cells are expensive and inefficient, so the cost of their electricity is high.
• Solar panels may only produce very hot water in very sunny areas, and in
cooler areas may need to be supplemented with conventional boilers.
• Although warm water can be produced even on cloudy days, neither solar
cells nor solar panels work at night.
Did you know?

In many African villages, people use the sun's energy to dry foodstuff like fish,
corn and cocoa for storage? That is raw solar energy in use. They spread the
foods on large mats and trays in the hot sun for days until the required moisture
content is attained. This is commonly practised in regions with a lot of sun.
Geothermal energy
Geothermal energy come in the form of hot steam from underground. It is clean
and sustainable. Resources of geothermal energy range from the shallow ground
to hot water and hot rock found a few miles beneath the Earth's surface, and down
even deeper to the extremely high temperatures of molten rock called magma.
The steam is used to generate electricity.
Rwanda has been active in exploring its prospective geothermal potential which is
manifested in form of hot springs in North-West region (Rubavu, Kalisimbi and
Kinigi) and volcanoes in the South-West region (Bugarama). As at the time of
writing this book, the exploration at Bugarama was at a reconnaissance stage and
was to be followed by geo-scientific survey to estimate the potential of the area.
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Exercise 4.1
1. What is renewable energy?

2. Why is renewable energy preferable?
3. Name:
(a) two renewable sources of energy.
(b) which component of sun's energy is responsible for drying clothes?
(c) two forms of energy usually used at homes.
(d) the radiation emitted from a hot source.
(e) two activities in our daily life in which solar energy is used.
(f) the kind of surface that absorbs maximum heat.
(g) the device that directly converts solar energy into electrical energy.
(h) the two elements which are used to fabricate solar cells.
4. What is the main cause of blowing of the wind.
5. State four characteristics of renewable sources of energy.
6. What are some ways that humans used renewable resources of energy
centuries or even a millennia ago?
7. Research what kind(s) of renewable energy our country produces. Why is
our country an optimal location for that form of renewable energy?
8. Choose a renewable energy resource. Brainstorm on five types of careers in
that field.
4.2.2 Non-renewable sources of energy
Activity 4.7 To find what non-renewable resources are and their

characteristics
1. Discuss with your classmates the meaning of non-renewable sources of energy.
2. Identify three (3) characteristics of non-renewable sources of energy.
3. Describe three examples of non-renewable sources of energy in Rwanda.
4. Note down your findings in your note books.
6. Discuss your findings as a whole class with the help of the teacher and note
them down in your notebooks.
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Non-renewable energy comes from sources that will run out or will not be
replenished in our lifetimes or even in many, many lifetimes.
4.2.2.1 Characteristics of non-renewable resources

• These sources are available only in finite quantities and hence are termed as
“stock resources”
• They cannot be regenerated easily.
• They are concentrated as minerals usually in the lithosphere of the earth in a
number of forms.
• They may be solids (coal, lignite, and minerals), liquids (petroleum) or gases
(natural gases).
Note: Most non-renewable energy come from fossil fuels. There is also
uranium, which is not a fossil fuel.
4.2.2.2 Examples of non-renewable sources of energy

Examples of non-renewable sources of energy include fossil fuels (coal, petroleum,
and natural gas) and nuclear energy.
(a) Fossil fuels
Activity 4.8 To find out the extent to which fossil fuels are used,
their advantages and disadvatages
Steps
1. Collect data on how many of you in your class use kerosene, gas or none of
these for cooking in your homes.
2. In a class discussion, identify advantages and disadvantages of using fossil
fuels as a source of energy.
3. Discuss how the use of fossil fuels contributes to global warming.
4. Compile your findings with your teacher and write down the notes in your
note books.
Fossil fuels are mainly made up of carbon. It is believed that fossil fuels were
formed over 300 million years ago when the earth was a lot different in its
landscape. It had swampy forests and very shallow seas. This time is referred to
as 'Carboniferous Period'.
The fossil fuels are coal, oil and natural gas. They are fuels because they release
heat energy when they are burned. They have chemical energy stored within them.
Fossil fuels are usually found in one location as their formation is from a similar
process. Let us take a look at the diagram in Fig 4.10 and see how fossil fuels are
formed under the sea.
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water dead organisms sedimentary impermeable

plant and animals rocks rocks
trapped fossils coal oil natural gas
Figure 4.10: Formation of fossils fuels
Millions of years ago, dead sea organisms, plants and animals settled on the ocean
floor and in the porous rocks. This organic matter had stored energy in them as
they had used the solar energy to make foods through photosynthesis.
With time, sand, sediments and impermeable rock settled on the organic matter, trapping
energy within the porous rocks. That formed pockets of coal, oil and natural gas.
Movements in the earth and rock create spaces that force these energy types to
collect in to well-defined areas. Using technology, engineers are able to drill down
into the seabed to tap the stored energy, commonly known as crude oil.
Fig 4.11 shows an energy transfer diagram for using generation of electricity from
a fossil fuel like coal.
Figure 4.11: Energy transformation diagram
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Advantages of using fossil fuels
Fossil fuels are relatively cheap and easy to obtain. This is the reason why most
people prefer using gas or kerosene for cooking over electricity.
Fig. 4.12 shows a (a)cooking gas cylinder and (b) a kerosene stove and kerosene
lamp.
Figure 4.12 (a). Gas cylinders at a petrol station in Kigali Rwanda
Figure 4.12(b). Kerosene stove and lamp
Disadvantages of using fossil fuels
• Fossil fuels are non-renewable energy resources. Their supply is limited and
they will eventually run out.
• Fossil fuels release carbon dioxide when they burn, which adds to the
greenhouse effect and increases global warming. Of the three fossil fuels, for a
given amount of energy released, coal produces the most carbon dioxide and
natural gas produces the least.
• Coal and oil release sulphur dioxide gas when they burn, which causes
breathing problems for living creatures and contributes to acid rain.
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In Rwanda, fossil fuels are widely used in many ways including kerosene and gas
for cooking. Diesel and petrol are used for powering locomotives, generators and
electricity generation. Substantial amounts of cooking gas (methane) is mined from
Lake Kivu, where an estimated of 120 to 250 million cubic metres of methane
is generated annually.
This resource can develop methane to power projects and other uses such as
fertilizer and gas-to liquids projects.
The methane in Lake Kivu is estimated to be sufficient to generate 700 MW of
electricity over a period of 55 years.
Exercise 4.2
1. Use the words provided below to fill in the blank spaces.

Impermeable, photosynthesis, energy, plants, matter, stored energy, sea
bed, movements, shifts, well defined, technology, dead sea organisms, sun’s
energy, animals.
Millions of years ago, _____, _____ and _____ settled on the ocean floor
and in the porous rocks. This organic matter had stored _______ in them as
they used the ______ to prepare foods (proteins) for themselves (_______).
With time, sand, sediments and _____ rock settled on the organic _______,
trapping its' energy within the porous rocks. That formed pockets of coal,
oil and natural gas. Earth ____ and rock _______ create spaces that force
to collect these energy types into ______ areas. With the help of ______,
engineers are able to drill down into the ______ to tap the ______, which is
commonly known as crude oil.
For questions 2 - 7, select the most suitable response from the options
given.
2. How are fossil fuels formed?
A. They are made by dead organisms buried in the ground and decomposed.
B. They are made by acid rain.
C. They are made by bread mold.
3. Fossil fuels are
A. Renewable
B. Non-renewable
C. sustainable
4. What are the three main types of fossil fuels?
A. Coal, natural gas and oil
B. Oil, solar and coal
C. Solar, coal and natural gas
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5. How do we use coal as an energy source?

A. Coal is not used as an energy source to generate electricity
B. The coal is melted to turn turbines that generate electricity
C. The coal is burned to produce steam to turn turbines that generate
electricity
6. What is the main element in fossil fuels
A. Oxygen
B. Water
C. Carbon dioxide
7. Why does the world continue to use fossil fuels?
A. Because they will never run out and we can continue to use them forever.
B. Because they are easily available and are cheap.
C. Because they are the only source of energy available for our use.
8. Match
1. Petroleum Black rock burned to make electricity.
2. Wind Energy from heat inside the Earth

3. Biomass. Energy from flowing water.
4. Uranium Energy from wood, waste, and garbage.
5. Propane Energy from moving air.
6. Solar Produces energy by splitting of atoms.
7. Geothermal Portable fossil fuel gas often used in grills.
8. Hydropower Fossil fuel for cars, trucks, and jets.
9. Coal Fossil fuel gas moved by pipeline.
10. Natural Gas Energy in rays from the sun.
9. Describe the formation of fossil fuels.

10. Explain why fossil fuels are called non-renewable sources.
11. How does the use of fossil fuels affect the environment?
(b) Nuclear energy

The main nuclear fuels are uranium and plutonium. These are radioactive metals.
Nuclear fuels are not burnt to release energy. Instead, the fuels are involved in
nuclear reactions in the nuclear reactor, which released heat.The rest of the process
of generating electricity is then identical to the process of using fossil fuels.
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The heat energy is used to boil water. The kinetic energy in the expanding steam
spins turbines, which then drive generators to produce electricity.
Fig 4.13 shows the structure of a nuclear reactor.
Fig. 4.13: A nuclear power plant
Advantages of nuclear fuels

• Unlike fossil fuels, nuclear fuels do not produce carbon dioxide or sulphur
dioxide.
Disadvantages of nuclear fuels

• Like fossil fuels, nuclear fuels are non-renewable energy resources hence are
depleted with time.
• If there is an accident, large amounts of radioactive material could be released
into the environment which would be dangerous to human beings, animals
and plants.
• In addition, nuclear waste remains radioactivate and is hazardous to health
for thousands of years. It must be stored safely.
For your safety!

Make sure you wear protective clothing when you visit sites with radioactive
substances.
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Exercise 4.3
For questions 1 - 6, select the most suitable response from the choices
given.
1. Which of the following is used to generate nuclear power?
A. Uranium B. Hydrogen
C. Carbon D. None of the above
2. How does nuclear fission occur?
A. Protons gather in the nucleus of uranium atoms and cause the nucleus
to explode and release energy in the form of heat.
B. Neutrons smash into the nucleus of uranium atoms. The neutrons are
then split and release the energy in the form of heat.
3. What are the advantages of nuclear power? Check all that apply.
A. Produces a small amount of waste
B. Nuclear power is very reliable
C. Nucleus power does not produce either smoke or carbon dioxide
D. Nuclear power is not expensive to make
4. Which are the main nuclear fuels?
A. Uranium and plutonium
B. Uranium and tritium
C. Plutonium and tritium
5. Which of the following sub-atomic particles is never found in the nucleus of
an atom?
A. Neutron B. Proton
C. Electron
6. Nuclear power is a
A. non-renewable source of energy.
B. renewable source of energy.
C. a stable energy source.
7. Fill in the blank spaces
Nuclear …………. is the ………. of a heavy, unstable nucleus into two
lighter nuclei, releasing vast amounts of energy.
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Nuclear ………….. is the process where two light nuclei ……… together
releasing vast amounts of energy.
8. Describe the production of energy from nuclear sources.
9. Justify why nuclear energy is a non-renewable source of energy
10. Describe two problems caused by the use of nuclear energy.
4.3 Energy Transformations

As we learnt in S1, an energy transformation is the change of energy from one
form to another. Examples of forms of energy include electrical, thermal, nuclear,
mechanical, electromagnetic, sound, and chemical energy.
All energy transformations obey the law of conservation of energy that states that
energy cannot be created nor destroyed but simply changes from one form to another i.e.
total energy in a closed system is conserved. A device that converts energy from one
form to another is known as a transducer.
4.3.1 Examples of energy transformations
4.3.1.1 Potential energy to kinetic energy and vice versa

Swinging Pendulum
Activity 4.9 To demonstrate transformation of P.E to K.E and

vise versa
Materials:
• Clamp and stand • 1 m of string • Pendulum bob
• Calculator • Weighing scale
Steps
1. Measure and record the mass, m, of the pendulum bob.

2. Tie the bob with a string and suspend it on a clamp with the string such that
the bob is just about to touch the ground when hanging freely (See Fig. 4.14).
3. Pick an arbitrary vertical height at which you will release their pendulum.
4. Measure and record this height, h. Preferably it should range from
15 cm - 40 cm from the floor.
5. Calculate the potential energy of the bob at this height.
6. Release the pendulum bob from this height. Observe and explain what
happens.
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Fig. 4.14. A swinging pendulum
7 Discuss the following questions in your group:

(a) Where will the pendulum have the greatest potential energy?
(b) Where will it have the greatest kinetic energy?
8. Calculate the kinetic energy at the bottom of the swing.
9. Calculate the theoretical velocity, v, at the bottom of the swing.
10. Make short notes from your discussion and read them to the whole class.
At the highest point of swing, potential energy is maximum while kinetic energy is
minimum (zero).When the pendulum bob is set to oscillate, the potential energy
is transformed to kinetic energy.
Fig. 4.15: Oscillating pendulum
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When the bob is swinging, its speed is highest at its lowest point thus it has the
maximum kinetic energy and zero potential energy due to its position. At the
highest point, it has the maximum potential energy due to its position and zero
kinetic energy.
4.3.1.2 Conversion of electrical energy into mechanical energy and
vice versa.
One device that converts electrical energy to mechanical energy is a motor. It
actually converts electrical energy to kinetic energy.
Activity 4.10 To make a simple motor

Materials:
• Safety pins, nails (screws) • Battery holder
• Wood block • Disk magnet
• Wire • Scotch tape
• Sharp knife/razor blade
Steps
Fig 4.16 Shows a simple electric motor that we are going to make following the
steps below.
Figure 4.16. An electric motor
1. Wind a wire to form a coil (solenoid) on a pen, making 6 or 9 turns and leave
some inches of wire free at each end.
2. Carefully, pull the coil off the pin (former) and make its shape permanently
by wrapping it around the loop.
3. Hold the coil at the edge of a table so that the coil is straight up and down(not
flat on the table) and one of the free wire ends lying on the table.
99
4. With a sharp knife, remove the top half of the insulation from the free wire
end. Be careful to leave the bottom half of the wire with the enamel insulation
intact. Do the same thing to the other free wire end.
5. Bend two safety pins from the middle.
6. Use nails (screws) to mount the bent safety pins on the wood block so that
the loops face each other and are about 1 inch apart.
7. Attach the wires from the battery holder to the supports(bent safety pins).
8. Swing the safety pins apart a little and insert the coil into both rings.
9. Insert the battery into the holder. Place the magnet on top of the wood just
underneath the coil. Make sure the coil can spin freely and it just misses the
magnet.
10. Spin the coil (armature) gently. What do you observe?
11. Discuss the energy transformation in the simple motor
A device that converts mechanical energy into electrical energy is a generator. It

uses the concept of electromagnet induction where electric current is induced
in a conductor moving inside a magnetic field or a conductor cutting through
magnetic field linking two points.
Activity 4.11 To demonstrate transformation of mechanical

energy to electrical energy
Materials:
• Galvanometer • Connecting wires
• Coil (solenoid) • Bar magnet
• Insulated copper wire
Steps
1. Make a coil (solenoid) using an insulated copper wire.
2. Connect the ends of the solenoid using connecting wires to a sensitive
galvanometer.
3. Quickly introduce (push) the bar magnet into the solenoid and stop (Fig. 4.17(a)).
4. Withdraw the magnet quickly from the coil and stop (Fig. 4.17(b)).
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(a) (b)
Fig 4.17: Mechanical energy being transformed into electrical energy.
5. Move both the bar magnet and the coil at the same speed and in the same
direction. Observe and explain what happens to the galvanometer.
Exercise 4.4
For questions 1 - 6, select the most suitable response from the choices
given
1. What energy transformation takes place in a toaster? Discuss.
A. Mechanical → electrical
B. Electrical → thermal
C. Thermal → chemical
2. What energy transformation takes place when an object burns?
A. Chemical → thermal
B. Electrical → mechanical
C. Electromagnetic → chemical
3. You notice that after you walk across a room, you feel a spark when you touch
the doorknob. What energy transformation must have taken place?
A. Thermal → electromagnetic
B. Mechanical → electrical
C. Electromagnetic → mechanical
4. In what state is chemical energy always in?
A. Kinetic B. Potential
5. When a turbine rotates to produce electricity, what energy transformation
takes place?
A. Thermal → mechanical
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(b) Electrical → electromagnetic

(c) Chemical → electrical
(d) Mechanical → electrical
6. What type of energy is produced and wasted during most energy
transformations?
A. Chemical B. Thermal
C. Electrical
7. Fig 4.18 shows a clock
Fig.4.18 : a clock
Use the correct words from the following to fill the blank spaces in the
paragraph below them.
Chemical, electrical, kinetic, light, sound, thermal
The battery stores energy. When the battery is first connected,

electrical energy is transferred to energy of the clock’s hands. Some
of the electrical energy is transferred to the surrounding as energy.
When the alarm bell rings, electrical energy is transferred to energy.
8. Match
Transducer Energy transformation

bulb Electrical to mechanical
Guitar wire Chemical to light and thermal
Athlete running Chemical to electrical
candle Chemical to mechanical
motor Thermal to mechanical
A battery in use Electrical to light
Hot air balloon Mechanical to sound
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9. A stone is thrown into the air. Explain the energy changes taking place from
the moment it is thrown to the moment it falls back to its starting point
10. Uwimana a student in Senior three has a bicycle with a motor lighting
system. Explain the energy transformation within the system.
11. Discuss the energy transformation in Rusumo hydroelectric power station.

• Energy sources fall into two categories: non-renewable and renewable.
• Non-renewable energy sources, like coal, nuclear, oil, and natural gas, are
available in limited supplies. They take a long time for them to be replenished.
• Renewable sources are replenished naturally and over relatively short periods
of time. The five major renewable energy sources are solar, wind, water
(hydro), biomass, and geothermal.
• Since the dawn of humanity people have used renewable sources of energy
to survive — wood for cooking and heating, wind and water for milling grain,
and solar for lighting fires.
• Fossil fuels include oil, natural gas, and coal. They formed from the buried
remains of plants and animals. They are non-renewable energy sources.
They can be burned to supply energy for generating electricity.
Other renewable energy sources
• Tidal energy, wind energy and geothermal energy can be converted into
electrical energy in generators.
Nuclear power plants
• Nuclear reactors use the energy released in the fission of U-235 to produce
electricity.
• The energy released in the fission reaction is used to make steam. The steam
drives a turbine that rotates an electric generator.
The risks of nuclear energy
• Nuclear power generation produces high-level nuclear wastes.
• Organisms could be damaged if radiation is released from the reactor.
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• Nuclear waste is the radioactive by-product produced by using radioactive

materials.
Energy transformation
• Energy can be transformed from one form to the other. For example in a
pendulum; potential energy → kinetic energy → potential energy,
in a motor; mechanical energy → electrical energy.
• Photovoltaic cells, or solar cells, convert radiant energy from the sun into
electrical energy.
• Hydroelectric power plants convert the potential enrgy in water to electrical
energy.
Unit Test 4
For questions 1 - 10, select the question that you think it is right.
1. Energy sources that once used can replenish themselves and can be used
again and again are termed as
A. Non-renewable B. Renewable
C. Finite D. Kinetic
2. Which of the energy sources listed is not a renewable source of energy
A. Oil B. Solar
C. Wind D. Tidal
(e) Geothermal
3. What is the other name for non-renewable
A. Non-renewable B. Finite
C. Infinite
4. Energy sources that once used cannot be replenished are called;
A. Non-renewable B. Renewable
C. Infinite D. Potential
E. Kinetic
5. What natural source is harnessed to generate hydro-electric power (HEP)?
A. Wind B. Water
C. Light D. Heat
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6. What is the other name of the renewable energy source generated from using
volcanic heat found under the earth’s surface?
A. Wind B. Hydro-electric power (HEP)
C. Tidal D. Solar
E. Geothermal
7. What is the name of the renewable energy supply generated by capturing
sunlight in panels that covert the sunlight into electricity?
A. Wind B. Hydro-electric power
C. Tidal D. Solar
E. Geothermal
8. Which statement below is not an advantage of tidal energy?
A. Tidal barrages have the potential to generate a lot of energy
B. Tidal barrages can double as bridges
C. Tidal barrages can help to prevent flooding
D. Tidal energy is renewable and once in use can be used for generations
9. What type of energy source comes from radioactive minerals such as uranium
and releases energy when the atoms of the radioactive minerals are split by
nuclear fission?
A. Biomass B. Natural gas
C. Geothermal D. Hydro-electric power
E. Nuclear
10. What type of energy source is formed from fossilized plants and is found
sandwiched between other types of rock in the earth?
A. Oil B. Coal
C. Geothermal D. Biomass
E. Nuclear
11. Describe the advantages and disadvantages of using fossil fuels to generate
electricity.
12. Describe how fossil fuels are formed.
13. Name three compounds that are formed from the chemical compounds in
petroleum.
14. If fossil fuels are still forming, why are they considered to be a nonrenewable
resource?
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15. Solar energy has provided almost all the sources of energy on the earth.
Explain.
16. Explain two advantages and two disadvantages of using solar energy.
17. Explain why geothermal energy is unlikely to become a major energy source?
18. Describe three ways solar energy can be used.
19. Explain how the generation of electricity by hydroelectric, tidal, and wind
sources are similar to each other.
106
Laws of Thermodynamics
UNIT 5 Heat Transfer and Quantity of heat
Key Unit Competence

By the end of the unit, the learner should be able to evaluate modes of heat transfer
and determine specific heat capacity of metal block.
Learning objectives

• Recall the differences between heat and temperature.
• Explain modes of heat transfer.
• Explain thermal expansion of solids.
• Explain applications of heat exchanges.
• Explain thermal expansions.
• Define the terms heat capacity and specific heat capacity.
• Describe experiments to determine specific capacity of a metal.
• Determine coefficient of expansion.
Skills
• Apply knowledge of heat capacity to predict the behaviour of different materials as
temperature is changed.
• Describe different modes of heat exchange.
• Calculate heat capacity, specific heat capacity, latent heat, specific latent heat
• Carry out an investigation that illustrates heat exchange.
• Analyse experiments on thermal expansion of solids.
• Differentiate linear surface and volume expansion of different object.
Attitude and value

• Appreciate application of modes of heat transfer.
• Acquire knowledge in analyzing the behaviour of materials under thermal expansion.
• Be aware of difficulties related to heat exchanges.
107
Introduction
Unit Focus Activity

Fig 5.1 below shows a person heating some liquid in saucepan over an electric coil.
Fig. 5.1: Heating some liquid in a saucepan
1. Identify the modes of heat transfer marked A, B and C.

2. Discuss how each of the modes of heat transfer takes place, citing the states
of matter through which the processes take place.
3. Describe one application of each type of the above modes of heat transfer in
real life.
4. Present your findings to the rest of the class in a class discussion.
108
In our environment, most interactions between systems involve transfer of heat

from one system to another. For example, when we bask in the sun, we feel
warmer, when we touch a hot sauce pan, we feel the heat. In this unit, we are will
discuss the different modes through which heat is transferred from one region to
another. We will begin by reviewing the difference between heat and temperature.
5.1 Heat and temperature

The following activity will enable us to understand the difference between heat
and temperature.
Activity 5.1 To investigate the difference between heat and

temperature
Materials
• A measuring cylinder • Two identified test tubes
• A beaker • A stirrer
• Cooking oil (about 200 g) • Water
• Bunsen burner (source of heat) • 2 clamps and stands
• 2 thermometers
Steps
1. Take equivalent masses of water and cooking oil in two identical test tubes
fitted with two identical thermometers. Place these tubes in a large beaker
containing water i.e water bath (Fig. 5.2).
thermometers
Stirrer
oil
water water bath
bunsen stirrer
burner
Fig. 5.2: Differences between heat and temperature
2. Record the initial temperature of both water and oil in the tubes.
3. Heat the water in the beaker and make sure that the heat is distributed uniformly
by stirring the water. After some time, note the temperature reading of water
and oil in the tubes. Are the two temperature reading the same? Explain.
109
When the same amount of heat energy is supplied to equal masses of two different
substances, that are initially at same temperature, they both gain equal amounts
of heat energy but their temperature rise to different values.
Heat is a form of energy that flows from a hot to a cold body while temperature
is the degree of hotness or coldness of a substance.
5.2 Modes of heat transfer
Activity 5.2 To describe the modes of heat transfer
Revisit the unit focus activity and differenciate between the three modes of heat
transfer.
5.2.1 Heat transfer by conduction

5.2.1.1 Demonstration of conduction of heat
The following experiment will illustrate conduction of heat in solids.
Activity 5.3 To investigate heat transfer in solids

Materials:
• A metal spoon • A beaker full of boiling water
• Bunsen burner • Wax
Steps
1. Take a metal spoon at room temperature. Dip the spoon (with the other end
waxed) into a beaker full of boiling water (Fig. 5.3). After a few minutes,
what do you observe? Touch the free end of the metal spoon outside water.
What do you feel? Explain.
waxed end
metal spoon
Fig. 5.3: A spoon inside boiling water
2. Discuss how heat is transferred by conduction.
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Solids transfer heat from one point to another. For instance, the free end of
the spoon outside the beaker in Fig. 5.3 has become hot. Heat energy has been
transferred from the inside to the outside through the metal spoon i.e. from a
region of higher temperature to a region of lower temperature.
This process of transfer of heat energy in solids is called conduction. Conduction
is the transfer of heat from one substance to another that is in direct contact with
it. In conduction there is no visible movement of the heated particles.
5.2.1.2 Mechanism of conduction of heat

We have already learnt that when temperature increases, the molecules have larger
vibrations. This knowledge can help us understand the mechanism of conduction
of heat. When the molecules at one end of a solid receive heat energy from the
heat supply, they begin to vibrate vigorously. These molecules collide against the
neighbouring molecules and agitate them. The agitated molecules, in turn, agitate
the molecules in the next layer and so on till the molecules at the other end of the
solid are agitated. Thus, the heat is passed from one point to another till the other
end becomes hot. Hence, in conduction, energy transfer takes place by vibration
of the molecules. There is no actual movement of the heated particles.
Activity 5.4 To demonstrate that heat energy flows due to a

temperature difference
Materials:
• An iron bar about a metre long with holes drilled at equal intervals
• Oil
• Thermometer
• Wooden screen
• Water bath
• A bunsen burner
• Hammer
• Metal punch
Steps
1. Fill the holes of the iron bar partially with oil and insert the bulb of the
thermometers into them. Note the readings of the thermometers.
2. Heat one end of the iron bar slowly and gradually (Fig. 5.4). Observe the
temperature increase in the thermometers and record the readings.
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Bunsen burner
Fig. 5.4: The higher the temperature difference the higher the energy transferred
3. After some time, note the temperature readings of the thermometers. What do
you observe? Explain.
The thermometer nearest to the hot bunsen burner registers the highest rise in
temperature, and the one farthest away registers the least temperature rise.
Initially the readings of all the thermometers were the same. When one end of
the rod was inserted into boiling water, a large temperature difference was set up
between the two ends and heat energy flowed from the region of higher temperature
to that of lower temperature. Hence heat energy flows due to temperature difference.
If the activity is repeated by replacing the hot water bath with a bunsen burner
flame (temperature of the bluish part of the flame is about 500˚C), the rise in
temperature registered by each thermometer is higher. Hence the higher the
temperature difference, the higher the energy transfer.
Heat energy flow in solids is due to temperature difference. The higher the
temperature difference, the higher the energy flow.
5.2.1.3 Comparing rates of conduction in metals
Activity 5.5 To show that heat transfer in solids depends on the

material
Materials:
• A copper rod • Iron rod • Aluminium rod
• 3 match sticks • Wax • Tripod stand
• A bunsen burner
Steps
1. Take three rods, copper, aluminium and iron of the same length and thickness.
Fix a matchstick (or a light metal pin) to one end of each rod using a little
melted wax.
2. Place the rods on a tripod stand and heat the free ends with a burner as
shown in Fig. 5.5. Observe what happens.
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copper aluminium
matchstick
bunsen burner
iron
Fig. 5.5: Comparing heat transfer through different conductors
The matchstick falls off from the copper rod first then aluminium and finally
from the iron rod.
When the temperatures of the other ends of the rods reach the melting point of
wax, the matchstick falls off. The matchsticks do not fall off at the same time,
because the energy transferred is not equal for all the rods. The matchstick from
the copper rod is the first one to fall off showing that of the three metals, copper
is the best conductor of heat followed by aluminium and then iron.
5.2.2 Heat transfer by convection

5.2.2.1 Convection in liquids
Activity 5.6
To observe convection current in water
Materials
• A long straw • A crystal of potassium permanganate
• A beaker containing water • A bunsen burner
Steps
1. With the help of a long straw,
drop a small crystal of potassium
permanganate to the centre of glass flask
the bottom of a flask or a beaker
containing water. What do you
observe? potassium
permanganate
2. Heat the flask gently at the centre crystals
of the flask (Fig. 5.6). Observe
what happens.
Fig. 5.6: Convection currents in water
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Coloured streaks are observed to rise from the bottom to the top.
The crystal dissolves and the hot water of less density starts rising displacing the
cold dense water down. The streams of physically moving warm liquid are called
convection currents.
Heat energy is transferred by the convection currents in the liquid. The transfer
of heat by this current is called convection.
5.2.2.2 Convection in gases
Activity 5.7 To observe convection current in air
Materials:
• A box with a glass window, and two chimneys
• A candle
• Smouldering pieces of wick
Steps
1. Take a box with a glass window and two chimneys fixed at the top.
2. Place a lighted candle under one chimney and hold a smouldering piece of
wick above the other chimney as shown in Fig. 5.7. What do you observe?
smoke
A B
smouldering
wick
box
candle glass window
Fig. 5.7: Convection currents in air.
Smoke from the smoldering wick is seen to move down through chimney B then
to the candle flame and finally comes out through chimney A.
Air above the candle flame becomes warm and its density decreases. Warm air
rises up through chimney A and the cold dense air above chimney B is drawn
down this chimney and passes through the box and up the chimney A. The smoke
particles from the wick enable us to see path of convection current (Fig. 5.7).
Heat is transferred in air through convection currents. This convection current
passes energy as shown in Activity 5.8.
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Activity 5.8 To illustrate that convection currents possess energy
Materials:
• A thin circular disk • A card board • A candle flame
Steps
1. Take a thin circular disk of tin or cardboard and cut out six blades all round
(Fig. 5.8(a)). Pivot the disk on a bent needle (Fig. 5.8(b)).
2. Hold the disk above the candle flame for some time. Watch what happens.
disk pivoted
disc of tin with

blades cut
bent needle
(a) (b)
Fig. 5.8: A rotating disk.
The disk starts to rotate. The rotation is due to the convection current set up. If
a powerful electric bulb is available, you can make a rotating lamp shade.
hot air
thin cardboard
blade
thick wire wrapped around a
thin carboard bulb with a pointed pivot
cylinder
cold air
lamp stand to power supply
Fig. 5.9 : A rotating lamp shade

Convection currents possess energy. It is for this reason that steam is used to
rotate the turbine in geothermal electric plants.
5.2.3 Heat transfer by radiation
5.2.3.1 The concept of radiation

If you stand in front of a fireplace, you feel your body becoming warm. Heat
energy cannot reach you by conduction as air is a poor conductor of heat. How
115
about convection? The hot air molecules in and around the fireplace can only
rise and do not reach you by convection. How does the energy from the fireplace
then reach you? Heat energy must be transferred by a different mode other than
conduction and convection.
Activity 5.9 To demonstrate heat transfer by radiation
Materials:
• A thin tin lid painted black • A thumb tack
• Wax • A bunsen burner
Steps
1. Take a thin tin lid painted black on one side. Stick a thumb tack with melted
wax on the other side.
2. Keep the bunsen burner flame close to the painted side (Fig. 5.10). What
happens? Explain.
wax lid painted black
Thumb tack
wooden stand
Fig. 5.10: Radiation
As discussed in the case of the fireplace, the energy from the flame reaches the
tin lid and the wax by a different mode other than conduction and convection.
This third mode of heat transfer is called radiation. Radiation is the emission or
transmission of energy in the form of a wave or particles through a material or
space. Heat transfer from the sun travels through empty space (vacuum) and
reaches the Earth. This energy is transferred by radiation. The surfaces of all
luminous bodies emit radiation. A human face also emits some mild radiations.
While conduction and convection need a medium to be present for them to take
place, radiation can take place without a medium.
The amount of heat energy radiated depends upon the temperature of the body.
In Activity 5.9, if the bunsen burner is replaced by a candle flame, it will take
a longer time for the wax to melt. The temperature of the candle flame is lower
than that of a bunsen burner.
Heat transfer can take place without contact or in a vacuum. This method of heat
transfer is called radiation.
116
5.2.3.2 Good and bad absorbers of heat energy by radiation
Activity 5.10 To illustrate good and bad absorbers of heat

Materials:
• Two thin tin lid • A metal thumb tack (match stick)
• Molten wax • A bunsen burner
Steps
1. Take two thin tin lids, one with the inner side shiny and the other with the
inner side painted dull black.
2. Stick metal thumb tacks (or match sticks) on the outside of each lid using a
little molten wax.
3. Keep a bunsen burner flame midway between the lids as shown in Fig. 5.11.
Watch closely to and compare what happens to the two thumb tacks. Explain
your observation.
tin
lids wax
thumb tack
shiny surface
dull black
surface
supports for lids
Fig. 5.11: Good and bad absorbers
If a black and shiny surface receive the same amount of heat energy by radiation,
the black surface absorbs more heat than the shiny surface.
A dull black surface is a better absorber of heat radiation than a shiny surface.
Activity 5.11 To illustrate good and bad emitters of heat

Materials
• Three thermometers • Three identical empty cans
• Three cardboard
Steps
1. Take three identical empty cans of the same volume with their tops removed.
Apply clean and dry paints one white and the other black on two cans (both
117
inside and out surfaces) and leave the third can shiny.
2. Prepare three suitable cardboard covers with holes at the centre. Fill the cans
to the brim with hot water at 60˚C.
3. Cover the cans with cardboards and place a thermometer in each can through
the hole at the centre of the cardboard (Fig. 5.12).
cardboard
thermometer thermometer
thermometer
can can can
white black shiny
Fig. 5.12: Good and bad emitters
4. Record the temperature of water in the cans after a certain time interval.
Which can cools the water fastest? Which can takes the longest time to cool
the water? Explain the difference in the rate of temperate drop in the three
cans.
A shiny surface is a good emitter than a dull black surface
5.2.4 Applications of heat transfer
Activity 5.12 To describe the applications of heat transfer
Materials
• Internet • Reference books
Steps
1. You have learnt about heat transfer. Without referring to this book or any
other source, describe three ways in which heat transfer is important in our
daily lives.
2. Do a research from the internet and reference books on the applications of
heat transfer.
3. In your research, highlight clearly how the modes of heat transfer are applied
in vacuum flasks, construction of ventilations, domestic hot water system
and solar heating.
118
4. What other applications of heat transfer did you come across in your research.
5. Explain to your group members how natural phenomena like sea and land
breeze take place.
6. Make a presentation on your findings to the whole class through your group
secretary.
1. Vacuum flask
The vacuum flask popularly known as thermos flask, was originally designed by
Sir James Dewar. It is designed such that heat transfer by conduction, convection
and radiation between the contents of the flask and its surroundings is reduced
to a minimum.
A vacuum flask, Fig. 5.13 is a double-walled glass container with a vacuum in the
space between the walls. The vacuum minimises the transfer of heat by conduction
and convection. The inside of the glass walls, is silvered so as to reduce heat losses
by radiation. The felt pads on the sides and at the bottom support the vessel
vertically. The cork lid is a poor conductor of heat.
plastic cover cork lid
double-walled glass container

vacuum
silvered surface
felt pads outside case
vacuum seal
Fig. 5.13: Vacuum flask
When the hot liquid is stored, the inside shiny surface does not radiate much
heat. The little that is radiated across the vacuum is reflected back again to the
hot liquid, by the silvering on the outer surface. There is however some heat lost
by conduction through the walls and the cork.
2. Windows and ventilators in buildings

As shown in Fig. 5.14, warm exhaled air of less density goes out through the
ventilator and fresh air of high density enters through the windows at a lower
level. This refreshes the air in a room.
119
warm air
warm air
fresh air
Fig. 5.14: Ventilation in building
3. Natural convection currents over the earth’s surface

(a) Sea breeze
During the day, the temperature of the land rises faster than the temperature of sea water
and the air over the land becomes warmer than the air over the sea water. The warm air
of less density rises from the land allowing the cold dense air over the sea to blow to
the land. This creates a sea breeze in the daytime (Fig. 5.15).
warm air from
the land
cold air from

the sea
cold sea water

Fig. 5.15 Sea breeze
(b) Land breeze

During the night, the land cools faster than the sea water. Warm air from the sea rises
and the dense air from the land moves to the sea. This sets up a land breeze in the sea
(Fig. 5.16).
cold fresh air from the land
warm air above

the sea water rises
Fig. 5.16: Land breeze
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4. Electrical devices
An electric kettles has its heating coil at the bottom. A refrigerator has the freezing unit
at the top.
5. Domestic hot water system

A domestic hot water supply system works on the principle of convection current.
A schematic diagram of a hot water supply is shown in Fig. 5.17.
ball cock
expansion pipe C
main supply
of cold water
cold water storage tank
hot water
tap
pipe A
hot water storage tank
pipe B, cold water

boiler
heat
Fig. 5.17: Hot water system
Water is heated using fire wood, oil or electricity in the boiler. Hot water from
the boiler goes up to the hot water storage tank through pipe A. Cold water flows
down from the cold water storage tank into the boiler through pipe B (called
return pipe).
When the hot water is being drawn from the top of the hot water storage tank, it
is replaced by water from the main cold water tank built at the top of the house.
The expansion pipe C allows steam and dissolved air to escape. This ensures that
the tank does not explode due to the pressure created by the steam produced.
7. Solar heating
Flat plate collectors, called solar panels, are used to heat water. They can heat water
up to 70˚C. A solar panel consists of thin copper pipes, painted black, which carry
the water to be heated. These tubes are fitted in a copper collector plate which in
turn is fitted on to a good thermal insulator in a metal frame. A glass plate covers
the panel (Fig. 5.18). These panels can be fitted on the roof of houses.
121
Heat radiation from the sun falls on the tubes and on the collector plate through
the glass plate. The heat radiations trapped inside the panel by the glass plate heat
the water. The hot water is then pumped to a heat exchange coil in a hot water
tank which is connected to the domestic hot water system.
pipe 1
heat from the sun solar panel to domestic hot
1
water system
glass plate hot water
2
thin
pipe 2 e copper
plat
per tube
cop ctor
insulator colle pipe 2 water gains energy
in the exchanger
metal case
cold water
pump
heat exchange coil
cold
pipe 1 water insulated cold water tank
Fig. 5.18: Solar heating
8. Solar concentrations
In some heating devices, instead of a flat plate collector, curved mirrors (concave or
parabolic) are used to concentrate the heat radiations from the sun to a small area
at their focus. If the boiler is placed at the point of focus, very high temperatures
can be reached.
Exercise 5.1
1. Distinguish between heat and temperature.
2. What are the different modes of heat transfer? Explain clearly their difference
using suitable examples.
3. State three factors which affect heat transfer in metals. Explain how one of
the factors you have chosen affects heat transfer.
4. Describe an experiment to show that water is a poor conductor of heat.
5. Use particle behaviour of matter to explain conduction.
6. Describe a simple experiment to demonstrate that the heat radiated from a
hot body depends upon the temperature of the body.
7. With a suitable diagram, explain the working of a vacuum flask.
5.3 Thermal expansion

In general, nearly all substances increase in size when heated. The process in
which heat energy is used to increase the size of matter is called thermal expansion.
The increase in size on heating of a substance is called expansion. On cooling,
122
substances decrease in size. The decrease in size on cooling of a substance is called

contraction. Why is this so?
5.3.1 Thermal expansion and contraction in solids
When a solid (e.g. a metal) is subjected to heat, it:
(a) Increases in length (Linear Expansion).
(b) Increases in volume (Volume Expansion).
(c) Increases in area (Surface Expansion).
5.3.1.1 Linear expansion
(a) Demonstrations of linear expansion
Activity 5.13 To demonstrate expansion and contraction using the

bar and gauge apparatus
Materials
• A bar and gauge apparatus • Bunsen burner
Steps
1. Move the metal bar in and out of the gauge at room temperature (Fig. 5.19).
Observe what happens.
wooden handle
gauge metal bar
Fig. 5.19: Bar and gauge apparatus
2. Keep the metal bar away from the gauge and heat the bar for sometime.
3. Try to fit the bar into the gauge and observe what happens. Explain your
observation.
4. Allow the bar to cool and try to fit it into the gauge. What happens? Explain.
A bar and gauge apparatus consists of a metal bar with a suitable wooden handle
and a gauge. When both the metal bar and the gauge are at room temperature,
the bar just fits into the gauge.
On heating, the metal bar expands. There is an increase in length. It hence expands
123
linearly and therefore, the bar cannot fit into the gauge.
On cooling the bar easily fits into the gauge due to contraction.
Solids expand i.e increase in length on heating and contract i.e reduced in length on cooling.
Activity 5.14 To demonstrate the bending effect of expansion and

contraction
Materials
• A bimetallic strip • Bunsen burner
Steps
1. Observe a bimetallic strip at a room temperature (Fig. 5.20).
brass
wooden
handle
iron
Fig. 5.20: A bimetallic strip
2. Take the bimetallic strip with the brass strip at the top and heat it with
a bunsen burner flame for sometime. Observe what happens. Explain the
observation.
3. Remove the flame and allow the bar to cool to a room temperature. Observe
and explain what happens.
4. Discuss with your friend what will happen if the bar is cooled below room
temperature. Sketch the strip at that temperature.
When the bimetallic strip is heated, it bends downwards with the brass strip on the
outer surface of the curvature, as shown in Figure 5.21(a). Why does this happen?
When the flame is removed and the strip left to cool to room temperature, the
strip returns back to its initial state (straight) as shown in Figure 5.20 above.
If the strip is cooled below room temperature, it bends upwards with the iron
strip underneath as shown in Figure 5.21 (b). What has happened?
brass
brass
iron
iron
(a) Heating the bimetallic strip (b) Cooling the bimetallic strip below room temperature
Fig. 5.21: Bending effect of expansion and contraction
124
As the bimetallic strip is heated, brass expands more than iron. The large force
developed between the molecules of brass forces the iron strip to bend downwards.
On cooling below a room temperature, the brass contracts more than iron and
the iron strip is forced to bend upwards.
The force developed during expansion or contraction causes a bending of the
metals.
(b) Comparison of rates of expansion of different solids
As we know from the kinetic theory of matter, the different states of matter expands
when heated but at different rates.
The following activity shows that different solids have different rates of expansion.
Activity 5.15 To compare rates of expansion and contraction of

different solids
Materials:
• Thin metal rods of different materials • Source of heat
• Rollers connected to a pointer • G - clamp
Steps
1. Clamp one end of a long thin metal rod tightly to a firm support, with the
end of the rod resting on a roller fitted with a thin pointer (See Fig. 5.22).
clamp pointer fixed to roller
deflection of
thin copper rod the pointer
roller
table heat
Fig. 5.22: Expansion and contraction of thin metal rods.
2. Heat the metal rod for sometime. Observe and explain what happens.
3. Remove the burner and allow the rod to cool. Observe and explain what
happens.
4. Repeat the activity with thin rods of different materials. Observe and explain
what happens, accounting for any differences.
The pointer deflects in the clockwise direction on heating and in the anticlockwise
direction on cooling.
The pointer deflects to different extents depending on the material.
125
On heating, there is an increase in length (linear expansion) of the rods. The

expanding rod moves the roller to the right making the pointer attached to the roller
to deflects in a clockwise direction. On cooling, the rod contracts and decrease in
length. The contracting rod moves the roller to the left hence the pointer deflects
in the opposite direction (anticlockwise direction).
When a different material e.g lead is used, the pointer deflects more to the right
(clockwise). When cooled, the pointer deflects more to the left (anticlockwise).
Different solids (e.g metals) expand and contract to different extents when heated
by the same quality of heat.
(c) Coefficient of linear expansion

Consider a thin metal of length l0. in Fig. 5.23.
l0 Δl
Fig. 5.23: A thin rod showing increase in length.
When the rod is heated, a temperature change of Δθ occurs and its length increases
by Δl.
The ratio of increase or decrease in length to original length ( Δll ) is directly
0
proportional to the change in temperature Δθ. Note
Δl ∝ Δθ ⇒ Δl Δl ∝ – proportionalitity sign
l0 = α Δθ and α = l Δθ
l0 θ α – alpha-constant symbol
where α is a constant called the coefficient of linear expansion. It is the value of
the increase in length per unit rise in temperature for a given material. The SI
units of α is K–1
Suppose: The temperature change = Δθ,
l0 represents the original length of the rod
l represent the new length for a temperature rise of θ
Then, Δl = l – l0
The above expression may be expressed in terms of l0, l, θ and α as follows.
Δl l – l0
α= = Rearranging l – l = l0αΔθ
l Δθ l Δθ
l = l0 + l0 αΔθ
l = l0(1 + αΔθ)
126
Example 5.1
A copper rod of length 2 m, has its temperature changed from 15 °C to 25 °C.
Find the change in length given that its coefficient of linear expansion
α = 1.7 × 10–5 K–1.
Solution
Δθ = (25 – 15) oC = 10 oC
Δl = l0 α Δθ = 2 × 1.7 × 10–5 × 10
= 3.4 × 10–4 m
= 0.34 mm
5.3.1.2 Area and volume expansion
(a) Demonstrations of area and volume expansions
Activity 5.16 To demonstrate volume and surface expansion and

contraction using the ball and ring apparatus
Materials:
• A ball and a ring • Bunsen burner • A bowl of cold water
Steps
1. Move the ball in and out of the metal ring at room temperature
(see Fig. 5.24). What do you observe?
2. Keep the metal ball away from the ring and heat it for sometime.
3. Try to pass the ball through the ring. What do you observe this time? Explain.
4. Cool the metal ball in a bowl of cold water and try to pass the ball through
the ring again. What do you observe now? Explain the observation.
metal ball
metal ring
Fig. 5.24: Ball and ring apparatus
127
A ball and ring apparatus consists of a ball and ring both made of the same metal.
At a room temperature, the ball and the metal ring have approximately the same
diameter, thus the ball just passes through the ring. On heating, the metal ball
expands. There is an increase in volume and surface area of the ball. As a result,
the ball cannot pass through the ring. On cooling, contraction occurs and the
original volume is regained. The ball can now pass through the ring again. This
activity shows volume and surface area expansion and contraction in solids.
Most solids expand on heating and contract on cooling.
Why solids expand on heating?

In Senior I, we learnt that molecules of a solid are closely packed and are
continuously vibrating about their fixed positions. When a solid is heated, the
molecules vibrate with larger amplitude about the fixed position. This makes
them to collide with each other with larger forces which pushes them far apart. The
distance between the molecules increases and so the solid expands. The reverse
happens during cooling.
(b) Coefficients of area expansion of solids

Consider a solid whose surface area is A0.
When the surface of the solid is heated or cooled to a temperature change of Δθ,
its surface area increases or decreases by ΔA to a new value A.
Experiments have proved that the ratio of the change in surface area to original
area i.e ΔA
A is directly proportional to the change in temperature (Δθ)
0
ΔA ∝ Δθ ⇒ ΔA = βΔθ (β is a constant called coefficient of area expansivity)
A0 A0
ΔA A – A0
Hence β = A Δθ Or β= (since ΔA = A – A0)
0 A Δθ
⇒ A –A0 = AθβΔθ
A = A0 – A0βΔθ
∴ A = A0(1 - βΔθ)
Note:
Coefficient of area expansivity = 2 × coefficient of linear expansivity
β = 2α
Example 5.2
A round hole of diameter 4.000 cm at 0 °C is cut in a sheet of brass (coefficient of
linear expansion is 0.0000017(Co)-1. Find the new diameter of the hole at 40 °C.
128
Solution
ΔA = βA0 (θ2 - θ2 )
Given: α = 0.00017/K, (θ2 – θ1) = 40°C, D = 4.000 cm so r = 2.000 cm, β = 2 α
then
Area (A0) = πr2 = 22 × 2.000 × 2.000 cm2
7
= 12.571 cm2
New area A = A0 + ΔA = (12.571 + 0.017) cm2
= 12.588 cm2
Since A = πr2, the new radius r = A = 12.587

π 3.141
= 2.002 cm
(c) Coefficients of volume expansion in solids
When a solid is heated or cooled to a temperature change of Δθ, its volume
increases or decreases by ΔV to a new value V.
The ratio of the change in volume to original volume i.e ΔV
V0 is directly proportional
to the change in temperature (Δθ)
ΔV α Δθ ⇒ ΔV = ϒΔθ (ϒ is a constant called coefficient of volume expansivity)
V0 V0
ΔV V – V0
Hence ϒ = V Δθ Or ϒ= (since ΔV =V – V0)
0 V Δθ
⇒ V –V0 = VθϒΔθ
V =V0 – V0ϒΔθ
∴ V = V0(1 - ϒΔθ)
Note:
Coefficient of volume expansivity = 3 × coefficient of linear expansivity
ϒ = 3α
Example 5.3
A metal vessel has a volume of 800.00 cm3 at 0 °C. If its coefficient of linear
expansion is 0.000014/K, what is its volume at 60 °C?
Solution
Given: V0 = 800.00 cm3, (θ2 – θ1) = 60 °C and α = 0.000014/K = 0.000014/ °C
129
Change in volume, (ΔV) = 3 α V0Δθ

= 3(0.000014/°C) × 800.00 cm3 × 60°C
= 2.016 cm3
New volume (at 60°C) = V0 + ΔV
= (800.00 + 2.016) cm3
= 802.016 cm3
Exercise 5.2
1. What do you understand by the phrase 'coeficient of linear expansion'?

2. A vertical steel antenna tower is 400 m high. Calculate the change in height
of the tower hence its new height that takes place when the temperature
changes from –15 °C on winter day to 35 °C on a summer day.
(Take α = 0.00000645/K
3. A 8 m long rod is heated to 50 °C. If the rod expands to 10 m after some time,
calculate its coefficient of linear expansion given that the room temperature
is 32 °C.
4. A rectangular solid of Brass has a coefficient of volume expansion of
56 × 10-6 /°C. The dimensions of the rectangle are 5 cm × 6 cm × 8 cm at
10 °C. What is the change in volume and the new volume if the temperature
increases to 90 °C?
5. A solid plate of lead of linear expansion 29 × 10-6 /°C is 8 cm × 12 cm at 15 °C.
What is the change in area and the new area of the lead if the temperature
increases to 95 °C?
6. A cup made of pyrex glass has a volume of 200 cm3 at 0 °C. If the coefficient
of linear expansion is 0.000003/K, what will be its volume if it holds hot water
at 92 °C.
5.3.2 Thermal expansion and contraction in liquids

Like solids, liquids expand on heating i.e volume increases and contract i.e Volume
reduces on cooling. But liquids expand more than solids since they have relatively
weak intermolecular forces. Activity 5.17 will help us to understand expansion
and contraction in liquids.
130
Activity 5.17 To demonstrate expansion and contraction in liquids
Materials:
• A glass flask • Coloured water • Tripod stand
• A rubber stopper • Bunsen burner • Wire guaze
• Long glass tubing
Steps
1. Fill a glass flask with coloured water.
2. Fit the flask with a rubber stopper carrying a long narrow glass tubing.
3. Note the initial level of water in the glass tube before heating (Fig. 5.25).
C
thin tube A
B
glass flask
coloured
water
wire gauze
heat
Fig. 5.25: Expansion of liquid
4. Heat the water in the flask. Observe what happens to the level of water at
A immediately the heating starts and after a few minutes. Explain your
observation.
In a similar activity, it was observed that at first the level of the coloured water in
the tube drops to level B and then rises to level C.
On heating, the glass flask is heated first and expands i.e its volume increases.
The level of water immediately drops from A to B. On continuous heating, water
starts to expand hence water level rises up the tube from B to C.
If the setup is allowed to cool below room temperature, the water level drops to
a point lower than A and B.
This shows that liquids expand on heating and contract on cooling.
Why liquids expand on heating?
Molecules are loosely packed in liquids. The force of attraction between the
131
molecules is weaker than in solids. The molecules move freely in the liquid. On
heating, the speed of the molecules increases.The collisions between the molecules
increases the distance between them causing the liquid to expand.
5.3.3 Thermal expansion and contraction in gases
Just like solids and liquids, gases expand on heating and contract on cooling.
Gases expand more than liquids and solid because their molecules move furthest
on heating. The following activity will help us to study expansion and contraction
in gasses.
Activity 5.18 To demonstrate expansion of gases

Materials:
• A thin glass flask • A rubber stopper
• A long narrow glass tube
Steps
1. Take a thin glass flask with an open top.
2. Close the flask with a rubber stopper carrying a long narrow glass tube.
3. Invert the flask so that the glass tube dips into water in a container. What do
you observe? (Fig. 5.26).
air thin glass flask
tube
container
bubble
coloured water
Fig. 5.26: Expansion of air
4. Place your hands over the flask to warm it for sometime and observe what
happens. Explain your observation.
5. Remove your hands from the flask and wait for some time. Observe what
happens. Explain your observation.
When the flask is warmed by the warmth of the hands, the level of water in the
tube drops and some bubbles are formed due to air escaping from the flask
through the tube.
132
On removing the hands from the flask, water level rises up the glass tube again
due to contraction of air i.e volume of air reduces on cooling.This shows that
gases expand on heating and contract on cooling..
The volume of air increases in the flask due to expansion.
Why a gas expands on heating?
The force of attraction between the molecules of a gas is very small (almost
negligible) and the distance between the molecules is large compared to solids
and liquids. The molecules move freely in all directions. When a gas is warmed,
the molecules gain more energy and move far apart hence volume increases.
Different gases expand by the same amount when heated equally. Activity 5.19
will demonstrate further the expansion and contraction of gases.
Activity 5.19 To compare rates of expansion and contraction in

different gases.
Materials:
• A glass bulb B with air • Glass jacket
• A metre rule in vertical position • A reservoir R with mercury
• Steam
Steps
1. Set up the apparatus as shown in Fig. 5.27.
The apparatus consists of a glass bulb B containing mercury. Bulb B is
surrounded by the outer glass jacket through which steam can be passed.
Steam R
Gas
B
Fig. 5.27: Expansion of gas
133
2. Circulate water at 0º C through the jacket and adjust reservoir R so that the
level of mercury in B and R is the same.
3. Measure the volume of air (gas) in bulb B.
4. Pass the steam through the jacket until the temperature is constant.
5. Adjust the level of mercury in B and R until they are the same. Measure the
volume of air in B.
6. Repeat the experiment with different gases. Observe what happens. Compare
the volume of air in B for the different gases. Explain the differences if any.
On passing cold water at 0ºC through the glass jacket, the volume of the air in
bulb B reduces due to contraction of air.
The volume of air increases on passing the steam through the glass jacket due to
expansion of air.
Gases contract on cooling and expand on heating.
5.3.4 Applications of thermal expansion and contraction
Activity 5.20 To find out the applications of expansion and

contraction
Materials
• Internet enabled devices (lab computers or tablets)
• Reference books
Steps
1. You have now learnt about expansion and contraction. Suggest any three
applications of expansion and contraction in our daily lives.
2. Carry out a research from the internet on the applications of expansion and
contraction.
3. Report your findings to the whole class.
Thermal expansion and contraction, on one hand is a nuisance and on the

other hand is quite useful. The following are some of the applications of thermal
expansion and contraction.
1. Electric thermostats
A thermostat is a device made from a bimetallic strip that is used to maintain a
steady temperature in electrical appliances such as electric iron boxes, refrigerators,
134
electric geysers, incubators, fire alarms and the automatic flashing unit for indicator
lamps of motor cars. Fig. 5.28 show two such devices.
cell
bell iron
D
iron A
C
A B
brass brass resistance wire
(a) Fire alarm (b) Electric iron box

Fig. 5.28: Electric appliances with thermostat
The bimeltalic, as discussed earlier, bends on expansion and relaxes on cooling,

connectin and disconnecting the circuit to regulate temperature.
Be responsible and take care!

Conserve energy by switching off the socket after using electrical appliances.
Be careful when using electrical devices to avoid electric shocks.
2. Making of ordinary and pyrex glasses

You may have observed that when boiling water is poured into a thick-walled glass
tumbler it may break suddenly. This is because the inside of glass gets heated and
expands even before the outside layer becomes warm. This causes an unequal
expansion between the inside and the outside surfaces. The force produced by
the expanding molecules on the inside produces a large strain in the glass and the
tumbler breaks. This is the reason why pyrex glass tumblers are recommended
for use while taking hot liquids.
3. Rivets
In industries, steel plates are joined together by means of rivets. Hot rivets are
placed in the rivet holes and the ends hammered flat. On cooling the force of
contraction pulls the plates firmly together (Fig. 5.29).
rivet
rivet holes
steel plates
hot rivet rivet hammered flat
Fig. 5.29: Rivets
135
4. Expansion joint loops

Metal pipes carrying steam and hot water are fitted with expansion joint or loops.
These allow the pipes to expand or contract easily when steam or hot water passes
through them or when the pipes cool down. The shape of the loop changes slightly
allowing necessary movement of the pipes to take place (Fig. 5.30).
Expansion Expansion
Fig. 5.30: Expansion joint
5. Loosely fitted electric cables

Telephone and electricity cables are loosely fitted between the poles to allow room
for contraction in cold weather and expansion in hot weather.
6. Use of alloys
The measuring tape used by surveyors for measuring land is made of an alloy of iron
and nickel called invar. Invar has a very small change in length when temperature
changes.
7. Gaps in railway tracks

Gaps are left between the rails when the railway tracks are laid. The rails are joined
together by fish-plates bolted to the rails. The oval shaped bolt holes allow the
expansion and contraction of the rails when the temperature changes (Fig. 5.31).
rail gap rail
Bolts Oval shaped
bolts holes
fish plate
rigid supports
Fig. 5.31: Gaps left between rails
In very hot weather, the gaps may not be enough if the expansion is large. The rails
may buckle out. Modern methods use long welded lines rigidly fixed to the beds
of the track so that the rails cannot expand. Expansion for the rails is provided
by overlapping the plane ends (Fig. 5.32).
136
Fig. 5.32: Overlapping joints
8. Rollers on bridges
The ends of steel and concrete bridges are supported on rollers. During hot or
cold weather, the change in length may take place freely without damaging the
structure (See Fig. 5.33).
steel bridge
fixed point
wall
rollers
Fig. 5.33: Steel and concrete bridges are supported on rollers
9. Breakages.
Activity 5.21 To demonstrate causes of expansion and contraction
Materials:
• A beaker • Water
• An immersion heater • A measuring cylinder
• A thermometer • Stop watch
Steps
1. Take 200 g of water in a beaker and note its initial temperature θ1.
137
2. Heat the water with an immersion heater for 10 minutes (Fig. 5.34 (a)). Note
the final temperature θ2 and calculate the change in temperature, ∆θ = θ2 – θ1.
3. Repeat (2) above by taking 400 g of water in the same beaker and same initial
temperature θ1 (Fig. 5.34 (b)). Note the time taken to produce the same
change in temperature as before.
4. Compare the times taken to produce the same change in temperature in 200g
and 400g of water. What is your conclusion?
(a) (b)
Beaker Beaker
Thermometer Thermometer
Heating 200 g of Heating 400 g of

element water element water
Fig. 5.34: Relationship between heat energy and mass of the substance.
Sudden expansion and contraction can lead to breakages of things like glasses and
egg shells. This behaviour is mitigated against in the manufacture of glass items
such as the drinking glass. They are made of thin walls to allow even expansion
and contraction thus minimising chances of breakage.
Exercise 5.3
1. Use particles model to explain thermal expansion of solids.

2. Explain why:
(a) Steel bridges are usually supported by rollers on one loose side.
(b) Metal pipes carrying steam and hot water are fitted with loops.
3. Describe how shrink fitting is done.
4. State two applications of contraction of solids.
5. Name three physical properties that change when heating a solid.
5.4 Quantity of heat

As we have already learnt, heat is a form of energy that flow from a region of
high temperature to a region of lower temperature. Cold substances absorb heat
energy while hot ones lose heat energy. This section deals with measures of the
capacities of substances to gain or loose heat energy.
138
5.4.1 Heat capacity
Activity 5.22 To show that the heat energy required to produce a certain
change in temperature depends on the mass of the substance.
Materials
• Water bath • An egg
• Heat source
Steps
1. Place an egg in a water bath. Heat the water bath till the water boils.
2. Transfer the egg to a beaker of cold water and observe. Explain what happens to
the egg?
The larger the mass, the longer the time needed to change its temperature. This
means the larger the mass, the more heat is supplied to change the temperature by
one degree. Hence the quantity of heat energy, Q, gained by a substance through
a certain temperature change is directly proportional to its mass, m. Therefore,
Heat energy is proportional to mass that is;
Q α m, when temperature change is constant.
Activity 5.23 To show that the heat energy required by a substance

of a given mass depends on the change in temperature
Steps
1. Repeat activity 5.22 step 2 with 200 g of water, but this time, heat to produce
twice the change in temperature. Note the time taken for this to happen.
2. Which case takes more time, heating up to a given temperature or up to double
that temperature?
3. What relationship does this observation show between quantity of heat and
change in temperature?
The longer the time of heating, the more heat energy supplied and the greater
the temperature change.
Heat energy, Q is proportional to change of temperature, ∆θ, when mass of a
substance is constant. Q α ∆θ.
Example 5.4
200 J of heat energy is needed to change the temperature of a given mass of water
from 20 ºC to 34 ºC. How much heat energy is needed to change the temperature
of this mass of water from 20 ºC to 48 ºC.
139
Solution
Case 1: temperature change = (34 – 20) = 14 ºC, heat required Q1 = 200 J
Case 2: temperature change = (48 – 20)ºC = 28 ºC
200 J ––––––––––– 14 ºC
Q2 ––––––––––– 28 ºC
Heat energy needed, Q2 = 20014× 28
= 400 J
Heat capacity of a substance can be therefore defined as the heat energy required
to raise the temperature of a substance by 1 K.
Mathematically,
Amount of Heat supplied (Q)
Heat capacity (c) = J/K
Temperature change (∆θ)
The SI unit of heat capacity is joule per kelvin (J/K)
Example 5.5
Calculate the quantity of heat required to raise the temperature of a metal block
of capacity of 520 J/K from 9 ºC to 39 ºC.
Solution
Quantity of heat Q = Heat capacity × temperature change
Q = c × ∆θ
= 520 × (39 – 9)
= 15 600 J
Example 5.6
The quantity of heat required to raise the temperature of water from 10 ºC to 65 ºC
is 6 200 J. Calculate the heat capacity of water.
Solution
Q
Q = c∆ θ ⇒ C = ∆θ
6 200 J
= = 112.73 J/K
(65 – 10)K
The heat capacity of water is 112.73 J/K
140
Exercise 5.4
1. The heat capacity of water depends on the mass of the water being heated.
TRUE or FALSE? Justify your answer.
2. Calculate the heat capacity of tea when 400 J of heat are supplied to change
its temperature from 25 K to 40 K.
3. Calculate the amount of heat energy given out to lower the temperature of a
metal block of heat capacity 520 J/K from 60 ºC to 20 ºC.
5.4.2 Specific heat capacity

From activities 5.22 and 5.23 we learnt that.
Quantity of heat, Q α mass, m
Q α change in temperature, ∆θ
Q α m∆θ or
Q = mc∆θ where c is a constant
When the mass of the substance is 1 kg (i.e. m = 1 kg) and the change in
temperature is 1K (i.e. ∆θ = 1 K), then Q = c and c is referred to as the specific
heat capacity of the substance.
The specific heat capacity, c of a substance is defined as the heat energy required
to change the temperature of a substance of mass 1 kg by 1 Kelvin.
Q
c =
m∆θ
Therefore,
Quantity of heat = mass × specific heat capacity × temperature change
Q = mc∆θ
where, ∆θ = final temperature – initial temperature
The SI unit of specific heat capacity is joule per kilogram per Kelvin (J/kg K).
Example 5.7
Calculate the heat energy required to raise the temperature of 2.5 kg of aluminium
from 20 °C to 40 °C, if the specific heat capacity of aluminium is 900 J/kg K.
141
Solution
Heat energy required = mass × specific heat capacity × temperature change
Q = mc∆θ
= 2.5 × 900 × (40 – 20)
= 45 000 J
Example 5.8
18 000 J of heat energy is supplied to raise the temperature of a solid of mass
5 kg from 10 ºC to 50 ºC. Calculate the specific heat capacity of the solid.
Solution
Q
c=
m∆θ
180 000 J
=
(50 – 10)K × 5 kg
= 900 J/kg K
Example 5.9
Find the final temperature of water if 12 000 J of heat is supplied by a heater to
heat 100 g of water at 10 ºC.
(Take specific heat capacity of water and 4 200 J/kg K)
Solution
Q
Q = mc∆θ ⇒ ∆θ = m × c
12 000 J
=
(0.1 × 4 200) J/K
= 12 000 J
420
= 28.57ºC
∆θ = θf – θi, where θf – final temperature, θi – initial temperature
⇒ θf = ∆θ + θl = 28.57 ºC + 10 ºC
θf = 38.57 ºC
The final temperature is 38.57 ºC.
142
Exercise 5.5
1. 45 000 J of heat are supplied to 5 Kg of aluminium initially at 25ºC. What

is its final temperature? (Take the specific heat capacity of aluminium as
900 J/kgk).
2. What is the difference between heat capacity and specific heat capacity?
3. 24 000 J of heat energy is supplied to raise the temperature of a substance
of mass 6 kg from 12 ºC to 48 ºC. Calculate the specific heat capacity of the
substance.
Comparison of specific heat capacities of the three states of matter
Activity 5.24 To show that different substances have different

specific heat capacity
Materials
• Two thermometers
• A lid with two holes
• Two boiling tubes (one containing cooking oil and the other water)
• A hot water bath
Steps
1. Pour equal volume of liquids (cooking oil and water) into two identical test
tubes. Place identical thermometers in each test tube (Fig. 5.35).
Thermometer
Water bath
cooking oil
Water
Fig. 5.35: Set-up to show different specific heat capacities
2. Heat the test tubes in a hot water bath for the same time. Observe and
compare the temperature changes in the two cases. Explain the difference if
any.
Different substances have different specific heat capacities. Solids require more
heat energy to melt than liquids and gases. This means that solids have higher
143
specific heat capacity than liquids and gases. Gases have the lowest specific heat
capacity.
Two different substances of the same mass when subjected to the same quantity
of heat, acquire different changes in temperature. Table 5.1 shows that different
substances have different specific heat capacities. This is true for solids and liquids
but not in gases
Table 5.1:Values of specific heat capacities of metals
Substance Specific heat capacity (c) J/kg K

Aluminium 900
Brass 370
Copper 390
Cork 2000
Glass 670
Ice 2100
Iron 460
Lead 130
Silver and tin 230
Activity 5.25 To determine the specific heat capacity of a solid by

the electrical method
Materials
• Electric circuit • Heating element
• Metal cylinder • Thermometer
• Variable resistor • Cotton wool
• Aluminum foil • Wooden container
• Solid metal blocks in the form of a cylinder, with 2 holes.
Steps
1. Measure and record the mass, m, of the metal cylinder.
2. Insert an electrical heater in position in the metal block through the larger
hole and a thermometer through the other hole.
3. Note the initial temperature of the metal block θ1.
4. Cover the solid with cotton wool or felt material and wrap a aluminium foil
around cotton wool.
5. Place the set up a wooden container. Complete the electrical circuit as shown
in Fig. 5.36.
144
+ – – +
A
S
Variable
+ –
V resistor
Thermometer
Cotton wool Cardboard lid
aluminium
foil Metal block
(a) Metal solid cylinder Heating element

Wooden
container
(b) Circuit diagram
Fig. 5.36: Specific heat capacity of a solid by electrical method.
6. Close the switch S and start a stop watch at the same time.
7. Use the variable resistor to maintain a steady current passing through the
heater.
8. Note the current I through the heater with the ammeter and the potential
difference, V across the heater with the voltmeter.
9. Pass this steady current for some time so that the rise in temperature in the
solid is about 8 ºC.
10. Note the time t, when the final temperature of the solid is θ2.
11. Calculate the change in temperature ∆θ = θ2 – θ1.
12. Show the relationship between electrical energy used and the heat energy gained
by the metal and hence calculate the specific heat capacity of the metal cylinder.
13. How much electrical energy has been spent in this time? What has happened
to this energy? What is the purpose of cotton wool or felt material, aluminium
foil and the wooden container?
Electrical energy E, spent by the heater in a time, t , is given by E = VIt. This

energy is converted into heat energy and has been absorbed by the metal solid
cylinder. Heat energy gained by the metal
Q = mc∆θ
Assuming no energy from the heater is lost to the surrounding,
electrical energy used = heat energy gained by the metal cylinder.
∴ VIt = mc∆θ
from which the specific heat capacity, c, of the solid can be calculated.
VIt
∴ c =
m∆θ
145
Example 5.10
The following data was obtained from an experiment similar to that of
Activity 5.25. Mass of copper metal block = 200 g, initial temperature of the block
= 22°C, ammeter reading = 0.5 A, voltmeter reading = 3.0 V, final temperature
of the block = 30 °C, time of heating = 7 minutes. Use the data to calculate the
specific heat capacity of copper. What does this value mean?
Solution
Electrical energy spent is given by, E = VIt.
Assuming no energy from the heater is lost to the surrounding,
Heat energy gained by the metal block = mc∆θ.
mc∆θ = VIt
VIt 3.0 × 0.5 × (7 × 60)
∴ c = m∆θ = 0.200 × (30 – 22)
3.0 × 0.5 × 420
= 0.200 × 8
= 393.75 J/kg K
∴ specific heat capacity of copper = 394 J/kg K
This means that to raise the temperature of 1 kg of copper by 1 K(or by 1°C),
394 Joules of heat energy are required.
Example 5.11
Calculate the heat energy required to raise the temperature of 2.5 kg of aluminium
from 20 °C to 40 °C, if the specific heat capacity of aluminium is 900 J/kg K.
Solution
Heat energy required Q = mc∆θ
= 2.5 × 900 × (40 – 20) J
= 45 000 J
Activity 5.26 To determine the specific heat capacity of water by

the method of mixtures
Materials
• A solid of known specific heat capacity (cs)
• Weighing machine • Water bath
• Thermometer • Beaker
• Stirrer • Heating source
• Tripod stand
146
Steps
1. Take a solid of known specific heat capacity (cs) and measure its mass (ms).
2. Heat it in a water bath till the water starts boiling, as shown in Fig. 5.37 (a).
3. In the meantime, take an empty, clean and dry container of known specific
heat capacity (cc) and measure its mass (mc).
4. Put water into the container, say to half of the container, and measure the
total mass.
5. Calculate the mass of water (mw) whose specific heat capacity (cw) is to be
determined.
6. Find the initial temperature (θ1) of water and the container (Fig. 5.37 (b)).
7. When water in the water bath has started boiling, note the temperature of the
solid (θs) in the water bath.
8. Quickly transfer the hot solid into cold water in the container and observe the
temperature of the mixture.
Thermometer
Stirrer
String
Beaker
Container
Hot water Thermometer

Solid bath
Water
Heat
(b) Cold water in a container
(a) A solid being heated
Fig. 5.37: Specific heat capacity of water by method of mixture.
9. Stir the contents gently to distribute the heat uniformly throughout the
mixture and note the final maximum steady temperature of the mixture (θ2).
10. What happens to the cold water and the container when the hot solid is
transferred into the container?
11. Using all the data you have collected, calculate the specific heat capacity of
water using the equation:
Heat lost by solid = heat gained by water
12. What precautions have to be taken to ensure accuracy in the experimental
procedure?
13. Highlight the assumptions for this activity.
The temperature of the solid has decreased from θs to θ2, showing that the solid
has lost heat energy. The temperature of the cold water and the container has
147
increased from θ1 to θ2 showing that they have gained heat energy.

Quantity of heat energy lost by the hot solid = mscs(θs – θ2)
Quantity of heat energy gained by the container and cold water
= (mccc + mwcw) (θ2 – θ1)
Assuming no energy is lost to the outside;
Energy lost by the hot solid = heat energy gained by the container and cold water.
∴ mscs(θs – θ2) = (mccc + mwcw) (θ2 – θ1)
from which the specific heat capacity of water (cW) can be calculated.
As a precaution, the container has to be covered with wool or felt and aluminium
foil wrapped round the cotton wool. Note; the whole arrangement has to be placed
inside a wooden container before the hot solid is transferred into the cold water
in the container. These precautions ensure that minimum heat energy is lost from
the mixture to the surroundings.
Experiments show that specific heat capacity of water is 4 200 J/kgK. This means
that we need 4 200 Joules of heat energy to raise the temperature of 1 kg of water
by 1K. Note that this value is about 10 times more than that of copper or iron.
Once water is heated it retains the heat energy for a long time due to its high
specific heat capacity.
Example 5.12
In an experiment, to calculate the specific heat capacity of water, the following
data was obtained. Mass of the solid = 50 g, specific heat capacity of the solid =
400 J/kg K, initial temperature of the hot solid = 100 °C, mass of the container =
200 g, specific heat capacity of the material of the container = 400 J/kg K, mass
of water = 100 g, initial temperature of the water and the container = 22 °C.
When the hot solid was transferred into the cold water in the container, the
temperature of the mixture was 25 °C.
Use the data to calculate the specific heat capacity of water.
Solution
Let the specific heat capacity of water be cw
Heat lost by the hot solid = mc∆θ = 0.050 × 400 × (100 – 25)
= 1 500 J
Heat gained by the container and water = mc∆θ container + mc∆θ water
= 0.200 × 400 × (25 – 22) + 0.100 × cw (25 – 22) J
= 80 × 3 + 0.1 cw × 3
= 3 (80 + 0.1 cw) J
148
Assuming no energy losses to the surroundings

Heat lost = heat gained
1 500 J= 3 (80 + 0.1 cw )
500 = 80 + 0.1 cw (on dividing by 3 both sides)
420 = 0.1 cw
∴ cw = 4 200 J/kg K
Activity 5.27 To determine the specific heat capacity of a liquid by

electrical method
Materials
• Carolimeter • Stirrer • Themometer
• Heater • A liquid • Electrical circuit
• Variable resistor
Steps
1. Measure and record the mass, mc, of an empty, clean and dry copper container
with the stirrer of the same specific heat capacity, cc.
2. Gently pour the liquid of known mass, ml, into the container. Let the specific
heat capacity of the liquid be cl.
3. Note the initial temperature of the liquid and the container, θ1.
4. Complete the electrical circuit as shown in Fig. 5.38 with the heater fully
immersed in the liquid without touching the base or the sides of the container.
+ – – +
A
S
+ – VR
V
Thermometer
Cotton wool Cardboard lid
Aluminium foil
Copper container
Liquid
Stirrer Heating element
Wooden container
Fig. 5.38: Finding specific heat capacity of a liquid by electrical method.
149
5. Close the switch S and start a stop watch at the same time.
6. Use the variable resistor VR to maintain a steady current passing through
the heater.
7. Note the current I through the heater with the ammeter and a p.d V across
it with the voltmeter. Pass this steady current for some time so that the rise
in temperature of the liquid and the container is about 5 ºC.
8. Keep stirring the liquid gently throughout the experiment. Note the time,
t, taken when the final temperature of the liquid and the container is θ2.
Calculate the change in temperature ∆θ = (θ2 – θ1).
9. How much electrical energy has been spent in this time?
10. What has happened to this energy?
11. Using all the data you have collected, calculate the specific heat capacity of
the liquid.
(Hint: Electrical energy supplied = heat energy gained by liquid)
12. What precautions have to be taken during the experiment?
Electrical energy spent = heat energy gained by the liquid, the container and the
stirrer.
∴ VIt = mccc(θ2 – θ1) + mlcl (θ2 – θ1)
VIt = (mccc + mlcl) ∆θ
From this equation, the specific heat capacity of the liquid cl can be calculated.
Different liquids have different specific heat capacities. Table 5.2 shows the specific
heat capacity of some liquid
Table 5.2
Specific heat Specific heat

Substance Substance
capacity (c) J/kg K capacity (c) J/kg K
Castor oil 2 130 Olive oil 2 000
Coconut oil 2 400 Paraffin oil 2 130
Glycerol 2 400 Sulphuric acid 1 380
Mercury 140 Water 4 200
Example 5.13
In an Activity similar to 5.27, the following data was obtained. Power of electric heater
= 30 W, mass of the container and the stirrer = 200 g, specific heat capacity of the
container and the stirrer = 400 J/kg K, mass of water in the container = 100 g,
specific heat capacity of water = 4 200 J/kg K.
150
Use the data to calculate the time taken by the heater to rise the temperature of water,
container and the stirrer from 20 °C to 23 °C. What assumption have you made in
your calculations?
Solution
Assuming all the electrical energy is absorbed by the container, stirrer and water,
Electrical energy used = Heat energy gained
VIt = (mc∆θ)container + stirrer + (mc∆θ)water
As electrical power P = VI and the time taken is t,
Pt = (mc∆θ)container + stirrer + (mc∆θ)water
∴ 30 t = 0.200 × 400 × 3 + 0.100 × 4 200 × 3
30 t = 3(0.200 × 400 + 0.100 × 4 200)
30 t = (80 + 420)3
∴ t = 50 seconds
The assumption made is that there is no heat used to the surrounding.
Exercise 5.6
Where necessary, take specific heat capacity of water = 4 200 J/kg K, acceleration
due to gravity g = 10 m/s2.
1. Define (a) heat capacity (b) specific heat capacity of a substance
2. Calculate the:
(a) heat energy required to raise the temperature of 200 g of gold of specific
heat capacity 130 J/kg K by 1 000 ºC.
(b) heat energy given out when a piece of hot iron of mass 2 kg cools
down from 450 ºC to 25 ºC, if the specific heat capacity of iron is
460 J/kg K.
3. Describe an activity to determine the specific heat capacity of a solid by the
method of mixtures. State the necessary precautions to be taken during the
activity.
4. Define specific heat capacity of water. How would you determine the specific
heat capacity of water by the method of mixtures?
5. An electric kettle rated 2 kW is filled with 2.0 kg of water and heated from
20 ºC to 98 ºC. Calculate the time taken to heat the water assuming that all
the electrical energy is used to heat the water in the electric kettle and the
kettle has negligible heat capacity.
151
6. A hot solid of mass 100 g at 100 ºC is quickly transferred into 100 g of water
in a container of mass 200 g at 20 ºC. Calculate the resulting temperature of
the mixture. Specific heat capacity of the solid and the container is 400 J/kgK.
7. An electric heater rated 1 500 W is used to heat water in an insulated
container of negligible heat capacity for 10 minutes. The temperature of
water rises from 20 ºC to 40 ºC. Calculate the mass of water heated.
8. A piece of metal of mass 200 g at a temperature of 150 ºC is placed in water
of mass 100 g and temperature 20 ºC. The final steady temperature of the
water and the piece of metal is 50 ºC. Neglecting any heat losses, calculate
the specific heat capacity of the metal.
9. Describe an experiment to determine the specific heat capacity of a liquid by
electrical method.
10. A piece of iron of mass 200 g at 300 ºC is placed in a copper container
of mass 200 g containing 100 g of water at 20 ºC. Find the final steady
temperature of the mixture, assuming no energy losses. The specific heat
capacities of copper and iron are 390 J/kg K and 460 J/ kg K. respectively.
11. A class of Physics students decided to determine the specific heat capacity of
water in a waterfall. They used a sensitive thermometer to find the difference
in temperature of water at the top and the bottom of the waterfall and
obtained the following results; height of the waterfall = 52 m, temperature of
water at the top = 21.54 ºC. Temperature of water at the bottom = 21.67 ºC.
Stating any assumptions made, calculate a value for the specific heat capacity
of water.
5.4.3 Latent heat and specific latent heat

5.4.3.1 Latent heat of fusion
(a) Definition of latent heat of fusion

Heat is either absorbed or given out at a constant temperature when a substance
is changing its state. When a substance changes from a solid state to a liquid state,
heat is absorbed. This heat is called the latent heat of fusion of a substance.
The latent heat of fusion of a substance is defined as the quantity of energy required to
change the substance from solid state to the liquid state without change in temperature
at a constant pressure.
When the mass of the substance undergoing change is 1 kg, the absorbed heat is
called specific latent heat of fusion. The specific latent heat of fusion of a substance
is defined as the quantity of heat energy required to change 1 kg of the substance from
solid state to the liquid state without change in temperature at a constant pressure.
152
The latent heat of fusion required in the (Q) is directly proportional to the mass
of the substance i.e.
Qαm
Q = lm
Where l is constant called specific latent heat of fusion of the substance.
Q
∴l= m
SI units of specific latent heat of fusion is Joule per kilogram (J/kg).
(b) Finding specific latent heat of fusion of ice

Since pure ice melts at 0 °C under standard atmospheric pressure, the specific
latent heat of fusion of ice (lice) is defied as the quantity of heat energy required to
change 1 kg of ice at 0 ºC to 1 kg of water at 0 ºC under standard atmospheric pressure.
Activity 5.28 To determine the specific latent heat of fusion of ice

by electrical method
Materials
• Pure ice • Filter funnel • Heater
• Stand & clamp • Electrical circuit • Stopwatch
Steps
1. Take some pure crushed ice at 0 °C in a filter funnel which is well insulated
from the outside with a cardboard lid and lagging material like felt or cotton
wool as shown in Fig. 5.39. Complete the electrical circuit as described in
Activity 5.27, with the electric heater completely covered by ice.
2. Close the switch. Take an empty, clean and dry beaker and find its mass m1.
3. Place the beaker under the funnel and start a stopwatch when the first drop
of water is collected.
4. Note the steady current, I, through the heater and the p.d V across it.
5. Collect enough water in the beaker and stop the stopwatch.
153
+ – – +
A
S
+
V
– VR
Cardboard lid
Ice
Heater
Stand Felt lagging
Filter funnel
Water drop
Beaker
Water collected at 0 °C
Fig. 5.39: Specific latent heat of fusion of ice
6. Note the time, t, taken to collect this water. Find the mass of the beaker with
the water m2.
7. Calculate the mass, m, of the water collected i.e. the mass of ice at 0 °C
which has melted. m = (m2 – m1).
8. Using the data you have collected, calculate the specific latent heat of fusion of
ice using the equation
Electrical energy supplied = latent heat gained by ice in melting
How much electrical energy has been spent in time t? What has happened in
this energy?
Electrical energy spent = VIt
Heat energy gained by ice at 0 °C in time t is given by Q = mlice , where lice is the
specific latent heat of fusion of ice. Assuming no electric energy is lost from heater,
VIt = mlice
From which lice can be calculated.

Experiments show that the specific latent heat of fusion of ice is 3.36 × 105 J/kg.
This means that we need 336 000 joules of energy to convert 1 kg of ice at 0 °C
to water at 0 °C under standard atmospheric pressure.
154
Example 5.14
An electric heater rated 1.5 kW is used to melt 1.5 kg of ice at 0 °C. Calculate
the specific latent heat of fusion of ice, if it takes 5.6 minutes for the heater to
melt all the ice at 0 °C.
Solution
Electrical energy spent in time, t = VIt
= Pt (since electrical power, P = VI)
Heat energy gained by ice to change its state Q = mlice
Assuming no energy losses,
Pt = mlice
P×t 1 500 × (5.6 × 60)
lice = m = 1.5
= 336 000 J/kg
Activity 5.29 To determine the specific latent heat of fusion of ice

by the method of mixtures
Materials
• Container with known specific heat capacity • Heat capacity
• Warm water • Thermometer
• Ice • Weighing machine
Steps
1. Take an empty, clean and dry container of known specific heat capacity (cc)
and measure its mass (mc).
2. Add some warm water at a temperature a few degrees above room temperature
to the container and note its temperature (θ1).
3. Measure the mass of the container with warm water and calculate the mass
of water (mw).
4. Dry small pieces of ice with a blotting paper and gently immerse them into
the warm water, without splashing out any water.
155
Thermometer
Stirrer
String
Beaker
Container
Warm Thermometer
Ice water
Water
(a) Ice being added to hot water

(b) Mixture in a container
Fig. 5.40: Specific heat capacity of water by method of mixtures.

5. Keep adding the small pieces of dried ice till the temperature of the mixture
is a few degrees below room temperature.
6. Note the temperature of the mixture (θ2). Find the mass of the mixture.
7. Calculate the mass of ice (mice) which has been added.
8. Using the data you have collected, determine the latent heat of fusion of ice..
(Hint: Heat gained by ice to melt = heat lost by warm liquid water)
Assuming no energy losses;

heat energy given out by the container and warm water = heat energy used in
melting ice at 0 ºC + heat energy used in rising the temperature of the melted ice
from 0 ºC to θ2 ºC.
∴ mccc (θ1 – θ2) + mwcw(θ1 – θ2) = mice lice + mice cw (θ2 – 01)
where cw is the specific heat capacity of water and lice is the specific latent heat of
fusion of ice. From the above equation we can calculate lice.
5.4.3.2 Latent heat of vaporisation

(a) Definition of latent heat of vaporisation
When a liquid changes into vapour, the temperature remains constant till all
the liquid has changed its state to vapour. This shows that heat is absorbed to
change the state from liquid to vapour state. The heat is called the latent heat of
vaporisation of the liquid. The latent heat of vaporisation of a liquid is the amount of
heat energy required to change the substance from the liquid state to vapour state at a
constant pressure without change in temperature.
When the mass of the substance undergoing change is 1 kg, the heat is called specific
latent heat of vaporisation. The specific latent heat of vaporisation of a substance is
the quantity of heat energy required to change 1 kg of the substance from the liquid state
to the gaseous state at a constant pressure without change in temperature.
156
The quantity of heat energy required to change a liquid of mass, m, into vapour
is given by,
Q = ml
where l is called the specific latent heat of vaporisation of the liquid
(b) Finding specific latent heat of vaporisation of water or steam

Specific latent heat of vaporisation of water (lwater) is the quantity of heat
energy required to change 1 kg of water at 100 ºC to 1 kg of steam at
100 ºC under standard atmospheric pressure.
Activity 5.30 To determine the specific latent heat of vaporisation

of water by electrical method.
Materials
• Beaker • Heater electrical circuit • Cold water
• Insulated container • Tube T • Thermometer
• Stopwatch • Stopper
Steps
1. Take some water in a glass container + – – +
A
which is well insulated from the S
outside with a rubber/cork stopper at +

V
– VR
the top and bottom and a lagging
material like cotton wool around it.
Rubber/cork stopper
2. Insert a long glass tube T and a Glass container
condenser as shown in Fig 5.41,
with the electric heater fully Lagging
immersed in water. Heater
Water
3. Complete the electrical circuit as Tube T
shown. Rubber/cork stopper
Warm water
4. Close the switch S. Let a steady
current, I, flow through the heater Condeser
for sometime till water boils. Cold water
5. Adjust the variable resistor so that Water drop
a steady current flows through the Beaker
heater.
Water collected at 100
6. Take an empty, clear and dry beaker °C in a time t
and find its mass, m1. Fig. 5.41: Specific latent heat of vaporisation
157
7. Place the beaker under the condenser, where cold water is being circulated,
and start a stopwatch when the first drop of water is collected.
8. Collect enough water in the beaker. Note the time taken, t, to collect this
mass of water.
9. Find the mass of the beaker again with the water collected, m2 and calculate
the mass m of water collected, i.e the mass of steam at 100 ºC which has
condensed m = (m2 – m1).
10. Assuming that there is no energy loss,calculate the latent heat of fusion using
the data you have obtained.
How much electrical energy has been spent in this time, t? What has happened
to this energy?
Electrical energy spent = VIt
Heat energy gained by water at 100°C = energy used to form steam at 100˚which
condenses into water.
The heat energy is given by the equation;
Q = mlw
Where lw is the specific latent heat of vaporisation of water or steam.
Assuming no energy losses,
VIt = mlw
VIt
Hence lw =
m
Experiments show that the specific latent heat of vaporisation of water lw is
2.26 × 106 J/kg. This means that 2.26 × 106 Joules of energy are needed to convert
1 kg of water at 100 ºC to 1 kg of steam at 100 ºC under standard atmospheric
pressure.
Example 5.15
An electric kettle rated 1 500 W is used to boil 500 g of water into steam at
100 °C. Calculate the time required to boil off the water, if the specific latent heat
of vaporisation of water is 2.26 MJ/kg. Why is the correct time likely to be longer
than your calculated time?
Solution
Let the time taken to boil off water be t (s)
Electrical energy spent = VIt = Pt …(i)
Heat energy required to boil off 500 g of water = mlwater…(ii)
From (i) and (ii), mlwater = Pt
158
mlwater
∴t=
P
0.500 × 2.26 × 106
= 1 500
= 753 s
The correct time is likely to be more than the calculated time, as no allowance
has been made for the energy loss from the heater to the outside or the energy
gained by the kettle itself.
Example 5.16
The graph in Figure 5.42 shows how temperature varies with time when
1 kg of ice at –10 °C is converted into 1 kg of steam at 100 °C under standard
atmospheric pressure. Calculate the heat energy required to convert ice into
steam given: Specific heat capacity of ice = 2 100 J/kg K, specific heat capacity of
water = 4 200 J/kg K, specific latent heat of fusion of ice = 3.36 × 105 J/kg, specific
latent heat of vaporisation of water = 2.26 × 106 J/kg.
100
80
temperature/°(C)
60
40
20
–20
time(s)
Fig 5.42: Temperature – time graph
Solution
Heat energy required to:
1. raise the temperature of ice from –10 °C to 0 °C is given by
Q1 = (mc∆θ)ice= 1 × 2 100 × 10 J = 21 000 J
2. convert ice at 0 °C to water at 0 °C is given by
Q2 = (ml)ice= 1 × 3.36 × 105 J = 336 000 J
3. raise the temperature of water from 0 °C to 100 °C is given by
Q3= (mc∆θ)water = 1 × 4 200 × 100 J = 420 000 J
159
4. convert water at 100 °C to steam at 100 °C is given by

Q4 = (ml)water = 1 × 2.26 × 105 J = 2 260 000 J
∴ Total heat energy required to convert ice into steam is given by
Q = Q1+ Q2 + Q3 + Q4
= 21 000 + 336 000 + 420 000 + 2 260 000 = 3.04 × 106 J

Example 5.17
In Example 5.16, calculate the ratio of lw to lice, if the same electrical heater is
used throughout the experiment, the time taken to convert ice at 0 °C to water at
0 °C is 168 s and the time taken to convert water at 100 °C to steam is 1 130 s.
Solution
Assuming electrical energy spent = heat energy gained
VIt = mlice
∴ Pt1 = mlice …(i)
Similarly Pt2 = mlwater … (ii)
Pt2 mlwater
Dividing equation (ii) by (i) =
Pt1 mlice
lwater 1 130
∴ = 168 = 6.73
lice
Exercise 5.7
Where necessary, take:

Specific heat capacity of ice = 2 100 J/kg K
Specific heat capacity of water = 4 200 J/kg K
Specific latent heat of fusion of ice = 3.36 × 105 J/kg K
Specific latent heat of vaporisation of water = 2.26 × 106 J/kg K
1. Define the terms: melting, boiling, melting point, boiling point.

2. Sketch a graph to show how temperature varies with time when ice at –8 ºC
is converted to water at 16 ºC.
3. (a) Define (i) the latent heat of fusion (ii) latent heat of vaporisation of a substance.
(b) Why do we use the term ‘latent’ heat?
160
4. (a) Define the term specific latent heat of fusion of ice.

(b) With the aid of a diagram, describe an experiment you could conduct in
order to determine the specific latent heat of fusion of ice.
5. Calculate the amount of heat energy required to convert 2.5 kg of ice at 0 ºC
to water at 0 ºC.
6. (a) How much heat energy is required to convert 2.0 kg of ice at –10 ºC to
water at 20 ºC?
(b) If an electric heater rated 1.0 kW is used, calculate the time taken to heat
ice at –10 ºC to water at 20 ºC? State the assumption you have made in
arriving at your answer.
7. (a) Define the term specific latent heat of vaporisation of water.
(b) With the aid of a circuit diagram describe an experiment you would
conduct to determine the specific latent heat of vaporisation of water.
8. An electric kettle rated 2.5 kW is used to boil 2.0 kg of water at 100 ºC into
steam. Calculate the mass of water converted into steam if the kettle was
used for 6 minutes.
9. An electrical kettle rated 1.5 kW is used in an experiment to convert
200 g of ice at –5 ºC to steam at 100 ºC under standard atmospheric
pressure. Calculate the time for which the kettle has to be used in the above
experiment.
5.4.4 Applications of specific heat capacity
Activity 5.31 To identify and describe the application of specific

heat capacity in our daily lives
Materials
• Internet
• Reference books
Steps
1. You have learnt about specific heat capacity, how is it useful in our daily lives
2. With the help of internet and reference books, conduct a research on
application of specific heat capacity. Identify those that are not highlighted
in this book. Note them in your notebook.
3. Represent your findings to the whole class
161
Specific heat capacity has many applications in our daily life. The following are
a few examples:
(a) A material with high specific heat capacity absorbs a lot of heat with only
a small rise in temperature. This accounts for the efficiency of water as a
coolant in a car radiator and of hydrogen gas in enclosed electric generators.
(b) Substances with low specific heat capacities are quickly heated up; they
experience a greater change in temperature after gaining a small amount of
heat energy. For this reason, they are used to make cooking utensils such as
frying pans, pots and kettles.
(c) Sensitive thermometers are made from materials with low specific heat
capacity in order to detect and accurately show rapid change in temperature,
even for small amounts of heat energy.
(d) Materials with high specific heat capacity are suitable for making handles of
heating devices like kettles, pans and oven covers. This is because they do not
get very hot easily when they absorb high amounts of heat energy.

• Heat is a form of energy which is transferred from a region of higher
temperature to a region of lower temperature.
• The SI unit of heat energy is Joule (J).
• Two substances of equal masses can be at the same temperature but contain
different amounts of heat energy and vice-versa.
• Heat energy can be transferred by three different modes: conduction,
convection or radiation.
• Solids are heated by conduction and fluids by convection. Radiation can take
place through vacuum.
• We get heat energy from the sun by radiation.
• The quantity of heat transferred depends on the following factors:
(a) The temperature difference. (b) The nature of the materials.
(c) The cross-sectional area. (d) The length of the material.
(e) The time taken to transfer heat.
• Heat capacity of a substance is the quantity of heat energy required to change
the temperature of the substance by 1 K.
• Specific heat capacity of a substance is the quantity of heat energy required
162
to change the temperature of 1 kg of the substance by 1 K. Whenever there is

a change in temperature of a substance, the quantity of heat energy involved
is given by, Q = mc∆θ.
• Latent heat of fusion of a substance is the quantity of heat energy required
to change the substance from the solid state to the liquid state without any
change in temperature at a constant pressure.
• Specific latent heat of fusion of a substance is the quantity of heat energy
required to change 1 kg of a substance from the solid state to the liquid state
without any change in temperature at a constant pressure.
• Specific latent heat of vaporisation of a substance is the quantity of heat
energy required to change 1 kg of a substance from the liquid state to the
vapour state without any change in temperature at a constant pressure.
• Whenever there is a change of state of a substance at constant temperature,
the quantity of latent heat involved, Q = ml.
Unit Test 5
For questions 1 – 5, select the correct answer from the choices given.
1. Radiation in a thermos flask is minimized by
A Cork B Vacuum
C Felt pad D Silvered glass water
2. A dull black surface is a good
(i) Absorber of heat energy
(ii) Emitter of heat energy
(iii) Reflector of heat energy
A (i) only B (i) and (ii) only
C (ii) and (iii) only D (i), (ii) and (iii)
3. Radiation is the transfer of heat _______
A in a liquid which involves the movement of the molecules.
B from one place to another by means of electromagnetic waves.
C through a material medium without the bulk movement of the medium.
D through a fluid which involves the bulk movement of the fluid itself.
4. The mode of transfer of heat between the boiler and the storage tank of a hot
water supply system is
A radiation B conduction
163
C convention D evaporation
5. The transfer of heat by the actual movement of molecules of matter takes
place
A only in liquid B only in gases
C in solids and liquid D in liquids and gases
6. Match each heat transfer mechanisms to its description
Conduction Electromagnetic waves
Evaporation Transfer of vibrational energy from particle to particle
Radiation Escaping of particles from the surface of a liquid
Convection Movement of particles due to changes in density
7. Fill in blanks
(a) The specific latent heat of fusion is the energy required to _______ a kg
of solid.
(b) The specific latent heat of vaporization is the energy required to
_________ a kg of liquid.
(c) The specific heat capacity is the energy required to increase the _______
of ______ kg of material by 1°C.
8. Define the terms:
(a) heat capacity (b) specific heat capacity of a substance
9. Calculate the heat capacity if 8 000 J of heat is used to cool a solid of mass 1 kg
from 80 ºC to 20 ºC.
10. Calculate;
(a) the heat energy required to raise the temperature of 200 g of gold of
specific heat capacity 130 J/kg K by 1 000 ºC.
(b) the heat energy given out when a piece of hot iron of mass 2 kg cools
down from 450 ºC to 25 ºC, if the specific heat capacity of iron is
460 J/kg K.
11. In experiment requiring storage of heat energy, water is preferred to other
liquids. Give two reasons for this.
12. Calculate the heat energy required to raise the temperature of 400 g of water
from 25 ºC to 45 ºC. Specific heat capacity of water = 4 200 J/kg ºC.
13. Find the initial temperature of aluminium if 2 400 J of heat is used to raise
the temperature of 50 g of aluminium to 62 ºC. Specific heat capacity of
aluminium is 900 J/kg K.
14. 620 000 J of heat energy is supplied to raise the temperature of a solid of mass
164
10 kg from 40 ºC to 75 ºC. Calculate the specific heat capacity of the solid.

15. Calculate the heat required to heat 0.5 kg of ice at –8 ºC to steam at 100 ºC.
(Specific heat capacity of ice = 2 100 J kg/K. Specific heat of fusion of ice is
3.34 × 105 J/kg and specific latent heat of vaporization of water = 2.26 × 106 J/kg K.
16. Explain the following statements:
(a) A metallic seat seems to be hotter during the day and colder during the
night than a wooden seat under the same conditions.
(b) The bottom of cooking vessels are usually blackened.
(c) It is safer to hold the other end of a burning match stick.
17. A cross-section of a solar heater panel fitted to the roof of a house is shown
below (Fig. 5.43).
P
Q
radiation from
the sun
f
roo
glass
sheet
f water pipes
roo
Fig. 5.43
(a) Surface P is dull black whereas surface Q is covered in shiny aluminium

foil. Give a suitable reason for each choice.
(b) What is the purpose of the front glass sheet?
18. A mass of 700 g of copper at 98 ºC is put into 800 g of water at 15 ºC contained
in a copper vessel of mass 200 g and the final resulting temperature of the
mixture is found to be 21 ºC. Calculate the specific heat capacity of copper.
19. A refrigerator converts 1 kg of water at 25 ºC into ice at –5 ºC in 2.5 hours.
Show that the rate at which heat is extracted from the refrigerator is about
50 J/s.
20. It takes 15 minutes for an electric kettle to heat a certain quantity of water
from 0 ºC to 100 ºC. It requires 80 minutes to convert all the water at
100 ºC into steam. Calculate the specific latent heat of vaporisation of water.
21. The graph in Figure 5.44 shows how 400 g of ice at –10 ºC would change
with time, if it were to be heated at a steady rate of 600 W.
165
(a) Explain in molecular terms, what has happened to the energy

being supplied at 0 ºC.
(b) Use the graph to determine the specific latent heat of fusion of ice.
40
temperature (°C) 30
20
10
0
10 20 30 40 50 60
–10
time(s)
Fig. 5.44
166
UNIT 6 Laws of thermodynamics
Key Unit Competence

By the end of this unit the learner should be able to describe the internal energy
of a system by applying laws of thermodynamics.
Learning objectives

• State the laws of thermodynamics.
• Describe applications of thermodynamics laws.
• Explain the functioning of the refrigerator.
• Describe heat exchanges.
• Explain the work done on a system when heat energy increases or reduced
frequency.
• Explain the functioning of a transformer .
• Relate peak and r.m.s values.
Skills
• Show how heat energy increase with an increase of external work done.
• Determine the quantity of heat using the method of mixtures.
• Demonstrate the change of state as a result of internal energy.
• Carry out an investigation on how one state is transformed into another state.
• Differentiate internal energy and external energy as an increase in energy due to
work done on a system.
Attitude and value

• Appreciate the importance of internal energy in increasing the temperature of a body.
• Be aware of the rate at which bodies loose energy.
• Adapt scientific method of thinking
• Recognise the need of acquiring knowledge of analysing and modelling physical
processes using laws of thermodynamics.
167
Introduction
Unit Focus Activity
Materials
• Beaker • Bunsen burner • Tripod stand
• Wire gauze • Thermometer • Water
Part 1
1. Set up the apparatus as shown in Fig.
6.1.
2. Record the initial temperature of
water in the beaker.
3. Heat the water for 2 min and record
its final temperature.
4. Stop heating and wait for the water
to cool for 1 min. Record its final
temperature.
5. Explain how heat is transferred from
one particle to another in the water.
6. Describe the effect of heat gain on the
internal energy of a particle.
Fig. 6.1: Heating some water in a beaker
168
Part 2
1. (a) Define the term entropy.
(b) How does entropy change in the water during:
(i) heating? (ii) cooling?
2. (a) State the first and second laws of thermodynamics
(b) Explain how these two laws govern the changes in the thermal energy of
the water during:
(i) heating (ii) cooling
3. Describe how the principle of heat exchange is demonstrated during the
cooling of the water.
6.1 Introduction to themodynamics
Activity 6.1 To define thermodynamics and distinguish between

the three types of systems
Materials
• Water in a beaker • Bunsen burner
• Thermometer • Solid ice
Steps
1. From knowledge you acquired in unit 5, how does heat travel from one end
to another in different states of matter?
2. How do particles behave in each state of matter during heat transfer?
3. What is thermal dynamics as applied to heat transfer?
4. Heat the water in the beaker till it boils. Measure and record the temperature.
5. Dip the solid ice into the boiled water, measure the temperature again.
Compare the reading with that of step 4. What do you observe? Explain
6. Distinguish between open, closed and isolated system. Classify the behaviour
the beaker and its contents in steps 3 and 4 as one of these systems giving
reason for your answer.
8. Discuss your findings as the whole class with the help of the teacher and note
them down in your note books.
169
The term thermodynamics is composed of two words 'thermo' and 'dynamics'.

Dynamics is the study of why things move the way they do. For instance, in the
unit on Newton’s laws, we looked at what compels bodies to accelerate and how
they move. The prefix 'thermo' means heat.
Therefore, thermodynamics is the study of systems involving energy in the form
of heat and work. A good example of a thermodynamic system doing work is a
gas confined by a piston in a cylinder. If the gas is heated, it will expand, doing
work on the piston by pushing it.
Since thermodynamics deals with the flow of heat in and across systems, it is
important to first remind ourselves the different types of systems.
There are three types of systems in thermodynamics: open, closed, and isolated.
• An open system is a system that exchange both energy and matter with its
surroundings. An open heating pot is an example of an open system, because
heat and water vapour are lost to the air.
• A closed system is a system that can exchange only energy with its
surroundings, but not matter. A heating pot with a very tightly fitting lid
would be approximated to be a closed system.
• An isolated system is a system that cannot exchange either matter or energy
with its surroundings. Perfect isolated systems are very rare, but a completely
insulated drinks cooler with a lid would ideally represent an isolated system.
The items inside can exchange energy with each other, but they exchange
negligible or no heat energy with the outside environment.
6.2 Internal energy of a system
Activity 6.2 To demonstrate and explain internal energy of a

system
Materials:
• Reference books • Internet
• Transparent container • Marbles of two different colours
Steps
1. Open the transparent container and drop the marbles of one colour followed
by by those of the other colour.
2. Shake the container and observe the marbles. What do you see? Did the
marbles remain at the same position?
3. Suppose the container is the system and the marble is the internal energy.
Explain how internal energy flows inside a substance.
170
4. Now conduct a research from internet or reference book on the definition of

internal energy.
5. Compare and discuss your findings with other groups in class.
6. Write down your findings in your note book.
The internal energy of a system is identified with the random, disordered motion
of molecules. The total (internal) energy in a system includes:
• translational kinetic energy
• vibrational and rotational kinetic energy
• potential energy from intermolecular forces
The symbol for internal energy change is ΔU.
Note that same quantities of substances at the same temperature do not

necessarily possess the same amount of internal energy. For example,
one gram of water at 0°C compared with one gram of copper at 0 °C
do not have the same internal energy. This is because, though their
kinetic energies are equal, water has a much higher potential energy
causing its internal energy to be much greater than the copper's internal
energy.
6.3 First law of thermodynamics
Activity 6.3 To demonstrate the first law of thermodynamic
Materials:
• Beaker • Bunsen burner
• Water • Thermometer
Steps
1. Put some water in a beaker and dip the thermometer in the water. Note the
temperature of the water.
2. Place the beaker on bunsen burner and observe what happens as you
continue to heat.
3. Observe what happens to the water particle as the water boils.
4. Describe energy transformation taking place in the set up during heating
and at boiling point.
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5. State the law that governs the energy transformations that govern the heat
energy exchange in this activity.
We have already learnt about the law of conservation of energy which states that
energy can neither be created nor destroyed but can be converted from one form
to another. This law is obeyed by all systems including thermodynamic systems.
In thermodynamic systems, the law of conservation of energy governs the energy
transformations involving applied heat and internal energies. In such systems for
example, the heat applied (external energy) and the work done by the environment
onto the system are converted to internal energy. Specifically, for thermodynamic
systems, the law of conservation of energy is summarized into what is known as
the first law of thermodynamics.
The law states that the change in the internal energy (∆U) of a system is equal to the
sum of the heat (Q) that flows across its boundaries and the work (W) done on the
system by the surroundings.
The law is mathematically represented as

Change in internal = heat dissipated or + Work done by or
Energy of a system absorbed by the system to the system
∆U = Q + W
Note:
(a) The absorption of heat by the system tends to raise the energy of
the system and vice versa.
Therefore, in the equation, we take heat (Q) to be positive if it is
supplied to the system and negative if heat is dissipated (removed)
from the system.
(b) The performance of work by the system requires use of energy by
the system hence lowers the energy of the system, and vice versa.
Therefore, in the equation, we take work done as positive if it is
done on to the system and negative if it is done by the system.
Example 6.1
A gas in a system has constant pressure. The surroundings lose 49 J of heat to the
system and does 316 J of work onto the system. What is the change in internal
energy of the system?
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Solution
Let us first consider the relationship between the system and the surroundings.
The surrounding loses heat and does work onto the system. Therefore, we take
Q and W as positive in the equation.
Therefore,
ΔU = Q + W
ΔU = 49 J + 316 J
= 365 J
Example 6.2
In a certain process, 450 J of work is done on the system. If the system gives off
124 J of heat, what is the change in internal energy of the system?
Solution
We take the heat (Q) as negative because it is given off by the system.
We take the work done (W) as positive because it is done on the system.
Therefore,
ΔU = Q + W
= -124 J + 450 J
= 226 J
Example 6.3
In a certain process, 523.6 J of heat is added to a system. The system does work
equivalent to 78I.4 J by expanding against the surrounding atmosphere. What is
the change in internal energy for the system?
Solution
We take the heat (Q) as positive because it has been added the system.
We take the work done (W) as negative because it is done by the system
Therefore,
ΔU = Q + W
= 523.6 J – 781.4 J (In this case, W is negative)
= -257.8 J
= -257.8 J
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Exercise 6.1
For questions 1 - 3 select the most appropriate response from the choices
given.
1. Which has more internal energy?
A. Bathtub full of cool water
B. Cup of hot water
2. If a piece of aluminum is placed in contact with both a bathtub full of cool
water and a cup of hot water, which way will heat flow?
A. From the cup to the bathtub
B. From the bathtub to the cup
3. A system has 30 J of heat added to it and in doing so, it does 25 J of work.
Its change in internal energy will be:
A. +25 J B. + 5 J C. -5 J
D. -30 J E. -25 J
Fill in the blank spaces.

4. Besides kinetic energy, molecules have rotational kinetic energy, potential
energy due to forces between molecules and more. The total of all energies
inside a substance is called ___________?
5. If heat flows from a cup of hot water to a bathtub full of cool water, the water
in the cup will have a ___________ in internal energy while the water in the
tub will have an ___________ in internal energy.
6.4 Second law of thermodynamics
Activity 6.4 To define and explain the term entropy
Materials:
• 20 red marbles and 20 green marbles • a tray
Steps
1. Place the red marbles on one side in the tray and the green ones on the
opposite side in an orderly manner.
2. Shake the tray with your hands for some time then stop.
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3. Observe the marbles. In what state are they, orderly as at the beginning or
disorderly?
4. Based on your observations in step 1(a) to (c), describe how the order of
particles in a substance is affected, when the internal energy of the substance
is increased by heating.
5. Research the meaning of the term entropy from the Internet or reference
books.
Activity 6.5 To demonstrate and state the Second law of

thermodynamics
Materials:
• A beaker • Ice
• Burner water • Thermometer
Steps
1. Warm some water in a beaker.
2. Remove the beaker from the heat and place it on the bench. Place the
thermometer in the warm water and record its temperature.
3. Place the piece of ice into the warm water in the beaker. Observe what
happens to the temperature of the water as the ice eventually melts.
4. Discuss with your classmate whether the two process that took place in step 3
would be reversed in the same time if after taking place, the beaker is heated
again.
Entropy is the measure of a system’s thermal energy per unit temperature that is
not available to do useful work.
Since work arises from ordered motion of molecules, the amount of entropy is
also a measure of the molecular disorder, or randomness of a system. The concept
of entropy helps us to understand the direction in which spontaneous processes
happen in systems. It enables us to understand which processes are possible or
impossible, without violating the law of conservation of energy.
For example, when a block of ice is placed on a hot stove, it surely melts and at
the same time the stove cools. Such a process is called irreversible because after
taking place, heating the stove again will not make the melted water to freeze
back to ice again.
A German physicist called Rudolph Julius Emanuel Clausius in 1850, summarised
the concept of entropy in what is now known as the Second law of thermodynamics
in 1850.
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The law states that the spontaneous change for an irreversible process in an isolated
system always proceeds in the direction of increasing entropy. In other words, the entropy
of any isolated system always increases.
For example, the block of ice and the stove mentioned earlier constitute two parts
of an isolated system for which total entropy increases as the ice melts.
Different scientists have come up with different formulations/statements of this
law to describe the same phenomena. The simplest of all these formulations states
the second law as follows:
‘Heat flows spontaneously from a hotter object to a colder one, but not in the opposite
direction; the reverse cannot happen without the addition of energy’.
Exercise 6.2
Select the most appropriate response from the once given.

1. The second law of thermodynamics would back up this statement: For
natural processes, in the long run, entropy decreases.
A. True
B. False
2. Which of the following is the best definition of entropy?
A. Increase in the concentration of a substance in a solution
B. An energetically favorable reaction
C. Increase in disorganization within a system
D. Increase in organization within a system
3. In terms of energy changes that take place daily, the universe tends to move
towards:
A. the north B. the sun
C. order D. disorder
4. ______________ is the measure of randomness or disorder.
A. Metabolism B. Synthesis
C. Entropy D. A coupled reaction
5. The second law of thermodynamics will back up this statement: Heat will
never flow from a cold object to a hot object!
A True
B False
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6.5 Heat exchange
Activity 6.6 To demonstrate heat exchange using cold and hot

water
Materials:
• Water • Beakers A and B • Thermometers
• Tripod stand • Bunsen burner • Wire gauze
• Stirrer
Steps
1. Pour equal amounts of water into beakers A and B. Measure and note down
the temperature of water in beaker B.
2. Place the wire gauze on the tripod stand and place the Bunsen burner below
it. Light the Bunsen burner.
3. Place beaker A on the set up in step 2 and allow the water to heat upto
above the room temperature. Measure and record its temperature using a
thermometer. Compare the temperature of water in beaker A and B. Which
one is higher?
4. Put off the Bunsen flame and mix the water in B with that in A and stir well.
5. Measure and record the final temperature of the mixture.
6. Compare the final temperature of the mixture and initial temperature of
water in beaker B. Which one is higher? How has the initial temperature of
water in B changed in the final mixture, increased or decreased? Explain.
7. During the mixing up of the water in A and B, in what direction did the heat
move? A to B or vice versa?
8. In one statement, summarize the direction of heat transfer between cold and
hot regions.
Heat transfer is the exchange of thermal energy between physical systems. It moves from
one system to another because of temperature difference in the given systems.
The direction of heat transfer is always from the region of high temperature to that
of lower temperature, and is governed by the Second law of thermodynamics. The
region with higher temperature is known as the heat source and the one with lower
temperature as the heat sink.
In other words, heat transfer occurs in a direction that increases the entropy of the
collection of systems and changes the internal energy of the two systems involved in the
transfer. This is known as the principle of heat exchange.
The rate of heat transfer is dependent on the temperatures of the systems and
the properties of the intervening medium through which the heat is transferred.
A thermal equilibrium in the two system is reached when all involved bodies and
the surroundings reach the same temperature.
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The heat will continue flowing till the two systems are in thermal equilibrium (at
equal temperatures).
Activity 6.7 To show the quantity of heat using the principle of

heat exchange
Materials:
• Water • Beakers • Thermometer
• Tripod stand • Bunsen burner • Wire gauze
• Stirrer and beam balance • Beam balance
Steps
1. Pour water into beakers A and B, measure and note down the mass using a
beam balance of the water in both beakers i.e. m1 and m2 respectively and
temperature, t1 of water in beaker B.
2. Place the wire gauze on the tripod stand and place the Bunsen burner below
it, light the Bunsen burner.
3. Place beaker A on the set up in step 2 and allow the water to heat up above
the room temperature, then measure and record its temperature, t2.
4. Put off the Bunsen flame and mix the water in B with that in A and stir well.
5. Measure and record the final temperature, t3 of the mixture.
6. What conclusion can you make from your findings?
From the principle of heat exchange,

Heat lost by a hot object = heat gained by a cold object.
The specific heat capacity of water being a standard figure 4200 J/kg °C.
We can find either the quantity of heat lost by the hot water or the quantity of heat
gained by the cold water using the formula Q = mc∆θ, where Q is the quantity of
heat given out or absorbed, m, the mass of the substance, c, is the specific heat
capacity of a substance and ∆θ, is the temperature change.
Example 6.4
In an experiment similar to Activity 6.7, the following data was obtained. Mass of
the solid = 50 g, specific heat capacity of the solid = 400 J/kg K, initial temperature
of the hot solid = 100 °C, mass of the container = 200 g, specific heat capacity
of the material of the container = 400 J/kg K, mass of water = 100 g, initial
temperature of the water and the container = 22 °C.
When the hot solid was transferred into the cold water in the container, the
temperature of the mixture was 25 °C. Calculate the specific heat capacity of
water. Use the data to calculate the specific heat capacity of water.
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Solution
Let the specific heat capacity of water be cw
Heat lost by the hot solid = mc∆θ = 0.050 × 400 × (100 – 25)
= 1 500 J
Heat gained by the container and water = mc∆θ container + mc∆θ water
= 0.200 × 400 × (25 – 22) + 0.100 × cw (25 – 22)
= 80 × 3 + 0.1 cw × 3
= (80 + 0.1 cw)3
Assuming no energy losses to the surroundings
Heat lost = heat gained
1 500 = (80 + 0.1 cw )3
500 = 80 + 0.1 cw
420 = 0.1 cw
∴ cw = 4 200 J/kg K
6.6 Change of state and kinetic theory of matter
Activity 6.8 To describe the change of state and kinetic theory of

matter
Materials:
• Solid ice • Thermometer
• Bunsen burner
Steps
1. State the three states of matter.
2. Name the process of the change of states shown with arrows in the Fig. 6.2.
Fig. 6.2: Diagram of change in state
179
3. Highlight the key facts in the kinetic theory of matter.

4. With the help of your teacher harmonize your findings, write down the notes
in your books.
When a pure solid is heated, its temperature rises until it starts to melt. At its
melting point, any additional heat supplied does not change its temperature.When
the pure solid becomes a pure liquid (a change in state), further heating raises
the temperature of the liquid until it starts to boil.
At its boiling point, any additional heat supplied causes boiling without any
temperature rise. When the pure liquid becomes a pure gas (a change in state),
further heating will again raise the temperature of the gas.
When a gas is cooled, it reaches a temperature where it condenses to a liquid
at a constant temperature. When the liquid is cooled, it reaches a point where it
freezes to a solid at a constant temperature.
Important!
During the changing of state, the temperature of the gas/liquid/solid is constant.
The properties of the molecules of the substances vary with the amount of thermal
energy they possess.
The changes of state can be explained by using the kinetic theory of matter. The
kinetic theory of matter is theory that tries to explain the properties and behaviours
of the three states of matter. It makes the following three assumptions:
1. Matter consists of small particles
The first assumption is that matter consists of a large number of very small
particles either individual atoms or molecules.
All matter (solid, liquid, and gas) is made up of tiny particles called atoms,
or atoms that are joined to form molecules.
2. Large separation between particles

The second assumption describes the separation of the particles:
In a gas, the separation between particles is very large compared to their size,
such that there are no attractive or repulsive forces between the molecules.
In a liquid, the particles are still far apart, but are close enough that attractive
forces confine the material to the shape of its container.
In a solid, the particles are so close that the forces of attraction confine the
material to a specific shape. Fig 6.3 shows the separation between particles
of matter.
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Solid Liquid Gas

Figure 6.3. Separation between particles of matter
3. Particles in constant motion

The third assumption is that each particle in matter is in constant motion.
Particles of a solid are held very close to each other by strong attractive
forces. These forces hold the solid particles in place and gives a solid fixed
size and shape.
Liquid particles have more energy than solids allowing them to move around
more freely, and spread out more than those of a solid and therefore increasing
the space between the particles. This is why a liquid will take the shape of its
container up to its surface. However, liquids have definite volume.
Gas particles have more energy than liquids hence their particles move more
freely. Therefore, gases expands to fill its entire container. The particles of
a gas are so far apart that the attractive forces between them are assumed
to be negligible. The particles are viewed as independent from each other,
meaning that the gas is the opposite of a solid and has neither a fixed size nor
shape.
The velocity of each particle in matter determines its kinetic energy. There
is an exchange or transfer of energy between particles both atoms and
molecules during collisions between them.
Fig 6.4 shows movement of particles in matter.
Solid Liquid Gas

Figure 6.4. Movement of particles in matter
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Based on these assumptions, the kinetic theory states that matter is composed of a
large number of small particles that are in constant motion.
The theory helps to explain:
1. The flow or transfer of heat and the relationship between pressure,
temperature and volume properties of gases.
2. Why substances change state under certain conditions .e.g. solid to liquid
to gas and vice versa. The change of state occurs when energy is added to or
taken away from matter usually in the form of heat.
Exercise 6.3
Select the most appropriate answer from the choices given.

1. Which one of the following is an assumption of the kinetic theory of matter?
A. Gases are the same in all matter.
B. Matter is the same as kinetic energy.
C. Matter consists of tiny particles.
D. None is correct.
2. What happens when particles collide?
A. Molecules break into atoms.
B. They exchange or transfer kinetic energy.
C. They change the state of the material.
D. None is correct.
3. What determines how close molecules are to each other?
A. The diameter of the molecules.
B. There is no way to tell.
C. The state of phase of the material.
D. The diameter of the container holding them.
4. What leads to a change of state in matter?
A. A change in the number of particles.
B. A change in the amount of internal energy.
C. A change in the direction of particle movement.
D. A change in the shape of the container.
5. Matter consists of tiny particles termed as ____________?
A. Matter B. Atoms
C. Ions D. Elements
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6. _____________ is the reverse of evaporation.

A. Boiling B. Freezing
C. Melting D. Condensation
7. A temperature of 0 K is known as __________.
A. The freezing point of air B. 0°c
C. Absolute zero D. The melting point of water
8. The temperature of a substance __________ as its heat of fusion is supplied
to melt it.
A. Remains at 0 K B. Remains constant
C. Decreases D. Increases rapidly
9. Which of the following is vapour?
A Oxygen B Hydrogen
C Steam D Helium
10. The constant, random motion of tiny chunks of matter is called __________.
A linear motion C Brownian motion
B Kinetic motion D Parabolic trajectory
6.7 Applications of the principle of thermodynamics

6.7.1 Functioning of refrigerators
Activity 6.9 To show the basic working principle of a refrigerator

Materials
• Tomato • Ether.
Steps
1. Feel the temperature of tomato with the back of your palm.
2. Pour a little ether on the tomato and once again feel its temperature.
3. Compare the temperatures of the tomato in steps 1 and 2. Which one is
high?
4. Suggest a reason for your answer in step 3.
5. Basing on the reason you have given in Step 4, explain to your classmates
how the refrigerator operates.
6. Report your findings to the whole class with assistance of your teacher, make
short notes from the discussion.
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When a volatile liquid is poured onto a surface, it quickly evaporates by absorbing

heat from the surface, due to the temperature difference between the two. As a
result, the surface is cooled. This is why we say evaporation causes cooling. This
simple fact is applied in the working of a refrigerator. Let us now discuss its
working in details.
Fig. 6.5(a) shows a photograph of a commercial refrigerator. Fig 6.5 (b) shows
the main parts of a refrigerator.
The main parts of the refrigerator are the; the compressor, condenser, evaporator
and the expansion valve. The compressor and the condenser are outside the
refrigerator while the evaporating coil and the expansion valve are inside it. The
refrigerant, i.e. the liquid used in the refrigerator’s circulatory system is Freon
(dichloro-diflurol- methane) which is non-poisonous and odourless.
Fig. 6.5 shows the refrigerator and its parts.
Freezing compartment
Expansion valve
Evaporating
coil
Liquefied refrigerant
Food compartment
Vaporized refrigerant
Valves
Motor Fan
To mains Compressor Condensing

power supply coil
Figure 6.5(a) A picture of refrigerator (b) Picture of parts of a refrigerator
The refrigerants vapour is drawn in by the compressor. It is then compressed and

passed into the condensing coil, which is air cooled by a fan. The gas gets liquefied,
losing its latent heat of vaporization. This liquid refrigerant is made to enter the
evaporating coil, where the pressure is relatively low. The liquid at once evaporates
at a lower pressure and the latent heat of vaporization required for the change
of state is taken from the air in the freezing compartment and other food items
stored in it. As a result of this, the temperature of the air in the refrigerator cabinet
184
is lowered to about 7 °C, an ideal temperature for storing most food items. The
temperature quite close to the evaporating coil falls below the temperature of the
ice due to large surface area of the coil. The gas goes back into the compressor,
gets liquefied and the cycle repeats itself.
The refrigerator cabinet is well insulated so that no heat is conducted from
outside into it. The temperature inside the cabinet is controlled by a thermostat
(a bimetallic strip). When the temperature rises a few degrees above 7 °C, the
thermostat makes electrical contact and starts the motor. When the temperature
inside falls by the desired amount, the circuit is broken by the thermostat and
the motor stops working.
6.8 Melting and solidification
6.8.1 Melting
Activity 6.10 To show the heating curve of ice
Materials:
• Thermometer • Glass beaker • Crushed ice
• Tripod stand • Bunsen burner and wire gauze.
Steps
1. Take pure crushed ice at about -10 °C and
put it in a beaker placed on wire gauze on a
tripod stand as shown in Fig. 6.6.
2. Note the initial temperature of the ice.
3. Light a Bunsen burner and adjust the blue
flame to a small low temperature.
4. Note the temperature of ice at 30 seconds
interval until the temperature of the container
is about 10 °C.
5. Record your results as shown in Table 6.1
6. What happens to the amount of ice as heating
continues? Fig. 6.6: Set up to show heating
7. Plot a graph of temperature against time of ice
8. Explain the shape of the graph.
Table 6.1
Time (s) 0 30 60 90 120

Temperature (°C)
185
Melting is the change of state from a solid to a liquid. Melting of a pure substance
occurs at a particular constant temperature called melting point.
Fig 6.7 shows a graph of temperature against time for ice.
10 D
Water from melted ice
Temperature (°C)
Melting ice
0 Time (s)
B C
Solid ice
–
10 A
Fig. 6.7: Melting process of ice
The graph shows that

1. The temperature of ice rises steadily from -10°C to 0°C. During this time,
along AB, the ice remains as solid.
2. At 0°C, along the line BC, the temperature remains constant for a period of
time. During this period, the ice is melting. During the melting process, solid
and liquid exist in equilibrium.
Step by step process of what happens during melting:

(i) Heat energy is absorbed by the solid particles.
(ii) Heat energy is converted to kinetic energy.
(iii) The kinetic energy of the particles increases and the particles in the
solid vibrate faster.
(iv) At melting point, the particles have gained enough energy to overcome
the attractive forces between particles.
(v) Particles starts to move away from their fixed position.
(vi) Liquid is formed.
(vii) The cause for constant temperature during melting: The absorbed heat
energy during melting is used to weaken the attractive forces between
particles and not the kinetic energy of the particles.
3. After all the ice has melted, the temperature of water starts rising again as
seen along the line CD of the graph.
4. The line AB of the graph is steeper than the line CD of the graph showing
that the time taken to raise the temperature of ice by 10 °C is less than the
time taken to raise the temperature of water by 10 °C. it is easier to raise the
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temperature of ice than water. This means that the specific heat capacity of
ice is less than that of water.
Activity 6.10 shows that when a substance is changing its state from solid to
liquid, heat is required. Thermal energy absorbed during the melting process is
called latent heat. There is no change of temperature, as shown in the part BC of
the graph, until all the ice has melted. Conversely if water at 0 °C freezes to ice
at 0 °C, it must give out the same heat energy.
If pressure remains constant, a solid substance melts or freezes at a specific
temperature. The melting point of ice is 0°C under standard atmospheric pressure.
6.8.2 Solidification
Fig 6.8 Shows the cooling curve for water.
temperature (0C)
A
liquid
liquid + solid
00 C B C
time(s)
solid(ice)
D
Figure 6.8. A cooling curve for water
The graph shows that:

1. The line AB represents water still in liquid state where molecules are free to
move in random directions, colliding with each other and the walls of the
container. The temperature of water falls steadily between points A and B.
2. At 0 °C, along the line BC, the temperature remains constant for a certain
period of time. During this time, there is a mixture of a liquid and solidi.e.
water is observed to freeze (solidify). At C all the water has completely
(frozen) changed to ice.
3. After all the water has solidified, the temperature of ice starts reducing again
as seen along the line CD of the graph.
Note: Activities with other substances show a similar behaviour. This

shows that kinetic theory can be applied to all substances.
Solidification is the change of state from a liquid to a solid. It is also known as
187
freezing. A pure substance freezes at a temperature equal to its melting point.

Freezing of a substance occurs at a particular constant temperature called the
freezing point.
6.8.3 Factors that affect melting/freezing point

(a) Pressure
Under standard atmospheric pressure, a pure substance always melts at a definite
temperature.The melting point however changes with the change in pressure acting
upon a substance. The change in the melting point due to change in pressure is
however not large.
Activity 6.11 To show the effect of pressure on the melting point of

ice.
Materials:
• Wire • 2 Equal weight • Block of ice
• 2 cement blocks • Soft pad
Steps
1. Rest a large block of ice at 0 ºC on two stools or two cement blocks.
2. Hang a thin copper wire around the block and attach two equal heavy
weights to the ends of the wire as shown in Fig. 6.9. Observe and record
what happens to the wire and the block of ice as time progresses.
Ice
Wire block
Cement
blocks
Weights
Fig. 6.9: Effect of pressure on melting point of ice
The wire completely cuts through the block of ice and the weights fall to the soft
pad on the floor. Is the ice in one piece or two pieces?
In the above experiment, it is observed that initially as the pressure of the wire
on the ice increases, the melting point of ice decreases and so the ice melts. The
water flows above the wire. The latent heat of fusion required for the melting of
ice comes from the copper wire. The water above the wire is no longer under
pressure. As the pressure is released, the water which is at a temperature below
zero freezes again binding the two pieces of ice together. During freezing heat
188
is given out by water and this heat is conducted down through the copper wire.
This provides heat for further melting of the ice under the wire. At some point,
the wire cuts right through the block of ice and falls to the floor, leaving ice still
in a solid block.
This phenomenon in which ice melts when pressure is increased and again solidifies
(freezes) when the pressure is reduced is called regelation (re-again: gelare-freeze)
Ice contracts on melting. An increase in pressure would help it in its contraction
and hence we should expect a decrease in the melting point of ice as pressure on
its surface is increased. The melting point of ice decreases with the increase in pressure.
For substances like wax, gold, silver etc. which expand on melting, an increase in
pressure would make its expansion difficult. These substance have to be heated
more in order to melt. As a result, we should expect an increase in the melting
point, as pressure is increased. For such substances, the melting point increases with
the increase in pressure.
Impurities
Experiments show that impurities decrease the melting point of a substance. Though
pure water freezes at 0 ºC, salty water remain as water even at –1 ºC. The extent
to which the freezing point is lowered depends on the concentration of impurities
dissolved into the liquid. For example when salt is added to ice, its melting point
is reduced to a value as low as –10 ºC. This method is used to defreeze roads in
cold countries during winter. Antifreeze material is added to the water in the car
radiators to stop water from freezing when temperature falls below 0 ºC.
Exercise 6.4
For questions 1 - 3, select the appropriate response from the choices

given.
1. Freon group of refrigerants are
A. Inflammable B. Toxic
C. Non-inflammable and toxic D. Nontoxic and non-inflammable
2. In a refrigeration system, the expansion device is connected between the
A. Compressor and condenser B. Condenser and receiver
C. Receiver and evaporator D. Evaporator and compressor
3. The bank of tubes at the back of domestic refrigerator is
A. Condenser tubes B. Evaporator tubes
C. Refrigerant cooling tubes D. Capillary tubes.
189
4. Explain the functions following parts in a refrigerator:

(a) Compressor (b) Condensing coil
(c) Evaporating coil (d) Thermostat
5. Why is it not advisable to put a refrigerator on a wall or near the wall?
6. Explain the following processes:
(a) Melting (b) Solidification
7. Using a sketch graph describe the process of heating a solid ice till it becomes
water vapour.
8. Use the correct words from the following to answer questions that follows:
decreases, increases, heat, specific heat, insulator, condensation, entropy, internal
energy.
(a) is the process of changing from gas to liquid.
(b)
_____________ is the measure of the average kinetic energy of the
particles in a material.
(c)
When an object absorbs heat energy, it entropy ____________.
(d)
_____________ is the name for energy that is transferred only from a
higher temperature to a lower temperature.
(e) The measure of the amount of disorder in a system is called
_____________?

• Thermodynamics is the study of systems involving exchange of energy in the
form of heat and work.
• An open system is a system that exchange both energy and matter with its
surroundings.
• A closed system is a system that can exchange only energy with its
surroundings, not matter.
• An isolated system is a system that cannot exchange either matter or energy
with its surroundings.
• The internal energy of a system is identified with the random, disordered
motion of molecules.
• The first law of thermodynamics states that the change in the internal energy
(∆U) of a system is equal to the sum of the heat (Q) that flows across its
boundaries and the work (W) done on the system by the surroundings.
190
∆U = Q + W
• Entropy is the measure of a system’s thermal energy per unit temperature
that is not available to do useful work. Entropy is also a measure of the
molecular disorder, or randomness of a system.
• The Second law of thermodynamic states that the spontaneous change for
an irreversible process in an isolated system always proceeds in the direction
of increasing entropy. In other words, the entropy of any isolated system
always increases.
OR
‘Heat flows spontaneously from a hotter object to a colder one, but not in
the opposite direction; the reverse cannot happen without the addition of
energy’.
• Melting is the change of state from a solid to a liquid. Melting of a pure

substance occurs at a particular constant temperature called melting point.
• Solidification is the change of state from a liquid to a solid it is also known as
freezing. A pure substance freezes at a temperature equal to its melting point
Unit Test 6
For questions 1 - 9, select the appropriate response from the choices given.
1. The laws of thermodynamics dictate the specifics for the movement of

_____________ and ______________ within and across systems.
A. energy, motion B. heat, work
C. light, heat D. none of the above
2. The mathematical representation to the first law of thermodynamics is ____.
A. W + Q = U B. Q = U + W
C. U = Q – W D. none of the above
3. The first law of thermodynamics states that the total energy output is equal
to the amount of heat supplied in a system. True or False?
A. True B. False
4. Which of the following best illustrates the Second Law of Thermodynamics?
A. A cold frying pan heats up faster when placed in the microwave rather
than the stove
191
B. A hot frying pan cools down when it is taken off the kitchen stove.
5. The second law of thermodynamics states that energy _____________ and
disperses rather than staying concentrated.
A. spreads out B Stays
6. ______ is the change from liquid to solid.
A. boiling B. melting
C. evaporation D. freezing
7. ''Absolute zero'' is the bottom point on the ____________ temperature scale.
A. Celsius B. Kelvin
C. Fahrenheit D. Zero
8. The following processes happen at the boiling point except ______.
A. change in density B. change in state
C. gain in heat energy D. change in temperature
9. Which of the following is the correct statement of the second law of
thermodynamics?
A. There is a definite amount of mechanical energy, which can be obtained
from a given quantity of heat energy.
B. It is impossible to transfer heat from a body at a lower temperature to a
higher temperature, without the aid of an external source.
C. It is impossible to construct an engine working on a cyclic process,
whose sole purpose is to convert heat energy into work.
D. All of the above.
10. State the first law of thermodynamics.
11. Describe the process of cooling a gas till it forms a solid.
12. A gas is enclosed in a metallic tube whose lid is a piston. Describe two ways
of increasing the internal energy of the gas.
13. Entropy is defined as the randomness of a system. Give two examples to
illustrate this.
14. Explain the term thermal equilibrium.
15. Distinguish between:
(a) Open and closed system.
(b) Reversible and irreversible process. Give examples of each.
192
16. In a certain process, 600 J of work is done on the system which gives off
250 J of heat. What is the change in internal energy for the system?
17. How much work does a heat engine do if it takes in 2 500 J of heat and expels
1 500 J?
18. A hot solid of mass 100 g at 100° C is quickly transferred into 100 g of water
in a container of mass 200 g at 20 °C. Calculate the resulting temperature
of the mixture. (Specific heat capacity of the solid and the container is
400 J/kg K.
19. A gas enclosed in a cylinder occupies 0.030 m3. It is compressed under a
constant pressure of 3.5 × 105 Pa until its final volume is exactly one-third
of its initial volume.
(a) What was the change in the gas volume? __________________
(b) How much work was done? _______________ ______________
(c) The gas lost 5.0 × 103 J as heat during the compression process. Did the
internal energy of the gas increase or decrease? By how much?________
20. A steel marble at room temperature was placed in a plastic-foam cup
containing ice and water at 0 °C. After thermal equilibrium was reached, the
temperature of the ice-water mixture and marble was 1 °C.
(a) Which object lost heat energy?
_______________________________________________
(b) Was any work done on the marble or by the ice?
_______________________________________________
(c) Did the internal energy of the marble increase or decrease? What was a
measurable effect of this change?
_______________________________________________
(d) Did the internal energy of the water-ice mixture increase or decrease?
How can this be observed?
_________________________________________________
(e) Did the internal energy of the system consisting of the water-ice mixture
and the marble increase or decrease?
_________________________________________________
193
Introduction to Electromagnetic
UNIT 7 Induction
Key Unit Competence
By the end of this unit the learner should be able to apply the principle of
electromagnetic induction.
Learning objectives

• State the Laws of electromagnetic induction.
• Recall the expression of induced EMF.
• Explain electromagnetic induction.
• Define the term alternating current.
• Explain EMF induced in a coil rotating in uniform magnetic field.
• Explain the functioning of alternating current generator.
• Explain the variation of induced EMF with change of generator frequency.
• Explain the functioning of a transformer
• Relate peak and RMS values.
Skills
• Explain the expression of induced EMF in a straight conductor moving in magnetic
field.
• Analyse induced EMF in a changing magnetic flux.
• Analyse applications of alternating current.
• Deduce an expression for induced EMF for a coil rotating in uniform magnetic field.
• Explain operation of basic alternating current generator.
• Evaluate variation in induced emf when generator frequency changes
• Relate peak and RMS values for sinusoidal currents.
Attitude and value

• Appreciate the application of electromagnetic induction.
• Adapt or acquire scientific techniques, reasoning and attitudes for analyzing theories
and expressions.
• Show the concern of possible risks involved in living and working near high- voltage
power lines.
194
ic
Introduction
Unit focus activity
Steps
Part 1
1. Physics student made the device shown in Fig. 7.1. What is the name of the
device?
Fig. 7.1: Picture of a device made by students
195
Introduction to Electromagnetic Induction
2. Discuss what you think the student observed when the wheel was rotated.
3. Briefly describes the working principle of the device.
Part 2
4. Study the picture in Fig. 7.2.
5. Have you seen the device in the picture
before?
Name it.
6. What is the function of the device?
7. Explain the working principle of the
device. Fig. 7.2: Picture of electronic device
8. Name the laws that govern the working
principle of the devices in Figures 7.1 and 7.2.
In 1819, Hans Oersted discovered that whenever an electric current flows through
a conductor, a magnetic field is produced around it. If such a conductor is placed
in a magnetic field, it experiences a magnetic force (motor effect). Can the reverse
effect hold true? That is, by moving a coil in a magnetic field, can an electric current
be generated? This question was answered by Michael Faraday and Joseph Henry
in the year 1831. The two scientists, although working independently, were able
to show that an e.m.f can be generated. This method of generating e.m.f is called
electromagnetic induction. The generated e.m.f is called induced e.m.f and hence
induced electric current flows in a ciruit. In this unit, we will discuss this concept
and apply it in making some devices.
7.1 Demonstrations of electromagnetic induction

The induced e.m.f can be produced in two ways.
(a) Relative movement (generator effect)

The following experiments will help us to understand electromagnetic induction.
Activity 7.1 To induce an electromotive force in a straight

conductor (wire)
Materials
• U-shaped magnet • Centre-zero galvanometer
• Copper wire (Conductor)
196
Steps
1. Connect a copper wire XY to a sensitive centre zero galvanometer (Fig. 7.3)
Centre zero galvanometer
N
U-shape magnet
X
Pull
Y
S
Fig. 7.3: A conductor in between the poles of a U-shaped magnet
2. Place a conductor in between the poles of a magnet as shown in Fig. 7.3 and
observe the galvanometer reading when the wire is stationary.
3. Pull the conductor horizontally away from the poles and stop. Observe and
explain what happens to the galvanometer pointer.
4. Re-introduce the wire in between the poles of the magnet and stop. Explain
what happens to the galvanometer pointer.
5. Repeat the experiment, keeping the wire stationary and moving the magnet.
Explain what happens to the galvanometer pointer.
6. Repeat the experiment by first moving the wire vertically up and down and
then repeat by moving the magnet. Explain what happens to the galvanometer
pointer.
Consider straight conductor XY connected to a galvanometer and placed in a
magnetic field. When the wire is stationary, the pointer does not move. When the
wire is being moved out, the pointer shows a deflection in one side (Fig. 7.4 (a)),
but returns to the zero position once the wire stops (Fig. 7.4 (b)).
G G G G
(a) (b) (c) (d)

Fig. 7.4: Deflection on a glavanometer
When the conductor is re-introduced in between the poles, the pointer deflects
but in the opposite direction (Fig. 7.4 (c)). However, when the conductor stops,
the pointer once again returns to the zero position (Fig. 7.4 (d)).
197
Similar effects are observed when the magnet is moved instead of the conductor.
However, no deflection is observed when the wire or the magnet is moved vertically
up or down.
From this activity, we can conclude that whenever there is relative motion between
a conductor and a magnet, an e.m.f is generated in the conductor. The generated
e.m.f. is called induced e.m.f. If the conductor forms a part of circuit in Fig.
7.3 where it is connected to a galvanometer, the e.m.f. produces a current. The
current produced is called induced current. This relative motion is such that the
wire ‘cuts’ the magnetic field lines of force (Fig. 7.5).
Fig. 7.5: A conductor cutting magnetic field lines of force
This phenomenon we are talking about is called electromagnetic induction.
Activity 7.2 To induce an electromotive force in a coil using a

magnet.
Materials
• Insulated copper wire • Bar magnet
Steps
1. Make a coil using an insulated copper wire.
2. Connect the ends of the coil to a sensitive centre-zero galvanometer. Write
down your observations.
3. Introduce a bar magnet into the coil and stop (Fig. 7.6).
N S Bar magnet
Coil
Galvanometer
Fig. 7.6: Moving a magnet into a coil
198
4. Withdraw the magnet from the coil and stop. Write down your observations.
5. Repeat the experiment by moving the coil and keeping the magnet stationary.
Observe and write down what happens to the pointer of the galvanometer in
each case.
6. Move both the coil and the magnet in the same direction at the same speed.
Observe and write down what happens to the pointer of the galvanometer?
7. Suggest explanations for all your observations in Steps 3 to 6.
When the magnet is introduced into the coil, the pointer of the galvanometer
shows a deflection in one side but returns to the zero position when the magnet
is brought to rest.
When the magnet is withdrawn from the coil, the pointer deflects but in the
opposite direction. However, when the magnet stops, the pointer once again
returns to the zero position.
Similar effects are observed when the coil moves instead of the magnet.
No deflection is observed when both the coil and the magnet are moved at the
same speed in the same direction.
From this activity, we can conclude that an electromotive force is induced whenever
there is relative motion between the coil and the magnet hence a current flows in the coil.
(b) Changing a magnetic field (transformer effect)
Activity 7.3 To induce an electromotive force in a coil using

another coil
Materials
• Insulated copper wire • Galvanometer
• Connecting wires • Source of current
Steps:
1. Make two coils using the insulated copper wire. Connect the coils as shown
in Fig. 7.7. Keep the coils close together.
Coil 1 Coil 2
Fig. 7.7: Electromagnetic induction using two coils

2. Close the switch in coil 2. What happens to the pointer of the galvanometer?
Explain.
3. Open the switch. Observe what happens to the galvanometer pointer. Explain
your observation.
199
When the switch is closed, the pointer of the galvanometer momentarily deflects
to one side but returns to the zero position when the switch is left closed. When
the switch is opened, the pointer momentarily deflects again but in the opposite
direction. However, when the switch is left open, the pointer once again returns
to the zero position.
From this activity, we can conclude that an e.m.f is induced in coil 1 the moment
the switch in coil 2 is closed or opened.
Electromagnetic induction is as a result of one of those actions-at-a-distance
effects. Michael Faraday explained this action using a model based on magnetic
field lines of force. He explained that an electromotive force is induced in a
conductor which is a part of a closed loop or circuit when there is a change in the
number of magnetic field lines (also known as the strength of magnetic field B)
passing through this loop or when the conductor ‘cuts’ the field lines.
Consider a coil moving away from a magnet as shown in Fig. 7.8.
Motion of the coil
1 2
Position A Position B
3
S N
5 4
0
6 0
Fig. 7.8: Electromagnetic induction using a magnetic and a coil
The number of magnetic field lines linking or threading the coil decreases from
6 to 4 as the coil is moved from position A to position B. We can also say that the
coil cuts two lines as it moves from position A to position B. Similar reasoning
may be applied when the magnet is moved away or towards the coil.
In case of switching on and off of the circuit, the fields builds up to a certain
strength and reduces to zero respectively as shown in Fig. 7.9.
Number of field lines linking Number of field lines linking the coil is 4
the coil is zero
Fig. 7.9: Building the magnetic field
200
7.2 Factors affecting the magnitude of emf induced

The magnitude of the induced electromotive force depends on a number of
factors. Before we discuss these factors, we recommend that you refresh your
knowledge of quantities and measurements associated with magnetism that we
have summrised in Appendix 1 at the end of this book.
The following experiments will help us to investigate what the magnitude of the
induced e.m.f depends on.
Activity 7.4 To investigate factors affecting the magnitude of the

induced e.m.f. in a coil
Materials:
• Insulted copper wire • Bar magnet

• Iron rod • Ceramic magnet
Steps
1. Make a coil of a few turns using the insulated copper wire. Connect the coil
to a centre-zero galvanometer.
2. Slowly introduce the magnet into the coil as shown in Fig. 7.10(a).
S N S S N N
0 0 0
(a) Magnet moving slowly (b) Magnet moving fast

Fig. 7.10: Moving a magnet into a coil at different speed
3. Repeat the activity by quickly introducing the magnet into the coil. Note and
explain what happens to the pointer of the galvanometer.
4. Repeat the activity but wind the coil around a soft iron rod as shown on
(Fig. 7.13(b)). What do you observe? Explain.
5. Repeat the activity using a coil with more turns. What do you observe?
Explain.
6. Repeat the activity but use a stronger magnet e.g. ceramic magnets. What do
you observe? Explain.
201
When the magnet is slowly introduced into a coil, the induced e.m.f (ε) is less
than when the magnet is moved quickly. Same effect will be observed when the
coil is moved slowly or quickly towards the magnet.
When a stronger magnet is used, the induced electromotive force increases, .
When a coil with more turns is used, the induced emf is found to increase. The
induced electromotive force is found to increase when a soft iron core is used.
Fig. 7.11 (a) shows one line of force linking the coils while in (b) there are three
lines of force.The soft iron core concentrates the flux lines onto the coil producing
a higher rate of change of flux when there is relative movement.
S N
(a) (b) (c)

Fig. 7.11: Inducing an e.m.f using a coil wound on a soft iron core
We can conclude that, the magnitude of the induced emf is directly proportional to:
• the strength of magnetic field i.e the stronger the magnet the higher the induced emf.
• the number of turns in the coil i.e. the more turns the higher the induced
emf.
• the induced emf is much higher in the presence of a soft iron core.
• Area of the coil.
• The rate (speed) of cutting the magnetic flux
7.3 Laws of electromagnetic induction
Activity 7.5 To state the laws of electromagnetic induction

Materials
• Magnet • Insulated copper wire
• Centre-zero-galvanometer • Connecting wires
Steps
1. Revisit activity 7.4, change the motion of the magnet in the coil; start slowly
then increase the speed. What do you observe on the pointer?
2. Now explain, how the rate of change of magnetic flux relates to the force induced.
3. Conduct a research from the internet and reference books on the laws of
electromagnetic induction..
202
7.3.1 Lenz’s law

When an electromotive force is induced in a circuit, a current flows in the circuit.
What is the direction of the induced current?
Consider a magnet moving towards or away from a coil as shown in Fig. 7.12
(a) and (b).
S S
N N
N S
0 0
(a) (b)
Fig. 7.12: Direction of the induced electromotive force
When a north pole is moved towards a coil, the current flows in such a way as to
oppose the introduction of the north pole. A north pole (N) is therefore induced
at the top end of the coil to repel the incoming north pole of the magnet. Similarly
a south pole is induced at the top end of the coil to resist the withdrawal of north
pole of the magnet.
The direction of the induced current can be determined using a law that was
developed by a German scientist called Lenz.
Lenz’s law states that the direction of the induced current is such that it opposes the
change producing it.
7.3.2 Fleming’s right hand rule

In case of straight conductors moving in a magnetic field, Fleming’s right hand
rule gives the relationships between the directions of the field motion and induced
current.
It states that,
If the thumb and first two fingers of the right hand are mutually perpendicular to each
other, the First finger pointing in the direction of the Field and the thuMb in the direction
of Motion of the conductor, then the seCond finger points in the direction of the induced
Current see Fig. 7.13.
203
First Finger
i
e
l
SeCond d
u
r ThuMb
o
r First finger
t
e i Thumb
n o
t n
Second
finger
Fig. 7.13: Fleming’s right hand rule
7.3.3 Faraday’s law
7.3.3.1 Definition of terms

Magnetic flux (Φ): is a measure of the quantity of magnetism, being the total
number of magnetic lines of force passing through a specified area in a magnetic
field. It represents the strength of the magnetic field over a given area. The SI unit is
weber (symbol Wb).
Magnetic flux density (B) is the amount of magnetic flux through a unit area taken
perpendicular to the direction of the magnetic flux, It is a vector quantity. Its SI unit is
tesla (symbol T), or N/(A·m) expressed in SI base unit.
Magnetic flux through a close wire loop is the product of the magnetic flux density
and the area (A) of the loop.
Let us consider three cases of orientation of the place of the loop (A) in relation
to the direction of the magnetic field (B) .
Let:
Case 1
Φ=0
The loop of coil perpendicular to the B
magnetic field (Fig. 7.14) Normal A
to the loop
i.e B and A (normal to loop) are parallel A
hence the angle between B and A is 00
Fig. 7.14
Φ1 = B × A cos 0 =BA 0
204
Case 2
The loop of the coil inclined at an angle
e
to the magnetic field (Fig. 7.15) o th
m al t
r
No op A
i.e Angle between B and A is θ lo
B
Φ2 = B × A cosθ A
Fig. 7.15
Case 3
Normal to the
The loop of the coil parallel to the A
loop
magnetic field (Fig. 7.16)
i.e B and normal A are at 900 Φ= 90 0
Φ3 = B × A cos 900 = 0 A
Fig. 7.16
7.3.3. Statement of Farady's law

The factors affecting the magnitude of the induced electromotive force were
summarised by Michael Faraday in what is known as Faraday’s law of
electromagnetic induction.
The law states that; the electromotive force induced in a conductor is directly proportional
to the rate at of change of the magnetic flux linked to the conductor
ε ∝ – ΔΦ ⇒ ε = – ΔΦ (since the constant k = 1)
Δt Δt
In case the conductor has N turns,
ε = – NΔΦ
Δt
Example 7.1
A small piece of metal wire is dragged across the gap between the pole pieces
of a magnet in 0.5 second. The magnetic flux between the pole pieces is
known to be 8 × 10-4 Wb. Calculate the emf induced in the wire.
205
Solution
The magnetic flux threading the coil has changed from 0 to 8 × 10-4 Wb in 0.5 s
ΔΦ 8 × 10-4
ε = Δt = 0.5 V = 1.6 × 10-3 V = 1.6 mV
Example 7.2
Calculate the emf induced in a 250 turn coil with a cross-section of 0.18 m2 if
the magnetic field through the coil changes from 0.10 Wb m-3 to 0.60 Wb m-3 at
a uniform rate over a period of 0.02 second.
Solution
E = ?, N = 250, A = 0.18 m2, B1 = 0.10 Wb m-1, B2 = 0.60 Wb m-2, Δt = 0.02 s
The magnetic flux tis changing due to change in B since A is constant.
Rate of change of magnetic flux,
ΔΦ NA ( B2 – B1) 250 × 0.18(0.60 – 0.10) V

Δt = Δt = 0.02
= 1 125 V
ΔΦ
E = Δt = - 1125 V
Exercise 7. 1
1. Describe an experiment to illustrate electromagnetic induction.

2. State the two laws of electromagnetic induction.
3. Describe an experiment to demonstrate Faraday’s law of electromagnetic
induction.
4. A long bar magnet is pushed into a coil with many turns as shown in
Fig. 7.17.
(a) What happens to the needle of the centre-zero-galvanometer when the
magnet;
S N
(i) slowly enters the coil,
(ii) remains at rest inside the coil, 0
(iii) is rapidly withdrawn.

(b) Explain Fig. 7.17: Bar magnet pushed into the coil
206
(i) why the magnet has to be long.

(ii) why the coil has to have many turns.
5. Explain the following terms:
(a) Electromagnetic induction. (b) Induced electromotive force.
(c) Induced current.
6. Explain how to determine the direction of the induced current in a coil,
when a magnet moves into the coil.
7. Describe energy conversions during the process of electromagnetic induction
in:
(a) A coil using a bar magnet.
(b) A coil using another coil carrying a current.
8. Two coils are placed near each other as shown in 0
Fig. 7.18. G
(a) What happens to the needle of the galvanometer

(i) On closing the switch?
(ii) If the switch is kept closed?
(iii) On opening the switch?
(b) Give three possible ways of increasing the
deflection in the galvanometer
Fig. 7.18: Two coil near each other
9. A wire cuts across a flux of 0.2 × 10-2 Wb in 0.12 second. What is the
magnitude of emf induced in the wire?
10. A conducting circular loop having a radius of 5.0 cm, is placed perpendicular
to a magnetic field of 0.50 T. It is removed from the field in 0.50 s. Find the
average emf produced in the loop during this time.
11. A coil of area 0.15 m2 and 100 turns is placed perpendicular to a magnetic
field. The field changes from 5 × 10-3 Wb m-2 to 2 × 10-3 Wb m-2 in a time
interval of 30 ms. Calculate the emf induced in the coil.
7.4 E.m.f induced in a straight conductor moving in a

straight field
Activity 7.6 To determine the magnitude of induced e.m.f in a
straight conductor
Materials
• Permanent magnets • Conductor
207
Steps
1. Slowly move the conductor to or away from the magnet observe what
happens to the galvanometer.
2. Repeat the experiment but move the
conductor quickly to or away from the
magnet.
3. Slowly move the magnet away or N
B
towards conductor. Observe what
happens to the galvanometer.
V
G
4. Repeat Step 3 by uniformly moving
the magnet instead of moving the
S
conductor.
5. Reduce the length of the conductor

in the magnetic field, and repeat the Fig. 7.19: A conductor cutting
experiment. What do you observe magnetic field lines of force
6. Repeat step 1 but use a strong magnetic

field. What do you observe? Comment.
7. Repeat the activity but tilt the conductor such that it makes angle Φ to that
field.
Consider a straight conductor of length l being moved through a magnetic field of

flux density B with a velocity v (Fig. 7.20). Since the conductor cuts the magnetic
field, an e.m.f. is induced into it.
Fig. 20: Straight conductor moving in magnetic field
208
(a) When the conductor is moving perpendicular to the magnetic field, it cuts the
magnetic field at right angles, It sweeps through an area A = vl every second in
the magnetic (see the vertical plane (a) in Fig 7.20). In other words, its rate of
ΔA
change of area with time is Δt = lv
ΔΦ ΔBA ΔA
From Faraday’s law, ε = Δt = Δt =B× Δt = Blv
In short, the induced emf in the conductor ε =Blv
(b) When the conductor cuts through the magnetic flux at an angle θ (see the
slanting plane (b) in Fig 7.20), where θ is the angle between the magnetic field
and the direction of motion, the induced emf is becomes
ε = Blvsinθ
If the conductor has N turns, then the induced emf is given by
ε =NBlvsinθ
From this equation, we can see that maximum e.m.f is induced when the conductor
moves at right angles to the field. ( i.e θ = 90o hence sin θ = sin 90o = 1).
Example 7.3
If a 8 m long metallic bar moves in a direction to the magnetic field with a
speed of 5 m s-1, 20 V emf is induced in it. Find the value of magnetic field
intensity.
Solution
ε 20
ε = Blvsinθ or B = or B = T = lT
lvsinθ 8 × 5 × sin30o

Example 7.4
A wire of length 80 cm moves with a speed 20 m s-1 perpendicular to a
magnetic field of induction 0.2 Wbm-2. Calculate the magnitude emf induced
in the wire.
Solution
l = 80 × 10-2 m, v = 20 ms-1, B =0.2 Wb m-2

ε = Blv = 0.2 × 80 × 10-2 × 20 V = 3.2 V
209
7.5 E.m.f induced in a coil rotating in uniform magnetic field
Activity 7.7 To investigate the emf produced in a rotating coil in

a uniform magnetic field
Materials
• A coil • Centre-zero galvanometer
• Two-permanent magnets • Connecting wire
Steps
1. Connect a centre-zero galvanometer to the coil using connecting wire to
form a closed loop.
2. Place two opposite sides (N-S) of the permanent magnet close to each other.
3. Move the coil in between the magnet, first vertically then horizontally. Observe
the deflection of the pointer each time. Comment on your observations.
4. Now move the coil diagonally and observe the changes (if any) on the pointer
of galvanometer. Comment on your observation.
5. Deduce how the area of the coil and magnetic field are related.
Consider a rectangular coil abcd rotating at linear velocity v m/s in a uniform

magnetic field B (Fig. 7.21).
v
a θ v = vsinθ
b
N Axis B S
c
l
v d
Fig. 7.21: A coil rotating in a magnetic field

Since it keeps on changing position during the rotation, the component of v that
is perpendicular to the magnetic field B at any position is v sinθ.
From Faraday’s law, the emf induced in a straight conductor at any position inside
a magnetic field is given by
ε = -Blvsinθ
Therefore, in the side ab of the coil, an emf of ε = -Blvsinθ is induced.
Similarly, in the side cd of the coil, an emf of ε = -Blvsinθ is induced, in a direction
opposite to that in ab.
210
Therefore, the total emf induced in the coil is ε = -2Blvsinθ.

From this expression, we can see that
(a) Maximum emf is induced in the coil when the coil is horizontal, i.e, when the
angle between v and B is 90o (since sin 90o =1). See Fig. 7.22 (a).
(b) Minimum emf is induced in the coil when the coil is vertical, i.e, when the
angle between v and B is 0o (since sin 0o =0). See Fig. 7.22 (b).
(a) (b)
ω
ω
Fig. 7.22: Horizontal and vertical positions of the rotating coil
We can express the induced emf in the coil i.e ε = -2Blvsinθ in terms of angular
velocity (ω)of the coil as follows.
From our knowledge of circular motion ( see the appendix at the end of book).
angular dispalcement (θ)
Angular velocity of the coil ω = time taken (t)
In symbols ω = θ ⇒ θ = ωt
t
angular velocity (ω)
In addition, linear velocity v =
radius of cirlcular path of rotation (r)
In symbols v = ωr But from the coil dimensions, r = ad
2
⇒ v = ωr = ωad
2
The length l of the coil in this case is l = ab
Substituting for l, v and θ in the expression ε = -2Blvsinθ we get
ε =2Blvsinθ = 2Bab ωad sin ωt
2
Since ab×ad = A (area of coil), the equation simplifies to
ε = -BAωsin ωt
If the coil has N number or turns the induced emf if is given by
ε = -NBAωsin ωt
Remember the negative sign is in accordance to lenz’s law.
Since ε0 = -NBAω (the peak or maximum value of induced emf)
Then we can write the general expression of induced emf at any position as
ε = ε0sin ωt
211
The graph of induced emf against

ε
time is a sine wave, shown in the
Figure 7.23. +ε0 Sint
ε(V)
Time(s)
–ε0
Fig. 7.23: Graph of emf induced against time
The current also follows producing a graph of the same shape as shown in fig.
7.24.
I 1 Cycle
Sinωt
Current (s)
Time(s)
–I0
Fig. 7.24: Graph of current induced against time
I = I0sinωt
The emf (hence current) is therefore an alternating one (a.c) which varies
sinusoidally with time i.e reverses direction at regular intervals, having a magnitude
that varies continuously in sinusoidal manner.
The emf is periodic i.e it continually repeats the same pattern in time with period,
T = 1
f
Therefore since ω = 2πf, where f is the frequency of the rotation of the coil the
induced emf is also given by ε = ε0 sin (2πf).
Example 7.5
A generator with a circular coil of 80 turns of area 3.0 × 10-2 m2 is immersed
in a 0.20 T magnetic field and rotated with a frequency of 50 Hz. Find the
maximum emf which is produced during a cycle.
Solution
The maximum emf for a generator is
ε =NBAω
212
We Know
N = 80 A = 3.0 × 10-2 m2, B = 0.20 T and f = 50 Hz
Since ω =2πf = 2π(50) = 314 radians/s
ε0 = (80)(3.0 x 10-2)(0.20)(314) = 150.72 V
Exercise 7.2
1. Write down an expression for the emf induced between the ends of a rod of
length, l moving with velocity v so as to cut a magnetic field of flux density
B normally.
2. A straight wire of length 60 cm and resistance 8 Ω moves sideways with a

velocity of 20 m/s at right angles to a uniform magnetic field of flux density
2.0 × 10-3 T. Find the current that would flow if it ends were connected by
leads of negligible resistance.
3. A rectangular coil 35.0 cm long and 20.0 cm wide of 30 turns rotates at

uniform rate of 3000 rev/min about an axis parallel to its long side to a
uniform magnetic field of flux density 5.00 × 10-8 T.
Determine
(a) The frequency of induced emf
(b) The maximum value of induced emf
4. The emf induced by an a.c generator is given by ε = ε0 sinωt

(a) Explain the meaning of each symbol used.
(b) Give the units for the symbols used.
(c) Draw diagram showing the position of the coil and magnetic find the
(i) t = 0
(ii) ε = ε0
5. (a) The Fig. 7.25 shows the instantaneous position of a rotating loop of
wire between two bar magnets. The loop is rotating clockwise when
viewed from P. The magnets and the loop all lie in the same plane. Copy
and indicate the direction of induced current flow in the loop.
213
Axis of rotation
Axis of rotation
N S
P
Fig. 7.25: Rotating loop of wire between magnets
(b) The electromagnetic induction can be summarized by two laws namely:
Faraday’s and Lenz’s laws of electromagnetic induction. State the laws.
6. A wire of length 0.1 m moves with a speed of 10 m s-1 perpendicular to a
magnetic field of induction 1 Wb m-2. Calculate the induced emf.
7. A straight conductor 1 m long moves at right angles to both its length and
a uniform magnetic field. If the speed of the conductor is 2 m s-1 and the
strength of the magnetic field is 1 T, find the value of induced emf in volt.
8. A horizontal straight wire 10 m long extending east and west is falling with a
speed of 5.0 m s-1 at right angles to the horizontal component of the earth's
magnetic field 0.30 × 10-4 Wb m-2. Calculate the emf in the wire.
9. An air plane with 20 m using spread in flying at 250 m s-1 straight south
parallel to the earth's surface. The earth's magnetic field has a horizontal
component of 2 × 10-5 Wb m-2 and the dip angle is 60o. Calculate the induced
emf between the plane tips.
10. A rectangular coil of 100 turns has dimensions of 10 cm by 15 cm. It rotates
about an axis through the midpoint of the short sides. The axis of rotation is
perpendicular to the direction of the magnetic field of strength 0.50 T, and
it is rotating at 600 rpm.
(a) When is the emf induced in the coil a maximum?
(b) Is the induced emf ever zero? If so, when?
(c) What is the maximum induced emf?
214
7.6 Alternating current (a.c) generator
Activity 7.8 To observe and describe the working of an a.c

generator
Materials
• Model of a.c generator • Manila paper
Steps
1. With the help of a model provided by your teacher, describe the working of
an a.c generator. Illustrate the working on a manila paper.
2. Present your description to the rest of the class during the class discussion.
7.6.1 Structure of a simple a.c generator

Fig. 7.26 shows a diagram of a simple a.c generator.
Annular magnet
N S
Coil
Carbon brush Slip rings
Resistor Axis
Springs
Fig. 7.26: A simple ac generator
The generator consists of a rectangular coil of conductor wire whose ends are
connected to two slip rings. The slip rings make contact with carbon brushes
which connect them to an external circuit. Two light springs are used to make the
carbon brush press lightly on the slip rings thus making a good contact between
the carbon brushes and the slip rings. The coil is placed in between two poles
of a permanent magnet. The poles are annular in shape so as to concentrate the
magnetic field lines on the coil.
7.6.2 Working of simple a.c generator

When the coil is rotated about its axis, an electromotive force is induced depending
on the position of the coil. Let us start with the coil in a vertical position as shown
215
in Fig. 7.27 (a). In this vertical position the wires XY and WZ are moving along
the magnetic field lines. The wires are therefore not cutting the magnetic field lines
resulting in no electromotive force being produced (Fig. 7.27(a)). As the coil is
rotated from this position, it starts to cut across the magnetic field lines of force
and an electromotive force is induced (Fig. 7.27 (b)). During the first quarter of
rotation, the induced e.m.f increases from zero to a maximum value (peak value)
when the coil becomes horizontal (Fig. 7.27(c)). During the second quarter of
rotation the induced electromotive force reduces and reaches zero again when
the coil is in vertical position (Fig. 7.27(d)).
X X
W W
Y Y
N S N S
Z Z
0 0
(a) Coil vertical zero electromotive force (b) Coil starts to rotate
W
W X
X
Z
N S S
Z Y N
0
0
(c) Coil horizontal electromotive force (d) Coil vertical zero electromotive force
is maximum
Fig. 7.27: Working of a simple generator
The induced e.m.f sets up a potential difference between the ends of the coil which are
connected to the two slip rings mounted on the axle on which the coil rotates. This
potential difference drives the current in the external circuit. This process is repeated
in the third and fourth quarter of rotation. However, the direction of the current in the
coil changes. The direction of the induced current can be determined by Fleming’s
right hand rule. In the first half of rotation the side XY is moving down. The current
therefore flows from X to Y and from Z to W. During the second half of rotation,
XY is moving up and so the current flows from Y to X and from W to Z (Fig. 7.28).
216
X Y W Z
W Z X Y
First half rotation Second half rotation
Fig. 7.28: Set-up showing the rotation of coil
The current changes direction after every half a rotation (cycle). Fig. 7.29 shows
how the current in the external circuit changes with the position of the coil. The coil
produces an electromotive force that changes in a manner similar to a sine curve.
This shows that the cutting of the magnetic lines of force is greatest whenever the coil
passes through its horizontal position. In this position, the induced electromotive
force is maximum. The current flows in the circuit, first in one direction and then
in the opposite direction. A current that flows back and forth in a circuit is called
an alternating current. The number of cycles it completes in one second is known
as the frequency of the alternating current and is measured in hertz (Hz).
current
Or e.m.f
cycles current out of paper

current into the paper
Y Z Y
Z Z Y Y Z
Y Z
Fig. 7.29: Variation of current or e.m.f with coil positions
7.7 Root-mean-square (r.m.s) value

The value of an alternating current (and emf) varies periodically with time in
magnitude (size) and direction. The problem then becomes "which value should
we use to measure it?"
217
Activity 7.9 To deduce an expression for root mean square

current
Material
• An ammeter 1 A • d.c. power supply 3 V
• A resistor 10 Ω • A bulb 2.5 V, 0.3 A
• A switch which selects between two connections
• Mains power supply 2 V, 50 Hz.
Steps
1. Connect the circuit as shown in Fig. 7.30.
3V
2V 2.5V
50HZ 0.3A
A A
Fig. 7.30: Set-up to determine root-mean-square
2. First switch on the a.c. circuit and note the brightness of the lamp.
3. Secondly switch on the d.c. circuit.
4. Adjust the currents such that the brightness is the same as when the a.c.
circuit was switched on i.e. the bulb is fully lit
5. Record the value of the d.c. current that produces the same brightness as the
a.c. current.
Since the lamp is fully lit by a d.c current of 0.3 A, the equivalent value of a.c. is
0.3 A a value called root-mean-square value (r.m.s).
The r.m.s value of an alternating current is the steady direct current which
converts electrical energy to other forms of energy in a given resistor at the same
rate as the a.c.
Consider energy in d.c and a.c circuit
d.c a.c
Power I 2
R = mean value I2a.c × R
d.c
218
I d.c = mean value of I2a.c

= square root of the mean value of the square of a.c the current
Id.c = Ia.c = Ir.m.s
If the a.c. is sinusoidal, then
I = I0Sinωt Where, I0 = peak value of a.c
∴Ir.m.s = mean value I02 sin2ωt
= I0 mean value sin2ωt

The graph of sinωt and sin2ωt is as shown in Fig. 7.31
+1
1
2
sin2ωt
1
sinωt
2
–1
Fig. 7.31: The sinusoidal graph of sinωt and sin2ωt
1
The value of sin2ωt = 2
I
Therefore, Ir.m.s = I0 1 = 0
2 2
I0
Hence, Ir.m.s = = 0.707I0
2
The same relationship is used for e.m.f and p.ds, that is
εr.m.s = 0.707ε0
Vr.m.s =0.707 V0, where, ε0 = Peak emf

V0 = Peak voltage
In a.c circuit the r.m.s value is the one that is usually quoted. Thus a power source
of 240V is the r.m.s value.
219
Example 7.6
The power source of a house is 240 V. Find the peak value.
Solution
ε0 = 240 V
εr.m.s = 0.707ε0
ε 240 V
ε0 = 0.707
r.m.s
= = 339 V
0.707
Note: a.c voltmeters and ammeters are calibrated to read r.m.s value.
Exercise 7.3
1. An a.c circuit an r.m.s current of 7.0 A. The current travels through a 12
ohm resistor.
(a) Calculate the peak current.
(b) What is the power dissipated in the resistor?
(c) Find the peak voltage drop across the resistor.
2. In a circuit whose total resistance is 8.54 Ω. Find the r.m.s value of the
current if the rms voltage of the source is 110 V.
7.8 Other applications of electromagnetic induction
Activity 7.10 To establish the application of electromagnetic

induction
Materials
• Small external speakers
• Two stereo cables
• Two stereo cables cut half with coils of magnet wire on each end
• Small radio
Steps
1. Turn on the radio and connect the external speakers using the cables till the
sound is heard.
2. Cut off the cable and replace each end with the modified cable (coils of
magnetic wire on each end).
220
3. Slowly bring the coil close together. What do you hear? Explain
4. In your daily live, you have come across a microphone. Explain how it works.
5. Discuss how the idea of electromagnetic induction is used in transformers.
6. Conduct a research from the internet and reference books on the applications
of electromagnetic induction and report to the whole class.
7. With the help of your teacher make short notes from your research and
discussion
1. Induction coil
An induction coil consists of two coils (secondary and primary coils) with one
wound over the other around a soft iron core. The secondary coil has a greater
number of turns (Fig. 7.32).
Spark
Metal electrodes
Soft iron core
Primary coil Secondary coil

Contact points
Cam
dc power supply
Fig. 7.32: Induction coil
An induction coil works like a step-up transformer but with a d.c power supply.
The direct current in the primary coil is switched on and off by a rotating cam.
The current in the primary coil produces a changing magnetic field which in turn
induces an electromotive force in the secondary coil.
Due to the large number of turns in the secondary coil and the rapid change of
current in the primary coil, a large potential difference is induced between the
metal electrodes. This large potential difference causes a spark between the metal
electrodes. This spark may be used in many ways. For example, the spark produced
is used in igniting the petrol-air mixture inside a car’s engine.
2. Moving coil microphone

A moving coil microphone is a device for changing sound energy into electrical
energy. It consists of a diaphragm with a light coil connected to it.This coil is placed
in between two poles of a strong cylindrical pot magnet as shown on Fig. 7.33.
221
Cylindrical pot magnet
A small alternating
S
Sound waves current
N
Coil To amplifier
Diaphragm
Fig. 7.33: A moving coil microphone

When a person speaks in front of a microphone, the sound energy set the
diaphragm into vibration. This moves the coil back and forth between the poles of
the magnet. A small alternating current is induced in the coil.When this alternating
current is made larger (amplified), it operates a loudspeaker.
Exercise 7.4
1. Explain the working principle of a microphone.

2. Fig. 7.34 shows a conductor-carrying a current between the poles of a
magnet.
B
N
S
A
Fig. 7.34: Conductor-carrying current in a magnetic field
(a) If the wire AB is allowed to move, in which direction will it move?

(b) What is the effect of
(i) reversing the current?
(ii) reversing the magnetic field?
(iii) reversing the field and current at the same time?
3. Fig. 7.35 shows a solenoid with a current flowing through it.

N S
• •
battery
Fig. 7.35: A solenoid
222
Determine
(a) The direction flow of the current in the solenoid.
(b) The polarity of the battery.
4 Look at figure 7.36 illustrating a magnet being moved towards a coil
S N S N
0 0
(a) Magnet moving slowly (b) Magnet moving fast

Fig. 7.36: Moving a magnet into a coil at different speed
(a) As the current is induced in the coil, what type of pole is formed at the
left end of the coil? Give a reason for your answer.
(b) In which direction does the conventional current flow through the meter?

• If a conductor cuts magnetic field lines, an electromotive force is induced.
This process of producing electricity is called electromagnetic induction.
• The magnitude of the induced electromotive force is given by the Faraday’s
law of electromagnetic induction.
• The direction of the induced electromotive force is determined by Lenz’s law.
• The magnitude of the induced electromotive force is affected by the rate of
change of magnetic field, strength of the magnet, the number of turns of the
coil and the material used for the core.
• Fleming’s right hand rule is used to show the direction of the induced current
in a straight conductor cutting the magnetic field.
• A generator converts mechanical energy to electrical energy.
• Slip rings in an a.c generator reverses the direction of the induced current in
the external circuit.
223
• Split rings or commutator help to maintain the flow of the current in one
direction in the external circuit of a d.c generator.
• In both the d.c and a.c generators, the induced electromotive force is maximum
when the coil is moving in a direction perpendicular to the direction of the
field. It is zero when moving parallel to the direction of the field.
Unit Test 7
1. (a) What do you understand by the term electromagnetic induction?
(b) State the applications of electromagnetic induction.
2. You have been provided with the following apparatus: a straight conductor,
connecting wires, a horse-shoe magnet and a centre-zero-galvanometer.
Describe an experiment to illustrate electromagnetic induction.
3. The following apparatus may be used to perform an experiment to demonstrate

one of the factors that affect the magnitude of the induced electromotive
force; two identical coils, rheostat, dry cells, a switch, connecting wires, a
centre-zero-galvanometer.
(a) Draw the set-up that you need to assemble to perform the experiment.
(b) Describe how the experiment is to be carried out.
(c) State the expected results.
4. Fig. 7.37 shows a trolley carrying a magnet moving at a high speed towards
the coil. The trolley enters and passes through the coil.
S N

Fig. 7.37: A trolley with a magnetic moving into a coil
224
(a) Explain what happens to the needle of the galvanometer when the trolley
(i) approaches the coil.
(ii) is moving inside the coil.
(iii) is moving away from the coil.
(iv) and the coil are made to move at the same
speed in the same direction. Motion
(b) State the energy changes that occur as the trolley

enters and leaves the coil.
5. (a) State Fleming’s right hand rule.
(b) Fig. 7.38. shows a conductor moving in a region Fig. 7.38: Conductor-
of a uniform magnetic field. carrying current in a
magnetic field
(i) State the direction of the magnetic field.
(ii) What is the direction of the induced current?
(iii) Give three ways of increasing the magnitude of induced current.
6. (a) Explain the working of a simple a.c generator.
(b) What changes would you make in the a.c generator in order to produce
a larger electromotive force?
7. Fig. 7.39 shows a simple generator.
W
X
Z
N S 1
Y
2
Fig. 7.39: A simple generator

(a) What is the function of the annular magnets?
(b) Name the type of the generator.
(c) What is the function of the parts labelled 1 and 2?
(d) Explain how the electromotive force is induced in the coil. Sketch the
output graph of p.d against time.
(e) Sketch the variation of the voltage from a.c. generator and use it to
define the term peak value and period.
225
8. Use the words provided below to fill in the blank spaces.

position, mechanical, dc, current, ac, direction, electrical
A generator is a device that converts ______ energy to ______ energy.
There are two types of generators commonly the ac and dc generators.
A ______ generator produces alternating voltage while ______ generator
produces direct voltage in both types of generators, the direction of the
______ induced in the coil depend upon the ______ of the coil and the
______ in which the coil is moved.
9. Fig 7.40 shows a wire placed in a uniform magnetic field. If the force acting
on the wire is into the paper.
(a) Indicate on the diagram the direction of the current through the wire.
(b) Explain what happens when the battery terminates connected to wire
AB are reversed.
A
Wire
N s
B
Fig. 7.40
10. (a) What is magnetic flux?
(b) Fig. 7.41 shows current flowing in a solenoid;
Sketch the magnetic field around the solenoid, clearly indicating the
palarities.
Fig. 7. 41
11. The direction of induced current in a conductor moving in a magnetic field

can be predicted by applying:
226
A Faraday's law B Maxwell's screw rule

C Fleming's left hand rule D Fleming's right hand rule
12. (a) A cable connected to a centre-zero galvanometer G as shown in Fig. 7.42.
Coil
Fig. 7. 42
(i) State what is observed when the N-pole of a bar magnet is moved
towards the coil.
(ii) State two ways in which the effect observation (a) (i) can be increased.
(b) With aid of a labelled diagram describe how a simple a.c generator works.
13. An AC circuit carries an rms current of 5.6 Amps. The current travels
through a 90 Ohm resistor.
(a) What is the peak current?
(b) What is the power dissipated in the resistor?
(c) What is the peak voltage drop across the resistor?
227
UNIT 8 Electrical Power Transmission
Key Unit Competence

By the end of the unit the learner should be able to analyse the transmission of
electrical power.
Learning objectives
• Explain the use of high-voltage step-up and down transformers and power
transmission.
• Describe the transmission of electrical power.
• Outline the reasons for power losses in transmission lines and real transformers.
• Explain the step-up and down transformers and power transmission.
• State dangers of staying near high- voltage power lines.
Skills
• Analyse the operation of an ideal transformer.
• Describe a transformer.
• Explain the operation of an ideal transformer.
• Analyse the transmission of electrical power.
• Discover reasons for power losses in transmission lines and real transformers.
• Show possible risks involved in living and working near high- voltage power lines.
Attitude and value

• Appreciate the applications of electromagnetic induction.
• Acquire scientific techniques, reasoning and attitudes in analysing theories and
expressions.
• Realize reasons for power losses in transmission lines and transformers.
• Develop scientific skills for analysing functioning of ideal transformers.
• Discover other reasons for power losses in transmission lines and real transformers.
• Show the concern of high-voltage step-up and down transformers and power
transmission.
• Realise dangers associate with living and working near high - voltage power lines.
228
Introduction
Unit Focus Activity

Materials
• Electricity transmission lines in or near your school
• A transformer
Steps
1. In unit 7, we learnt about the generation of electricity. How is it transmitted
for use?
2. Study the pictures in Fig. 8.1 (a) and (b).
Power plant Step-up Tower Transmission substation

transformer
Fig. 8.1(a): Electrical power transmission to an industry.
229
Electrical Power Transmission
Home Transformer Distribution substation

Fig. 8.1(b): Electrical power transmission
3. Describe how electricity is transmitted from a power plant to your home.

4. Rusumo power station produces 80 MW. The power received at a sub-station
was 79.9 MW. Explain the loss of 0.1 MW. If you were an electrical engineer
at the plant, what could you advice to be done to minimise power loss during
transmission.
5. Discuss the dangers associated with the transmission of electric power at
high voltage.
After electricity is produced at power plants, it has to get to the customers that
use it. Our cities, towns and entire country are crisscrossed by power lines that
transmit the electricity. In this unit, we will analyse in details how electrical power
is transmitted.
8.1 Structure and working of a transformer

8.1.1 Structure of a transformer
Activity 8.1 To describe the structure and working of a

transformers
Materials
• Internet
• Reference books
• Transformer in the school or near your school
Steps
1. Using a simple drawing, discuss how transformers are made and their
functions.
2. Conduct a research from Internet or reference books on transformers. In
your research, find out the different types, structure and operations
3. Write a short report on your findings.
4. Present your report to the whole class through your secretary.
230
As we learnt earlier, a simple transformer consists of two coils insulated from

each other and wound on the same soft-iron core. One coil contains a few turns
of thick wire and the other coil contains many turns of thin wire. The coil that is
connected to a.c mains is called the primary coil (p) while the one through which
the stepped up or stepped down electrical current output is delivered to the outer
circuit is called the secondary coil (s).
Fig 8.2 shows the structure of a transformer in which P is the primary coil and
S is the secondary coil.
Laminated soft
iron core
Output
a.c mains coil P coil S

Output
Input
Fig. 8.2: The working of a transformer (mutual individual)
8.1.2 The working of a transformer
Activity 8.2 To describe the working of transformers

Materials
• Insulated copper wires • Galvanometer
• Switch • Connecting wires
Steps
1. Make two coils using the insulated copper wire. Connect the coils in currents
as shown in Fig. 8.3 (a) and (b).
Coil 1 Coil 2
(a) (b)
Fig. 8.3: Electromagnetic induction using two coils
231
2. Close the switch in coil 2. What happens to the pointer of the galvanometer?
Explain.
3. Quickly open and close the switch severally. Observe what happens to the
galvanometer? Explain your observations.
A transformer is an electric device that transfers electrical energy from one circuit
to another by electromagnetic induction. In transferring this energy, a transformer
steps up or steps down the voltage or electromotive force from the source.
In Activity 8.2, you must have observed that by switching the current on and
off in one coil, an electromotive force is induced in another coil. The circuit that
induces the electromotive force is called the primary circuit, while the circuit
where the electromotive force is induced is called the secondary circuit. Although
the two coils are not connected, changes in current in the primary circuit induces
an electromotive force in the secondary circuit.
This effect is called mutual induction. Mutual induction occurs on switching
the current on and off in the primary circuit. The switching on and off of the
current can also be achieved by replacing the battery and the switch with an a.c
power supply as shown in Fig. 8.4 (a). Fig. 8.4 (b) shows how the induced current
varies with time.
Slide
current
Thick metal rod Current
2
time
ac power supply
1 (a) (b)
Coil of resistant wire (constantan) 3
Fig. 8.4: Inducing a current by mutual induction
The mutual induction is more pronounced when the two coils are wound round
a soft iron core. This was shown by Michael Faraday who used a soft iron ring
as shown in Fig. 8.5.
Secondary circuit
A.C source
Primary circuit Secondary circuit
Fig. 8.5: Faraday’s ring
The magnitude of e.m.f induced in the secondary coils also depends on the ratio
of the number of turns of secondary to primary coils.
232
Explaining mutual induction

An electric current creates a magnetic field around the conductor through which
it flows. When the current is switched on and off in the primary coil, the strength
of the field (magnetic flux) keeps changing from zero to maximum and back to
zero (alternating current does this automatically without being switched on and
off as it fluctuates from zero to maximum). The change in magnetic flux induces
a current in the secondary coin in a way that this current tends to oppose the
current in the primary coil, and also fluctuates from zero to maximum. Thus, an
a.c input in the primary coil induces an a.c output in the secondary coil (Fig. 8.6 (a).
Thus, the change in the number of magnetic field lines threading the primary
coil induces an electromotive force in the secondary coil in an opposite direction
in accordance with Lenz’s law as shown on Fig. 8.6 (a)).
Soft iron core increases the induced e.m.f because it can easily be magnetised and
easily be demagnetised. Soft iron also helps to concentrate the magnetic field lines in
the secondary coil (Fig. 8.6 (b)). The induced electromotive force in the secondary
circuit has the same frequency as the electromotive force in the primary circuit.
V V
(b)
(a)
Fig. 8.6: Effects of soft iron core on the number of field lines threading the secondary coil
8.2 Types of transformers
Activity 8.3 To describe types of transformer and their uses
Materials
• A transformer within or near the school compound
• Internet
• Reference books
Steps
1. Identify a transformer in or near your school. Move close to it. What type of
transformer it is? How many types do you know? Name them.
233
2. Describe some causes of transformer failures..

3. Now do a research on types and uses of transformers.
4. Discuss your finds with other members.
There are two types of transformers, namely, the step-up and the step-down
transformers. In a step-up transformer, the number of turns in the secondary coil
Ns is more than the number of turns in the primary coil Np (Fig. 8.7 (a)), while
in a step-down transformer, the number of turns in the secondary coil is less
than the number of turns in the primary coil (Fig. 8.7 (b)). Fig. 8.7 (c) shows a
commercial step-down transformer.
(a) Step-up transformer (b) Step-down transformer
(c) A commercial step-down transformer

Fig. 8.7: Types of transformers
The terms, step-up and step-down, apply to output voltages of the transformer.
When an alternating electromotive force is applied to the primary coil, a changing
magnetic field is produced. The soft iron core links this field to the secondary coil.
This alternating field produces an alternating electromotive force in the secondary
coil through mutual induction.
Fig. 8.8 shows the circuit symbols for the step-down and step-up of transformers.
234
Secondary
Primary Secondary Primary coil
coil coil coil
(a) Step-down transformer (b) Step-up transformer

Fig. 8.8: Circuit symbols for transformers
A transformer may have more than one secondary coil. Fig. 8.9 shows a transformer
with two coils in the secondary circuit. Such transformers can step-up and step-
down voltages simultaneously.
Fig. 8.9: A transformer with two coils in the secondary circuit
8.3 Transformer equations

Activity 8.4 To investigate the relationship between number of
coils and the induced e.m.f
Materials
• Connecting wires • Insulated copper wire
• Galvanometer • Source of current
• Switch
Steps
1. Make two coils (one with more turns than the other) using the insulated
copper wire.
2. Connect one coils to the galvanometer and the other coil to the source of
current and the switch as shown in Fig. 8.10.
Fig. 8.10: Transformer connected to a galvanometer
235
3. Bring the two coils close to each other.

4. Close the switch. What do you notice on the pointer of the galvanometer?
Record down the reading.
5. Repeat stops 1 to 4 with the galvanometer on the coil with lesser turns.
6. Compare the reading obtained in steps 4 and 5. Which one is larger? Explain.
7. Now discuss in your groups the relationship between the number of turns of
the coil and the magnitude of the emf produced.
Table 8.1 shows the results obtained in a similar activity.

Table 8.1
Primary e.m.f Turns in the Turns in the NS Deflection of the
(εp) (V) primary coil Np Secondary coil Ns NP galvanometer
1.5 10 5 0.5 Small

1.5 20 5 0.25 Smaller
1.5 30 5 0.17 Smaller
1.5 10 10 10 High
1.5 10 20 2.0 Higher
1.5 10 30 3.0 Higher
From the activity, we can conclude that the magnitude of the induced
electromotive force in the secondary circuit is directly proportional to the ratio
of the number of turns of coils used.
Let VP and VS represent the voltage in the primary coil and secondary coils
respectively.
NP and NS represent the number of coils in the primary and secondary coils
respectively.
Electromotive force induced Number of turns in secondary coil, NS
in the secondary circuit (εs) ∝ Number
of turns
coil, N
in primary p
Ns
i.e εs ∝
Np
When the experiment is done using an a.c power supply, it can be shown that
Secondary e.m.f ,Vs Number of turns in secondary coil, NS
Primary e.m.f ,Vp
=
Number of turns in primary coil, Np
Vs Ns
= ………… (1)
Vp Np
In an ideal case (no power loss), the electric power (P = VI) in the primary coil is
equal to that in the secondary. Thus, when the voltage is stepped up, the current
is stepped down and vice versa.
236
Hence in such a case,

IS Np
I = N
P s
Noted that, the e.m.f induced in the secondary coil is maximum when the two
coils are close together and when wound on a soft iron core.
When the emf in the transformer is maximized, it is transmitted to different places
through cables. Among the many risks involves when transmitting high voltage
include electric shock incase the poles collapse and fire outbreak on structures
and vegetation.
8.3.1 Efficiency of a transformer

As mentioned earlier, a transformer transfers electrical energy from one circuit to
the other. The energy per second, supplied to the primary coil is called the power
input, while the energy obtained per second from the secondary coil is the power
output. In an ideal transformer, the power output is equal to power input. The
term efficiency is used to indicate how effective a transformer is in transferring
the input energy to output energy.
The efficiency of a transformer is the ratio of the power output to power input
expressed as a percentage.
power output
Efficiency = power input × 100%
The term efficiency is used to indicate how effective a transformer is in transferring

the input energy to output energy.
Well designed practical transformers often have efficiency as high as 98%.

For an ideal transformer power output = power input (efficiency 100%)
power output
Efficiency = power input × 100% = 100%
power input = Vp Ip; power output = Vs Is

Vp Ip = Vs Is ;
Vs I
∴ = p ................... (2)
Vp Is
Where Ip and Is are the current in primary coil and in secondary coil respectively.
Hence from the equation (1) & (2) we obtain:
Vs IP Ns
Vp = IS = Np
237
And NpIp = NsIs ….......... (3)

Equations 2 and 3 can be used to calculate the current in transformers coils.
Be Safe!
Note that a transformer can be disastrous if tampered with. It can cause death,
fire on premises and surrounding vegetables.
Do not tamper with a transformer in your places.
Example 8.1
An alternating electromotive force of 240 V is applied to a step-up transformer
having 200 turns on its primary coil and 4 000 turns on its secondary coil. The
secondary current is 0.2 A. Calculate the
(a) (i) Secondary electromotive force. (ii) Primary current.
(iii)
Power input. (iv) Efficiency.
(b) Comment on the answer to (iv).
Solution
Vs Ns Is Np
(a) (i) V = N (ii) Ip =
p p Ns
V p × Ns Ns × Is
Vs = Np Ip =
Np
240 × 4 000 4 000 × 0.2
= 200 = 200
= 4 800 V = 4.0 A
Power output
(iii) Power input = Ip × Vp (iv) efficiency = Power input × 100%
IS × VS
= 4 × 240 = × 100%
960
0.2 × 4800
= 960 W = 960 × 100%
= 100%
(b) Since the efficiency of this transformer is 100%, then it is an ideal transformer.
238
Example 8.2
A step-down transformer is connected to a 240 V alternating current power supply.
The primary coil has 1000 turns. How many turns should the secondary coil have
so as to operate a 12 V alternating current toy car?
Solution
Vs Ns Ns
= ⇔ 12 = ⇒ Ns = 50 turns
Vp Np 240 1000

Example 8.3
A transformer has an input coil of 60 turns. When this coil connected to a 240 V
source, the output voltage is found to be 4 800 V. The output power is 3 600 W.
(a) Calculate the number of turns in the output coil.
(b) If the efficiency of the transformer is 80%, calculate the
(i) output current
(ii) input current.
Solution
(a) Np = 60; Vp = 240 V; Vs = 4800 V; Ns = ?
Vs N Ns
= s ⇒ 4 800 = Ns = 4 800 ×60
Vp Np 240 60 240
= 1 200 turns
(b) (i) Po = VsIs
3 600 = 4 800 Is ⇒ Is = 3 600
4 800
= 0.75 A
Pi V s Is
(ii) Efficiency (E) = × 100% = × 100%
P0 VPIP
80 = 4 800 × 0.75 × 100%

240 × Ip
= 360 000 × 100%

240Ip
Ip = 360 000 = 18.75 A

240 × 80
239
Example 8.4
An ideal transformer is used to operate a 16 V, 48 W lamp from a 240 V mains
supply. Its primary coil has 450 turns.
(a) Draw a well labelled sketch of the transformer.
(b) How many turns does the transformer have in its secondary coil?
(c) What is the current flowing in the mains?
Solution
lamp (48 W)
(a)
240 V 16 V

Fig. 8.11
(b) Np = 450; Vp = 240 V; Vs = 16 V; Ns = ?

Vs N Ns
= s ⇒ 16 =
Vp Np 240 450
Ns = 450 × 16
240
= 30 turns
(c) The current flowing in the mains is given by.
P
Is = = 48W =3A
V 16V
Ns I I
For an ideal transformer, = p ⇒ 30 = p .
Np Is 240 Is
Hence IP = 30 × 3A = 0.2A
450
8.4 Power losses in a real transformer
Activity 8.5 To find out the factors that contribute to power losses
in a transformer
Materials
Steps
1. In groups, suggest and explain some of the factors that contribute to power
loss in a transformer.
240
2. Now conduct a research from the internet and reference books on factors
that contribute to the loss of power in a transformer. Compare them with the
ones you suggested in step 1. How right were you?
3. In your research, also find out the application of transformers.
4. Give a presentation on your finding to the whole class through your group
secretary.
The energy supplied in the primary circuit of a transformer is lost in a number

of ways. The following factors contribute to the overall power loss and therefore
affects the efficiency of a real transformer.
1. Resistance of the coils
As the current flows in the coils, the wires heat up and energy is lost in form of heat.
Energy lost = I 2 × R × t
This method of losing energy is called joule-heating.
To minimise energy loss in this way, thick copper wires of low resistance are used where
large currents are to be carried.
2. Eddy currents
When the magnetic field changes, small amount of current called eddy currents,
are induced in the core of the transformer. This heats up the core and energy is
lost in form of heat.
To minimise this loss of energy, the core is laminated and insulated between the laminations.
This reduces the magnitude of the eddy currents.
3. Hysteresis losses
The magnetisation and demagnetisation of the core by the alternating magnetic
field requires energy. This energy heats up the core and is lost as heat energy. This
method of losing energy is called hysteresis loss.
To minimise this loss of energy, the core is made of a soft magnetic material that is easy
to magnetise and demagnetise e.g. soft iron.
4. Flux or magnetic leakage
Not all the magnetic field lines of force due to the primary coil may link the
secondary coil resulting in what is called flux leakage as shown on Fig. 8.12 (a)).
To reduce this loss, the core is designed in such a way that almost all the magnetic effect
due to the primary coil is transferred to the secondary coil e.g. using a loop.
241
(Fig. 8.12 (b)).
(a) Incomplete core (b) Complete core (loop)

Fig. 8.12: Magnetic or flux leakage
Since it is not possible to completely reduce energy losses in transformers, very

large transformers are oil-cooled to reduce overheating otherwise they will cause fire
and thus massive destruction to the surrounding.
8.5 Applications of transformers

1. Transformers are used in the National Grid system to step-up or down
voltage and current.
Electric power is usually transmitted over long distances at very high voltage e.g
11 000 V and at low current to minimize power loss due to internal heating (I²R).
Transformers are used to step-up voltage at the power station and step-down e.g
240 V for use in a home.
2. Transformers are used in electric welding
Electrical welding machines use electricity at high current to melt metals.
For example, a transformer with 800 turns in the secondary coil and only 5
turns as primary, steps-up current in the ratio of 160:1. In a perfect transformer
(step-up), the current is stepped up in the same ratio. A current of 1 A in the
primary might give a current of 160 A in the secondary. This large current heats
the metal until it melts. The current can be used to weld two metals together.
Exercise 8.1
1. (a) Name two type of transformers.

(b) Explain the structure of each type of transformer.
(c) Describe the working principle of transformers.
242
2. A student designed a transformer to supply a current of 10A at a potential

difference of 60V. If the efficiency of the transformer is 80%, calculate
(a) The power supplied to the transformer.
(b) The current in the primary coil.
3. Describe the energy changes in a transformer.
4. Give two factors that affect the efficiency of a transformer.
5. Fig. 8.13 shows a transformer with 400 turns in the primary coil and 1000
turns in the secondary coil.
A
4 cm
240 V B
ac 8 cm
C
Fig. 8.13
(a) What type of transformer is it?
(b) Find the potential difference across BC.
(c) What assumption(s) have you made?
6. A transformer is used to operate a 9 V ac shaving machine (Fig. 8.14).
240 V ac 54 turns Shaving machine 9 V
Fig. 8.14
(a) Explain why the primary coils should be made of thicker wire than that of the
secondary coils.
(b) How many turns are there in the primary coil?
(c) Explain what would happen to the transformer if a 240 V dc power supply is
used instead of 240 V mains.
(d) What happens to the primary current when the machine is being used?
7. A step-down transformer has a primary coil with 800 turns and secondary
coil with 100 turns. The primary coil is connected to 240 V supply.
(a) Find the voltage output.
(b) If the transformer has a primary current of 0.10 A and of secondary 0.72 A,
calculate its efficiency.
243
8.6 Electric power transmission
8.6.1 The grid power transmission system
Activity 8.6 To describe the transmission of electrical power
Materials
• Cables of different thickness made of different metals (e.g copper, aluminium
steel etc)
• Reference books
• Internet
Steps
1. Identify overhead wires used to transmit electricity near or in your school.
Tell your classmates the materials used to make the overhead wires you have
identified.
2. Observe the cables provided to you. By giving appropriate reason, identify
which one is suitable for the transmission of electricity. Which cable would
contribute to loss of more electric energy than the other? Explain.
3. Now, conduct a research from the Internet and reference books on
transmission of electrical power. In your research, find out:
(a) Ways in which electrical power transmitted is lost during transmission
and how to minimise it.
(b) How electricity is transmitted and the dangers it exposes to the people
in the surrounding.
4. Share your report on your findings to the whole class.
The electrical energy generated at a power station is delivered to consumers

through cables. It is distributed to consumers all over the country through the
National Grid System which consists of a network of transmission cables carried
over through structures called pylons.
Fig. 8.15 shows pylons and cables carrying electricity from a generating station
to the consumers.
244
Fig. 8.15: Pylons and electricity transmission cables
Electrical power is generated at a relatively high current (e.g at 100 A and 25 kV), its
voltage is immediately stepped up using a step-up transformer at the generating station,
automatically stepping down its current for transmission through the grid (e.g. at
6.25 A and 400 kV). On reaching the consumer, the voltage is stepped down to a low
value e.g. 240 V for use in a home by a step-down transformer placed near the home.
Fig. 8.16 shows a section of a typical National Grid System from the power
generating station to the factories, towns and villages.
Super grid Step down Grid
Step-up trans- transformers
Generator
formers
Town houses Heavy industries

Light industries
To village
Step down transformers

Fig. 8.16: The National Grid System
8.6.2 Power loss in transmission cables

Due to electrical resistance (R) of the transmitting cables, some electric power
(Given by P = I2R) is lost in form heat in the transmission cables.
Remember, the electrcal resistance of a a wire (conductor) is directly proportinal
to its length and inversely proportional to its cross-sectional area (A), that is;
245
L
(R = ρ A ).
This means that a long thin wire has high electrical resistance than a short thick
wire. As such, a very high quantity of electric power would be lost if electric power
is transmitted at high current and through thin wires in the National Grid.
To reduce power loss in transmission cables,

1. Very thick transmission wires are used.
2. The transmission wires are made of metals like copper that are very good
conductors of electricity hence have very low electrical resistance.
3. Electric current is transmitted at very high voltage and very low current.
8.6.3 Advantages of a.c over d.c. power transmission

1. An a.c. voltage can be easily and cheaply stepped up or down from one
voltage value to another by a transformer while d.c power cannot.
2. An a.c. voltage can be transmitted over long distances at high voltage and low
current with minimum power loss, while d.c. cannot be transmitted over a
long distance even at high voltage and low current because a lot of electrical
power will be wasted as internal energy warms the transmission cable.
8.6.4 Dangers of high voltage transmission
Activity 8.7 To find out the danger of high voltage transmission
Materials
Steps
1. Suggest some of the dangers of high voltage transmissions.
2. Conduct a research from the internet or reference books on the dangers
associated with high voltages power transmission.
3. In your research, find out the diseases likely to be caused in one living near
high voltage power lines.
4. Give summarized report on your finding through your group secretary.
Due to the high voltages in the transmission cables, a strong electric field is setup
between the cables and the earth. Air, an insulator under normal conditions, may
start to conduct electricity especially on rainy days. People or animals in the vicinity
may get electrocuted. To minimise this danger, transmission cables carrying high
voltages are supported high above the ground by pylons. When the cables enter
towns and cities, they are buried underground.
246
Caution
Avoid touching loosely hanging electric cables. You will be electrocuted.
8.6.5 Danger of living and working near high voltage power lines
Living near high-voltage power lines and towers exposs one to the electrical and
magnetic radiation produced by these high-voltage wires. Long-term exposure
is likely to cause several health problems.
Some of these include:
1. Risk of electric shock
There is high risk of electric shock involved when transmitting high-voltage
power. For example, if the pole collapses or cables hang too low, they can give
electric shock to human beings and animals when they come into contact. This
may result in death.
2. Risk of fire
When the high-voltage cables fall on structures and vegetation they cause fire.
This can lead to massive destruction of property and plants.
3. Childhood leukemia
A research conducted in 1979 indicates that children living near high voltage power
lines and towers are at high risks of suffering from leukemia than their counterparts
who live far away. However, no evidence has been provided to establish a direct
connection between childhood leukemia and electromagnetic fields produced by
high-voltage power lines.
4. Cancer
Long exposure to electromagnetic field radiation from high-voltage power lines
and towers, may result in incidences of cancer. Research has indicated that people
who live within a 50 m radius of power lines had 99% chances of developing
cancer as compared to those who are 500 m away.
5. Depression
A research conducted on the psychological effect of living close to high-voltage
power lines shows exposure to extremely low frequency electromagnetic fields
might contribute to the number of depression-related suicides in people living
close to high-voltage power sources.
In addition, many researchers have discovered a link between people living near
high-voltage power lines and a number of health concerns, including brain cancer,
miscarriage, breast cancer, birth defects, fatigue, hormonal imbalance, decreased
libido, sleeping disorder, heart disease and so on.
247
8.6.6 Environmental impact of power generation and

transmission
Activity 8.8 To find out the environmental impact of power

generation and transmission
Materials
Step
1. Brain storm among yourselves what impacts power generation and
transmission has on the environment.
2. Note down your points.
3. Conduct a research from the internet and reference books to verify your
suggestions.
Generation and transmission of power have both positive and negative impacts
on the environment. Before a power plant is constructed, one has to know the
environmental and health consequences of electricity generation and transmission.
Electric power is generated through sources like hydroelectric, nuclear reactions, fossil
fuels, solar, geothermal energy and biomass.
Hydroelectric power is one of the most commonly generated power in the world.This
method of power generation is cheaper, has low operating costs, compared to other
methods of generating electricity like electricity from fossil fuels or nuclear energy.
Some of the negative impact of hydroelectric power generation and transmission
on the environment are:
• Displacement of people living around the place where a dam has to be constructed.
• Releasing carbon dioxide during construction and flooding of the reservoir.
• Disrupting the aquatic ecosystems and animal life.
• Can be catastrophic if the dam wall collapses e.g can cause flooding.
• The dam becomes a breeding site of mosquitoes which carry and transmit malaria.
Environmental impact from the generation and transmission of power from other
sources include:
• Pollution from fossil fuels.
• Dangers of exposure to radioactive materials from nuclear generation.
248
Caution
Malaria is a killer disease. It can be prevented by keeping mosquitoes away
i.e sleeping under a mosquito net.
While planning to build a dam, for hydroelectric power, one has to make sure
that there are minimal negative effects in the environment.
Think about this!

How electricity produced through nuclear reaction and fossils fuel impact
the environment.
Exercise 8.2
1. (a) Briefly explain how electricity transmitted from Mukungwa power

station to your school.
(b) Discuss the risk involved in the high-voltage transmission.
2. The resistances of a length of power transmitting cables is 10 Ω and is used
to transmit 11 kV at a current of 1A.If this voltage is stepped-up to 16 kV by
a transformer, determine the power loss.
3. A generator produces 660 kW at a voltage of 10 kV. The voltage is stepped up
to 132 kV and the power transmitted through cables of resistance 200 Ω to
a step-down transformer in a sub-station. Assuming that both transformers
are 100% efficient:
(a) Calculate the
(i) current produced by the generator
(ii) current that flows through the transmission cables
(iii) voltage drop across the transmission cables
(iv) power lost during transmission
(v) power that reaches the sub-station
(b) Repeat (a)(i) to (v), but this time the 10 kV is stepped up only 20 kV
instead of 132 kV for transmission.
(c) Briefly explain three factors that contributes to power loses in a
transformer.
4. State the ways through which power loss in transmission cables is minimised..
249
Project work 8.1: Construction of a simple transformer
Materials needed
• Dry cells • Soft iron sheets or blades,
• Bulb • Connecting wires
• Insulated copper wires • Sheets of paper
• Masking tape or a cloth tape
Assembly
• Make a complete soft iron core by packing a number of soft iron blades
together. Use sheets of paper to separate the soft iron blades. (Fig. 8.17).
Paper to
separate
Soft iron
blades
In two dimensions In three dimensions
Fig. 8.17: A simple transformer

Fig. 5.39: A simple transformer
• Use the masking tape to hold the bundles of soft iron blades together. Make
20 turns of the primary coil and 40 turns of the secondary coil. Connect the
ends of the primary coil to three dry cells in series with a switch. To the ends
of the secondary coil, connect a small bulb.
• Switch the circuits on and off the circuit rapidly and note what happens to
the bulb.

• A transformer is an electrical device that transfers electrical energy between
two or more circuits through electromagnetic induction.
• There are two types of transformers; step down and step up.
• In a step down transformer, primary turns are many than secondary turns.
• In a step up transformer, secondary turns are main than primary turns.
• Power loses in transformers occurs through
(i) Resistance of the coils
(ii) Eddy current in transformers
250
(iii) Hysteresis
(iv) Flux or magnetic leakage
Ns V
• In transformers; N = s
p V p
Ip N
• For an ideal transformer, = s
Is Np
• The efficiency of an ideal transformer may be calculated from:

Power output
Efficiency = × 100%
Power input
• Power losses in a transformer occur due to resistance in the coils, eddy

currents, hysteriss losses and magnetic flux leakage.
• Electricity is transmitted through cables at high voltage and lower current
from power stations to the consumers.
• Power loss in transmission cables occurs due to resistance in the wires. It is
minimised by using thick wires of good conducting material and trnasmission
at high and low current.
• Dangers associated with power transmission include risk of shock to people
living near the lines.
Unit Test 8
For questions 1 to 4, choose the most appropriate response from the

choices given.
1. Conductors for high voltage transmission lines are suspended from towers
A. to reduce clearance from ground.
B. to increase clearance from ground .
C. to reduce wind and snow loads.
D. to take care of extension in length during summer.
2. The most main disadvantages of using high voltage for transmission is
A. The increased cost of insulating the conductors.
B. The increased cost of transformers, switch gear and the other terminal
apparatus.
C. Both (a) and (b)
251
D. There is a reduction in the corona loss

3. What is the main reason for using high voltage for the long distance power
transmission?
A To reduction in the transmission losses.
B To reduction in the time of transmission.
C To increase system reliability.
D None of the above.
4. Which of the following statements is true?
A. at higher voltage, cost of transmission is reduced
B. at higher voltage, cost of transmission is increased
C. at higher voltage efficiency decreased
D. all of the above
5. (a) Explain the term mutual induction.
(b) Why is a complete core (loop) used in transformers?
(c) Distinguish between step-up and step-down transformers.
(d) Explain why a transformer will only work with an alternating voltage.
6. A transformer has 400 turns in the primary winding and 10 turns in the
secondary winding. The primary electromotive force is 250 V and the
primary current is 2.0 A. Calculate:
(a) the secondary voltage.
(b) the secondary current assuming 100% efficiency.
(c) describe two features in a transformer design which help to achieve high
efficiency.
7. Explain two environmental impacts of generating and transmission of electric
power.
8. The secondary circuit of a transformer is connected to a bulb rated at
12 V, 40 W. The primary circuit has 5000 turns on the primary and the
secondary has 250 turns. If the bulb is operating normally, find:
(a) The input voltage to the transformer.
(b) The current flowing through the bulb
(c) The power taken from the supply if the efficiency of the transformer is 90%.
9. A power station has an output of 100 kW at a p.d. of 800 V. The voltage
is stepped up to 33 kV by transformer T1 and transmitted along a grid of
resistance 0.85 kΩ. It is then stepped down to a pd. of 500 V by transformer
T2 at the end of the grid for use in a light industry. Given that the efficiency
of T1 is 95% and that of T2 is equal to 90%, calculate to 2 decimal places the:
(a) power output of T1 (b) current in the grid
(c) power loss in the grid (d) input voltage of T2
(e) maximum power and current available for use in the industry.
252
Electric Field Intensity
UNIT 9 Electric Field Intensity
Key Unit Competence

By the end this unit the learner should be able to calculate intensity of electric
field due to one or more point charges
Learning objectives
• Illustrate electric field patterns due to two charges.

• State the principle of superposition for point charges in an electric field.
• Describe field patterns for two point charges.
• Explain the intensity of electrical field to the position of charge.
• Explain electric field intensity at different points due to a charge
• Explain superposition of parallel electric fields.
Skills
• Describe the electric field patterns due to like charges, and unlike charges.
• Relate the intensity of electric field to the position of charge.
• Differentiate electric force and electric field.
• Compare the electric field intensity at different points
• Determine the electric field intensity due to one or more charges.
Attitude and value

• Appreciate the applications of electrostatic force.
• Show concern about dangers of electric field and how to minimize hazards.
• Acquire ability to think logically and systematically as you pursue a particular line of
thought.
• Adapt scientific method of thinking applicable in all areas of life.
• Acquire knowledge for analysing and modelling physical processes.
253
Introduction
Unit Focus Activity

1. Draw the electric field pattern between:
(a) a positive and negative point charges,
(b) two positive charges.
2. (a) Define the term electric field intensity and state the factors that
determine its magnitude.
(c) Fig. 9.1 shows the electric field around three charges A, B and C.
Compare the electric intensity for the three charges.
A B C
Fig. 9.1: Three electric fields
254
3. Fig. 9.2 shows a small pith ball of charge Q0 = -2× 10-6 C placed between
positively charged metal sphere A with charge Qa = 5× 10-6 C and negatively
charged metal sphere B of charge Qb = -9× 10-6 C, 25 cm from each charge.
B
Qo
Qb
A
Qa
Fig. 9.2: Superposition of electric fields
(a) Determine the magnitude and direction of the resultant force on the pith ball.
(b) State the principle you have used to work out part (a)
(c) Present your findings to the rest of the class in a class discussion.
In lower classes we were introduced to electrostatics. We learnt among other
concepts the law of electrostatics and Coulombs law. In this unit, we will further
our knowledge by analysing the strengths of electric fields and the interaction of
electric fields. This interaction is known as superposition of electric fields.
9.1 Electrostatic force and Coulombs law
Activity 9.1 To demonstrate the electrostatic law between two

negatively charged polythene rods
Materials:
• Two identical polythene rods A and B • A silk cloth
• Clamp and a stand • Thread
Steps
1. Charge two polythene rods negatively by rubbing them with a silk cloth.
Suspend rod A on a stand.
2. Bring charged rod B (touching it using a stirrup cloth) near the suspended
polythene rod A.
3. While varying the distance of the rod A from B, note the magnitude of the force
of repulsion. Comment on the relationship between the distance and the force.
4. Charge the two rods with the silk cloth more vigorously while maintaining
the distances and repeat steps 2 and 3. What do you observe?
5. State the effect of the amount of charge on the force of repulsion.
6. Discuss the observation made with other members of your group.
255
In S1, we learnt that objects get charged by induction, friction and by contact
methods. Attraction occurs between unlike charges and repulsion occurs between
like charges.
In 1784, a French physicist Charles Augustine Coulomb actually did similar
experiments and came up with the following observations:
1. The electrostatic force between two charges varies directly as the product of
the two charges.i.e.
F ∝ Q1 Q2 ………….……… (i)
2. The electrostatic force between two charges varies inversely as the square of
the separation distance between the charges i.e.
1
F∝ ………………………. (ii)
d2
He summarized his observation in what is called the Coulomb's law.
It states that "the force of attraction or repulsion between two electrically charged particles
is directly proportional to the magnitude of their charges and inversely proportional to
the square of the distance between them".
The Coulomb force between two or more charged bodies is the force between
them due to Coulomb’s law.
Combining equation (i) and (ii),
Q1Q2 Q1Q2
Fα and F = k
d2 d2
The constant of proportionality k is called Coulomb's constant and depends on
the medium in which charged bodies are presented.
1
k=
4πε0
Therefore the equation becomes
1 Q1Q2
F= , where;
4πε0 d2
• F is the repulsion or attraction force between two charged bodies.

• Q1 and Q2 are the electrical charges of the bodies.
• d is distance between the two charged particles.
Example 9.1
Two balloons are charged with an identical quantity and type of charge:
-6.25 nC. They are held apart at a separation distance of 61.7 cm. Determine
the magnitude of the electrical force of repulsion between them. Take
k = 9.0 × 109 N.m2/C2
256
Solution
Q1 = -6.25 nC = -6.25 × 10-9 C
Q2 = -6.25 nC = -6.25 × 10-9 C
d = 61.7 cm = 0.617 m
1 Q1Q2
= 9 × 10 × 6.25 × 10 × 6.25 × 10
9 –9 –9
F=
4πε0 d2 0.6172
= 923.5 × 10-9 N

Note: The negative "-" sign was dropped from the Q1 and Q2 values
prior to substitution into the Coulomb's law equation. The use of
"+" and "-" signs in the equation would result in a positive force
value if Q1 and Q2 are like charged and a negative force value if Q1
and Q2 are oppositely charged. The resulting "+" and "-" signs on F
signifies whether the force is attractive (a "-" F value) or repulsive
(a "+" F value).
Example 9.2
Suppose that two point charges, each with a charge of +1.00 are separated by a
distance of 1.00 m. Determine the magnitude of the force of repulsion between
them.
Solution
Q1 = 1.00 C, Q2 = 1.00 C, d = 1.00 m
1 Q1Q2
= 9 × 10 × 1.00 × 1.00
9
F =
4πε0 d2 12
= 9.0 × 109 N
Exercise 9.1
Take k = 9.0 × 109 N.m2/C2 and use it where necessary to answer questions.
1. Two balloons with charges of +3.37 µC and -8.21 µC attract each other with
a force of 0.0626 N. Determine the separation distance between the two
balloons.
257
2. Determine the electrical force of attraction between two balloons with

charges of +3.5 × 10-7 C and -2.9 × 10-6 C when separated by a distance of
0.65 m.
3. Determine the electrical force of attraction between two balloons that are
charged oppositely type the with same quantity of charge. The charge on the
balloons is 6.0 × 10-5 C and they are separated by a distance of 0.50 m.
4. A helium nucleus has charge +2e and a neon nucleus +10e, where
e = 1.60 × 10-19 C. With what force do they repel each other when separated
by 3 × 10-9 m?
5. If two equal charges, each of 1 coulomb, were separated in air by a distance
of 1 km, what would be the force between them?
6. Determine the force between two free electrons spaced 10-10 m apart.
(e = 1.60 × 10–19 C)
7. What is the force of repulsion between two argon nuclei when separated by
10-9 m? The charge on an argon nucleus is +18e.
8. Two equally charged pith balls are 3 cm apart in air and repel each other
with a force of 4 × 10-5 N. Compute the charge on each ball.
9. Two small equal pith balls are 3 cm apart in air and carry charges of
+3 × 10-9 C and –12 × 10-9 C respectively. Compute the force of attraction.
10. A positive charge of magnitude 3.0 × 10–8 C and a negative charge of
magnitude 4.0 × 10-8 C are separated by a distance of 0.02 m. Calculate the
C force between the two charges and give its direction.
9.2 Superposition of forces
Activity 9.2 To find out the statement of the principles of superposition

of force and how it governs interactions of electric fields
1. What do you think superposition of force is?

2. Research from the Internet or text books, the statement of the principle of
superposition of forces and how it governs interactions of electric fields.
3. With the help of your teacher, harmonize your findings and write down the
notes in your note books
258
Coulomb’s law applies to any pair of point charges. When more than two charges
are present, the net force on any one charge is simply the vector sum of the forces
exerted on it by the other charges.
If three charges Q1, Q2, Q3 are present, the resultant force F2 experienced by Q2
due to Q3 and Q1 is given by
F2 = F12 + F23
Where
F12 is the force exerted on Q2 by Q1
F23 is the force exerted by Q3 on Q2
Note: Force is a vector quantity. Hence, the direction of the force

must be considered during vector addition. For forces between
charges, the direction of the force is determined by whether the force
is attraction or repulsion.
The superposition principle (superposition property) states that for all linear forces,
the total force is a vector sum of individual forces. It holds for linear systems.
The superposition principle is illustrated in the example below.
Example 9.3
Three charges are arranged as shown in Fig. 9.3. Find the force on the charge
Q2 given that, Q1 = 6.0 × 10-6 C, Q2 = 3 × 10-6 C and Q3 = -9 × 10-6 C when Q1
and Q3 are 2.0 m from Q2.
F23
Q3
Q1 Q2
F12
Fig. 9.3: Three charges
Solution
1 Q1Q2
= 9.0 × 10 × 6 × 10 × 3 × 10
9 –6 –6
F12 =
4πε0 d2 22
= 4.05 × 10–2 N
1 Q2Q3
= 9.0 × 10 × 3 × 10 × 3 × 10
9 –6 –6
F23 =
4πε0 d2 22
= 6.075 × 10–2 N
259
Since F12 and F23 are in the same direction,

F2 = F12 + F23 = 4.05 × 10-2 N + 6.075 × 10 -2 N
∴ F2 = 1.0125 × 10-1 N
Example 9.4
Three charges are arranged as shown in Fig. 9.4. Find the force on the charge Q3
assuming that, Q1 is -9.0 × 10-6 C, Q2 is 3 × 10-6 C and Q3 is -6 × 10-6 C when
Q1 and Q3 are 2.0 m from Q2.
Fig. 9.4: Three charges

1 Q1Q3
= 9.0 × 10 × 9 × 10 × 6 × 10
9 –6 –6
F13 =
4πε0 d2 42
= 3.0375 × 10–2 N
1 Q2Q3 9 × 0 × 109 × 3 × 10–6 × 6 × 10–6
F23 = =
4πε0 d2 22
= 4.05 × 10–2 N
∴ F3 = F23 – F13
= 4.05 × 10-2 N – 3.0375 × 10 -2 N
= 1.0125 × 10-1 N
Exercise 9.2
Take k = 9.0 × 109 N m2/C2 where necessary.

1. Three charges are arranged such that a charge of Q1 is 2.0 × 10-6 C,
Q2 is -1 × 10-6 C and Q3 is 1 × 10-6 C when Q1 and Q3 are 1.0 m from Q2.
find the force on the charge Q2.
2. Three point charges are fixed in a straight line. The quantities of charges
on each are Q1 = 4.0 × 10-6 C, Q2 = 1.0 × 10-6 C and Q3 = -5.0 × 10-6 C
respectively. The distance between Q1 and Q2 is 4.0 cm and the distance
between Q2 and Q3 is 3.0 cm. What is the approximate force on Q1 due to the
other two charges?
260
3. Calculate the force experienced by the ball of charge -2Q in each of the case
if d is 2.0 cm. (See Fig. 9.5)
Fig. 9.5: Ball of charges
9.3 Electric field
9.3.1 Definition of electric field
Activity 9.3 To define an electric field

Materials:
• Internet enabled computer • Reference books
Step
Research from reference books or Internet the definition of an electric field.
We are familiar with the observation that a charged body attracts small pieces of
paper, dust, hair etc. The basic law of electrostatics states that like charges repel
and unlike charges attract. So a charged body can affect other nearby objects
without touching them. This action at a distance can be explained by what is called
the electric field of a charged body.
Activity 9.4 To demonstrate the electric fields produced by charged

bodies
Materials:
• Glass dish • Castor oil
• Electrodes • Connecting wires
Steps
1. Assemble a pair of straight metal wires, called the electrodes, in a shallow
261
glass dish so that their ends are just Electrodes

covered by a layer of an insulating
liquid like castor oil or carbon
tetrachloride (Fig. 9.6).
Shallow
2. Apply a very high potential
glass dish
difference, from a suitable power
supply, to the two electrodes so
that they have opposite charges. Castor oil
3. Then sprinkle grass seeds or

semolina powder on the surface of Fig. 9.6: Arrangement to study the electric field
the liquid.
4. Observe what happens to the grass seeds
or powder and draw the resulting pattern.
5. Repeat the activity with different charges on electrodes and observe the
pattern formed.
6. Draw the various patterns and draw the various alignment of the grass seeds.
In the above Activity, the seeds acquire induced opposite charges at their ends
and align themselves in a particular pattern (Fig. 9.7 (a)). This pattern depends
upon the charge on the electrodes.
(a)
Two straight wires dipped into the liquid
and connected to the opposite polarities of
+ –
a high voltage power supply (Fig. 9.7 (a)).
(b) Two straight wires dipped into the liquid,

both connected to the same positive
+ +
potential terminals (Fig. 9.7 (b)).
(c) A straight vertical wire dipped into the liquid

+ having a high positive charge (Fig. 9.7 (c)).
Fig. 9.7: Electric fields due to different shapes of electrodes
262
+ +
(d)
Two parallel metal plates dipped into
the liquid and connected to the opposite
polarities of high voltage power supply
(Fig.9.7(d)).
– –
(e)
A straight horizontal metal wire electrode and
a point electrode dipped into the liquid and
+ + – –
connected to opposite polarities of high voltage
power supply (Fig. 9.7(e)).
(f) A straight vertical wire electrode and an

uncharged metal ring in the liquid. The wire
+ + electrode is connected to the positive polarity
of high voltage power supply (Fig. 9.7(f)).
Fig. 9.7 (d) to (f): Electric fields due to different shapes of electrodes
The above alignment of seeds depict the electric field produced in different
arrangements.
An electric field is as the region or space surrounding a charge. In this region,
another charged body may move away from or towards the charged body due to
the electric field. In Fig. 9.8, P is a positively charged body and N is a negatively
charged body producing an electric field. If another light charged body T is
introduced in this field, the body T may experience a force away from P or towards
N.
+ + +
+ +
+ +
T
+ P
+ Force
+ + +
+ +
++
+++
– – –
– –
– – T
– – Force
– N –
+
– –
– –
– – –
Fig. 9.8: Force between charges in an electric field
263
If a body is positively charged and has a charge of 1 coulomb, it is called a test

charge or a unit positive charge. A unit positive charge experiences a force in the
electric field.
An electric field is defined as the region around a charged body where electrostatic force
due to the charges is experienced.
9.3.2 Direction of electric field

Activity 9.5 To determine the direction of an electric field and the
line of force
Materials:
• Books or the Internet
Steps
1. Research from books and internet the definition of the direction of an electric
field line and electric line of force.
2. Discuss with your partner the properties of electric lines of force.
Fig. 9.9 shows the direction of force acting on the test charge, indicated by the
arrow head. The force is either away from or towards the charge which creates the
field. The direction of the electric field at a particular point is defined as the direction
in which a unit positive charge is free to move when placed at that point. It should
be noted that the force experienced by a negative charge will be in an opposite
direction to that of the electric field (Fig. 9.9).
+ + +
+ +
+ +
+ + Force – Electric field
+ +
+ +
++
+++
– – –
– –
– –
– – Electric field –
Force
– –
– –
– –
– – –
Fig. 9.9: Force acting on a negative charge in an electric field
9.3.3 Electric lines of force

In an electric field, the electric force exists at all points. The test charge, i.e., the unit
positive charge will be forced to move in a particular direction when placed at any
point in the field. The path along which a unit positive charge would tend to move
in the electric field is called the electric field line or the electric line of force. If the path is
a curve, the tangent at any point gives the direction of the electric field (Fig. 9.10).
264
Electric line of force Tangent
Fig. 9.10: An electric field line.

A line of force may be traced by placing a unit positive charge at any point and
allowing it to move throughout in the direction of the force acting on it. This is
similar to the magnetic line of force in the magnetic field created by a magnet.
Properties of electric lines of force

1. Lines of force start at 90° from the positive charge and end on the negative
charge at 90°.
2. No two lines of force can ever cross each other.
3. The field lines are ‘elastic’, i.e., the lines tend to contract or expand so that
they never intersect each other.
9.4 Electric field intensity
9.4.1 Definition of electric field intensity
Activity 9.6 To show the strength of an electric field varies with

quantity of charge and distance from the charge.
Materials:
• a clamp and a stand • polythene rod
• pith ball • a string
Steps
1. Suspend a pith ball with a string on a clamp ( Fig. 9.11). Charge it by rubbing
it with a dry cloth.
2. Charge a polythene rod by rubbing it with a cloth.
3. Hold the pith ball firmly with an insulating material and bring the charged
polythene rod close to the suspended pith ball.
265
Fig. 9.11: Demonstrating strength of electric field.
4. Measure the distance and observe what happens when you release the pith
ball.
5. Repeat steps 3 and 4 but this time bring the polythene rod more closely to
the pith ball. Compare your observations this time and your observations in
step 4.
6. Charge the rod more strongly and repeat steps 3 and 4, trying as much as
possible to maintain the same distance. Compare your observations in steps
3, 4 and 5. Comment on the composition.
7. Based on your observations in this activity, make a conclusion on how the
strength of the electric field varies with the
(a) Quantity of charge.
(b) Distance from the charge.
Electric field intensity is the measure of the strength of an electric field at a

specified point. It is defined as the electrostatic force per unit charge experienced by a
test charge placed at a specified point in an electric field.
Electrostatic force F
Thus, Electric field intensity, E = or E =
Charge Q
Its SI unit is Newtons per coulomb (N/C) or volts per metre (V/m).
9.4.2 Factors that determine the magnitude of electric field intensity

1. The distance of a point in the electric field from the charge
Electric field intensity decreases with the increase distance away from the charge.
For example, the electric field intensity at point A in the electric field of charge
Q is greater than that at point B in Fig. 9.12.
266
Fig. 9.12: Variation of electric field strength with distance.
2. The quantity of charge

Consider a unit charge, q at a fixed distance, d from charge, Q. The force
experienced by the unit charge due to charge, Q increases when the quantity of
charge, Q is increased.
Therefore, electric field intensity point at a point is directly proportional to the
quantity of charge.
Consider a point charge q in the field of charge Q, a distance d from Q (Fig. 9.13).
Fig. 9.13 A point charge in an electric field.
Applying Coulomb's law, the force between the charge (Q) and the unit charge
(q) is given by
Qq
F=
4πε0d2
267
Thus, the electric field intensity = force exerted on unit charge q by charge Q
given by
E = Force exerted
Charge
Qq
=
4πε0d2 × q
Q
= (on crossing out q)
4πε0d2
We can also get an expression for electric field intensity in terms of electric field
potential (V) as follows:
Electric field potential (V) is defined the work done in moving a unit charge
through a distance (d) in an electric field.
Electric field potential, V = Work done = Force (F) × Distance (d)

Charge (Q) Charge (Q)
But, Force (F) = Electric field intensity (E)

Charge (Q)
Thus, Electric field potential, V = Electric field intensity (E) × distance (d)
v
Hence, Electric field intensity (E)= Electric field potential (V) ⇒ E =
Distance (d) d
Thus, other units of electric field intensity are volts per metre.
Example 9.5
A force of 3 N is acting on the charge 6 μC at any point. Calculate the electric
field intensity at that point?
Solution
Given: Force F = 3 N, Charge q = 6 μC
The Electric field is given by, E = F = 3N

q 6 × 10–6C
= 5 × 105 N/C.
268
Example 9.6
Find electric field at a distance of r = 10-10 m from the nucleus of Helium atom?
Solution
Given: Charge in nucleus, q = 2 × 1.6 ×10-19 C = 3.2 × 10-19 C.
Distance d = 10–10 m
The formula of electric field is given by; E = kq = 9 × 10 × 3.2 × 10
9 -19
d2 (10–10)2
= 28.8 ×1010 N/C

= 2.88 × 1011 N C-1
Exercise 9.3
For questions 1 - 5, choose the correct response from the choices given.
Take k = 9.0 × 109N.m2/C2 where necessary.

1. Region around a charge q in which it exerts force on a test charge is called
A. Electric field intensity B. Electric force
C. Electric field D. Coulombs force
2. Field lines always emerge from.
A. negative charge B. can be both charges
C. the central point of both charges D. positive charge
3. Direction of free test charge will be
A. Direction of electric intensity B. Direction of coulombs force
C. Direction of magnetic intensity D. Direction of protons
4. Spacing between field lines shows
A. Their direction B. Their position
C. Both a and b D. Their strength
5. Electric field intensity is
A. a base quantity B. a scalar quantity
C. A and B both D. a vector quantity
6. A uniform electric field of magnitude 30 N / C is directed downward. What
is the magnitude and direction of the force on:
269
(a) +4.0 C charge placed in this electric field

(b) –3.0 charge placed in this electric field
7. A charge of magnitude -3.0 × 10-6 C is moved through a potential difference
of 80 volts. Calculate the work done on the charge.
9.5 Electric field patterns
Activity 9.7 To draw electric field patterns.
Materials
Steps
Research and draw electric field patterns of:
(a) A point positive charge and point negative charge.
(b) Two positive charges near each other and negative charges.
(c) Two equal unlike charges.
The electric field pattern around a charged body depends on whether the body is
completely isolated or is in the presence of other bodies. The following are some
examples of electric field patterns for isolated and non-isolated bodies.
1. Fig. 9.14(a) shows an isolated positive point charge. The field lines are
radially outwards from the positive charge.
2. Fig. 9.14(b) shows an isolated negative point charge. The field lines are
radially inwards towards the negative charge.
Fig. 9.14(a): Isolated positive point charge Fig. 9.14(b): Isolated negative point charge
270
3. Fig. 9.16 shows two equal positive point charges. The field lines start radially
outwards from each charge. The resultant field is due to the electric field
produced by each charge.
A point N lies midway between the two charges, on the line joining them. Here
the resultant force acting on the unit positive charge is zero and is called a neutral
point. A neutral point in an electric field is one where the resultant force acting on the
unit positive charge is zero (Fig. 9.15). Force due to A = force due to B. i.e. FA = FB
No field lines exist at the neutral point.
A B
N
FB FA
+ +
Fig. 9.15: Two equal positive point charges
4. Fig. 9.16 shows two equal unlike point charges. The field lines start from
the positive charge and end on the negative charge. In this case, there is
no neutral point as a unit positive charge placed at any point experiences a
force.
A B
Fig. 9.16: Two equal unlike point charges
271
5. Fig. 9.17 shows two unequal positive point charges. The neutral point N is
more closer to the weaker charge.
+Q +4Q
Fig. 9.17: Electric field between unequal positive charges
6. Fig. 9.18 shows a positive point charge and a straight metal plate with
negative charge.
–
Fig. 9.18: A positive point charge and a negative metal plate

7. Fig. 9.19 shows a positive point charge and an uncharged ring placed in the
electric field. The metal ring placed near the positive charge gets charged by
electrostatic induction. The field lines are as shown in Fig. 9.19. The field
lines do not pass through the conductor. The conducting ring acts as an electric
shield for the space enclosed by the ring.
– +
– +
– +
Fig. 9.19: A positive point charge and uncharged ring
272
8. Fig. 9.20 shows two parallel metal plates having opposite charges and placed
close together. In this case the field lines are parallel except at the edges. If
the field lines are parallel, the electric field is uniform.
+ –
+ –
+ –
+ –
+ –
Fig. 9.20: Two parallel metal plates having opposite charges

• The force between two point charges is…
(i) directly proportional to the magnitude of each charge Q1, Q2.
(ii) inversely proportional to square of the separation between their centers
(d). This relationship is known as Coulomb's Law. As an equation it is
usually written in one of two form.
Q1Q2 1 Q1Q2
F=k or F =
d 2 4πε0 d2
(i) When two charges have the same sign, their product is positive, which
means the action is repulsive.
(ii) When two charges have the opposite sign, their product is negative,
which means the action is attractive..
• An electric field is a region where a charged body experiences an electrostatic
force.
• The direction of the electric field is the direction in which a unit positive test
charge would move to when placed at that point.
• An electric line of force is the path along which a unit positive charge would
tend to move in the electric field.
273
• The electric field is uniform if the field lines are parallel.

• Electric lines of force do not cross each other.
• Electric lines of force start from positive charge and end in the negative
charge.
Unit Test 9
For questions 1 - 4, select the correct response from the choices given.
Take k = 9.0 × 109 N m2/C2 where necessary.

1. When distance from center is doubled then electric field strength will
A decrease by factor of four B increase by factor of four
C will be same D decrease by factor of two
2. Electric force on a dust particle having charge equal to 8 × 10-19 C2 when
plates are separated by distance of 2 cm and have a potential difference of
5 kV is
A 2 × 10-13N B 3 N
C 5 N D 20 N
3. Electric field strength can be defined as
A E = Q B E = W
F F
C E = F D E = P
Q Q
4. Which statement is correct about the conditions in a uniform electric
field?
A. All charged particles experience the same force.
B. All charged particles move with the same velocity.
C. All electric field lines are directed towards positive charges.
D. All electric field lines are parallel
274
5. Fig. 9.21 below shows electric field lines around a charged metal sphere (in air)
A•
Fig. 9.21:
Copy the diagram. Draw an arrow on each line to show the direction of the electric field if
a:
(a) positive point charge is placed at A.
(b) negative point charge is placed at A.
6. A small positive charge is placed in the electric field of a large charge . Both
charges experience a force F. derive the formula for electric field strength of
the charge at the position of charge.
7. A charge of magnitude - 4.0 × 10-6 C is moved through a potential differ-
ence of 70 volts. Calculate the work done on the charge.
8. What is the magnitude of the electric field intensity at a point in the field
where an electron experiences a 1.0 -newton force (electrons have a charge
of -1.6 × 10-19 C)?
9. Fig. 9.22 shows a metal ball with a charge of 1.6 × 10-19 C at rest in an
electric field with an intensity of 15.0 × 1020 N per coulomb. What is the
weight of the ball?
Fig. 9.22: Changed metal ball in electric field
10. If the intensity of the electric field at point P in Fig. 9.23 is -3.5 × 104
N/C, what is the magnitude of the electric force acting on an electron at P?
(Electrons have a charge of -1.6 × 10-19 C)
275
Fig. 9.23: Electric field
11. You have an electric field with an intensity of 18 N/C at a distance of 6.0 m.
What is the voltage?
12. Fig. 9.24 shows a point change in an electric field. If a point charge has
a charge of +3.2 × 10-19 C, what is the magnitude of the electric force
acting on the point charge located in an electric field with an intensity of
5.0 × 103 N/C?
Fig. 9.24: Point charge in electric field
13. Two parallel oppositely charged plates are 5.1 cm apart. The potential
difference, in volts, between the plates is 44.6 V. Find the electric field
strength between them.
14. If it takes 88.3 J of work to move 0.721 C of charge from a positive plate to a
negative plate, what is the potential difference (voltage) between the plates?
276
UNIT 10 House Electric Installation
Key Unit Competence

By the end of the unit, the learner should be able to analyse and carry out a simple
electric installation.
Learning objectives

• Describe electric circuit diagrams.
• Differentiate between surface wiring and conduit wiring.
• Identify components used in electrical installation.
• Explain protecting electric devices and their installation.
• Identify different lamps.
Skills
• Describe symbols used in electrical engineering drawing.

• Describe the cable by type and size used for lightning arrestor, lighting and socket
outlets.
• Carry out simple surface wiring for a residential house using appropriate tools.
• Evaluate different electrical protective devices.
• Explain the functioning of circuit breakers and fuses in electrical circuits.
Attitude and value

• Appreciate applications of electricity.
• Develop responsible behavior towards electrical installation.
• Develop responsible behaviour for protection against dangers associated with
electricity.
• Acquire skills for identifying problems in electric circuits.
277
Introduction
Unit Focus Activity

Materials:
• Manila paper • Geometrical set
Steps
1. Think of a house in the school compound or at home. Using a manila paper,
sketch a plan of how you can carry out a successful electrical wiring.
2. Highlight the electrical components you may need to complete the wiring.
3. In your view, do you need earth connection? Describe how it is installed.
4. Discuss how people can be safe from hazards caused by electricity.
Electrical installation in your home is a skill you can acquire. Knowing how circuits
work and what can be done with them is useful knowledge. Wiring in a residential
houses is not that complicated, but it can be dangerous. Proper understanding and
caution are required. In this unit, we are going to learn how electric installation
in houses is done.
278
House Electric Installation
10.1 Standard symbols for electrical installation
Activity 10.1 To identify standard symbols for electrical

installation
Materials
• Different fuses • A socket
• A switch • Bulb
• Chart showing standard symbols for electrical installation.
Steps
1. Take the electrical devices provided to you. Draw their symbols as used in
electrical circuit.
2. Tell your group members the uses of each electrical device provided.
3. Now, take the chart provided to you. Compare your drawings of the symbols
with the ones on the chart. How accurate were your drawings?
4. Take your group members through other electrical symbols shown on the
chart.
Most of the electrical devices are made in different styles, appearances and colours
according to users’ requirements. However, standard electrical symbols are used
to represent various electrical and electronic devices in schematic diagrams of
electrical or electronic circuits. They are easy to understand.
The following table lists some basic electrical symbols.
Table 10.1
Device Symbol Device Symbol
Cell Battery
Lamp ac supply
Ammeter A Voltmeter V
Galvanometer Transformer
279
Heating element Switch
2-way switch Bell
Fuse Fixed resistor
Variable resistor
Potentiometer
(rheostat)
10.2 Electrical lamps and fuses
Activity 10.2 To observe and describe different types of electrical

lamps used for lighting
Material
• Different types of electrical lamps • Reference books

• Internet
Steps
1. Take different electrical lamps provided to you and discuss their appearances
and structures. What differentiates them from each other?
2. Identify which electrical lamps are more efficient for use at homes.
3. Conduct a research on types of electrical lamps used for lighting.
4. In your research, find out the structure and the gases used in the lamps (if
any).
5. Present a summarized report of your findings to the whole class through
your group secretary.
An electrical lamp is a light emitting electrical device used in electric circuits,

mainly for lighting and indicator purposes. The construction of an electrical lamp
is quite simple. It has a filament surrounded by a transparent glass. The filament
of the lamp is usually made of tungsten since it has high-melting point.
When current flows through lamp, the tungsten filament glows without melting,
producing light energy.
280
10.2.1 Types of lamps

There are three main categories of electrical lamps namely: incandescent lamp,
LED lamps and gas-discharge lamps.
(a) Incandescent lamps
These are lamps which produce light from a filament heated white-hot by an
electric current. They are also known as tungsten lamps.
Fig. 10.1: An incandescent lamp
These lamps are aften considered the least energy efficient type of electric lighting.They
are commonly found in residential buildings. Although inefficient, incandescent
lamps posses a number of advantages: they are cheap, turn on instantly, are available
in a huge array of sizes and shapes and provide a pleasant, warm light with excellent
colour rendition.
An example of incandescent lamps is a vacuum
lamp. As the name suggest, the vacuum lamp
has the glass enclosing the tungsten filament
has no gas in it. It has a vacuum. The tungsten
filament is heated to a temperature at which
visible right is emitted. The light from the
low temperature lamps appear reddish yellow
while that from the high temperature lamps
has a white appearance. The filament acts as
an electrical filament resistor, that dissipates
power proportional to the product of the
voltage applied and the current through the
filament. When the power supplied is sufficient
to raise the temperature to above 1 000 K, visible
right is produced. As the power dissipated
Fig. 10.2: Vacuum lamp
is increased, the amount of light produced
increases.
281
(b) LED lamps

A LED lamp is made using light emitting diodes (LED). An LED consists of a
junction diode made from a semi conductor material usually gallium arsenide
phosphide. When current is passed through the diode, it emits light. LED lamps
are cheap and highly efficient because they emit almost no heat.
Fig. 10.3: LED lamp
(c) Gas-filled lamps

Gas-filled lamps produce light from an incandescent filament operated in an
inert gas atmosphere. The inert gas suppresses the evaporation of the tungsten
filament, increasing the lifetime of the lamp and allow the lamp to operate at
higher temperature. The commonly used gases are neon, argon xenon, sodium,
mercury. The cost of gas-filled lamps depends on the gases used. For instance,
the one filled with xenon is more expensive due to its low natural abundance.
The advantage of the higher atomic weight gases is that they suppress the
evaporation of the tungsten filament more effectively than the lower weight gases.
This allows the filament of gas filled lamps to run at temperature up to 3200 K
and achieve reasonable life times. The light from these lamps has a high blue
content giving the light a pure white appearance.
The disadvantage of a gas filled lamps is that they require more power to achieve the
same temperature than incandescent lamps.
Fig. 10.4: Gas filled lamps
282
Safe energy!
Use energy saver lamps on lighting system in your homes to minimise electricity
bills.
10.2.2 Fuses and circuit breakers

(a) Fuses
Activity 10.3 To find out the function of a fuse and interpret power
ratings of fuses
Material
• Fuses of different rates
Steps
1. Tell your class partner what a fuse is and its function in an electrical circuit.
2. Keenly observe the fuses provided to you. Check and record their voltage
and current ratings.
3. Determine the amount of the current allowed by each fuse to pass through
without breaking.
A fuse is a short thin piece of wire of low melting point. The wire melts as soon
as the current through it exceeds its rated value. Fig. 10.5 shows pictures of fuses.
Fig. 10.5: A fuse
Fuses are usually fitted in all electrical circuits to prevent dangerously large
currents from flowing. They melt or “blow off” and stops the flow of current
hence protecting the electrical appliances against the risk of fire caused by the
heat. The fuse should be therefore fitted on the live wire.
283
Fuse rating
Fuse rating is the current needed to blow (melt) the fuse. It is usually printed
on the side of the fuse. It is usually defined in ‘amperes’, which are the unit of
measuring electrical current (see Fig.10.6)
Fig. 10.6: Fuse rating
The fuse used in any electrical appliance should be of a value just slightly higher
than the normal current required by the appliance. The common standard values
of available fuses are 2 A, 5 A and 13 A, although 1 A, 3 A, 7 A and 10 A fuses
are also made. If the power rating of an electrical appliance is ‘2 000 W, 250 V’,
the required current through it is 8 A. The correct fuse to protect the appliance is
10 A. Similarly if the required current for an appliance is 4 A, the correct fuse to
be used is 5 A.
(b) Circuit breakers
Activity 10.4 To describe the working of circuit breakers

Materials
• School circuit breaker • Internet
• Reference books
Steps
1. With the permission of your teacher, switch on and off the circuit breaker.
What do you notice? Explain.
2. Tell your group members what a circuit breakers is and discuss how it works.
3. Do research from the internet or reference books on how a circuit breaker
works.
4. Share your findings to the whole class.
284
A circuit breaker (Fig. 10.7) is an automatically operated electrical switch designed

to protect an electrical circuit from damage caused by either excess current,
overload or short circuit.
Fig. 10.7: Circuit breaker
The basic function of a circuit breaker is to put off the circuit to discontinue current
flow after a fault has been detected. Unlike a fuse which operates once and then
must be replaced, a circuit breaker can be reset (either manually or automatically)
to resume normal current flow.
Exercise 10.1
1. Table 10.2 shows standard symbols for electrical installation. Fill in the
appropriate name or symbol.
Name Standard symbol
(a) Bulb/lamp
(b)
(c) Fuse
(d)
(e) Capacitor
(f) D.C Voltage
(g)
(h) Circuit breaker
Table 10.2
285
2. Explain why tungsten is used in lamps and not any other metal.
3. What is a fuse? Explain its function in an electrical circuit.
4. A microwave is rated 1 500 W, 240 V. What is the appropriate fuse used in its
circuit?
5. What is a circuit breaker? Explain how it functions.
6. Differentiate between a fuse and a circuit breaker.
10.3 Types of electrical cables and their sizes
Activity 10.5 To find out the types of electrical cables and their
standard sizes
Materials
• Electrical cables • Internet • Reference books
Steps
1. Remove the outer jacket of the cable provided to you. How many wire are
there? Write down their colours.
2. Suggest the name of each wire in the cable.
3. Now conduct a research, find out how the wire are connected to the electrical
application and the sizes of the cables that are there.
More often than not, the term wire and cable are used to describe the same thing,
but they are actually quite different. A wire is a single electrical conductor whereas
a cable is a group of wires covered in a sheath.
A cable usually has three wires namely the live wire (L), neutral wire (N) and
Earth wire.
Fig. 10.8 shows a cable with the live, neutral and earth wires.
Fig. 10.8: Electrical cable

The domestic supply in Rwanda is mainly 240 V ac with a frequency of 50 Hz.
This is supplied by two cables from a local sub-station (for example; Kigoma
Substation) to different homes, industries and offices for consumption. Let us
now learn these cables in details.
286
10.3.1 Live (L), neutral (N) and earth (E) wires

The live wire may be likened to the positive terminal of a cell or a battery and the
neutral wire to the negative terminal (Fig. 10.9 (a) and (b)).
Lamp Lamp
I I I
Neutral
I
L N
Live
I + –
(a) a.c (b) d.c
Fig. 10.9: Live and neutral wires
All electrical appliances need a live and a neutral wire to form a complete circuit
from the power supply through the appliance and back to the power supply. The
live wire delivers the current to the appliance.
II It is dangerous, because it is capable
of giving electric shocks, if handled carelessly. Switches in a circuit should be fitted in
the live wire, so that when the switch is off, the high voltage is disconnected from
the appliance. The current returns to the supply through the neutral wire. Some
electrical appliances have a third wire known as the earth wire (E) for safety (is
discussed later in the unit).
10.3.2 Colour codes for the wires used in house circuits
The insulation, usually of plastic, on the three wires of a cable is distinctively
coloured to denote the live, neutral and earth wires. The basic idea of using
different colours is to easily identify the wires so that correct connections are made
with care. The present international convention is brown for live, blue for neutral
and green with yellow stripes for earth. Electrical wiring should be checked to
ensure that the earth wire lead to (connected to) the metal case of the appliance.
10. 3.3 Standard size of electrical cables
Electrical cables come in different sizes. It is therefore important to select
appropriate sizes ( a process called sizing) for electrical power cable conductors.
The proper sizing of cables is important to ensure that the cable can:
1. Operate continuously under full load without being damaged.
2. Provide the load with a suitable voltage (and avoid excessive voltage drop).
3. Withstand the worst short-circuit current flowing through the cables.
Methods of cable sizing differ across international standards and some standards
emphasize certain things over others. However, the general principles that underpin
287
cable sizing calculation do not change. When sizing a cable, the following general
process is typically followed.
1. Determine the minimum cable size based on continuous current carrying
capacity.
2. Determine the minimum cable size based on voltage drop consideration.
3. Gather data about the cable, its installation condition, the load that it will
carry etc.
Table 10.3 shows cross-sectional area, maximum current capacity and voltage
drop of some electrical cables.
Table 10.3
An electrical cable can be fitted to a plug on one side and the other side to thee
electrical appliances like electrical iron, immersion heater or refrigerator.
Fig. 10.10 shows a 3-pin plug. It is usually marked with letters L, N and E to
stand for live, neutral and earth respectively.
Earth wire
E
(green/yellow) Earth pin
Fuse
Neutral wire L
Neutral pin
(blue)
Case clip Live wire Live pin

(brown)
Cable with 3 wires
(a) Inside view (b) Side view
Fig. 10.10: 3-pin plug
Note that the earth pin is slightly longer than the other two pins and that the
live pin is on the right hand side of the plug when connected to the socket.
288
10.3.4 Earth connection

The earth wire connects the metal case of an appliance (e.g. an electric iron) to
the ground and prevents it from becoming live, if a fault develops. If, for example,
the cable insulation wears out due to the heating effect of the current, there are
chances that a few fine strands of the bare live wire could touch the metal case.
When such a fault occurs, a current flows through the live wire and the earth wire
in series. The fuse in the live wire will blow and cut off the power supply. If on
the other hand, there was no earth wire connection, a person touching the metal
case would get an electric shock.
In appliances like television set, record player, etc. the outer case is not metallic
and hence 2-pin plugs are sufficient. It is dangerous to use the 2-pin plug with
any appliance which has an outer metal case.
10.3.5 Short circuits

If a few strands of the fine bare live wire touch, by chance, those of neutral wire, a
large current can flow between the live and the neutral wires of the supply cables.
This is due to the fact that current tends to take the path of least resistance. This
is called short-circuiting of the appliance.
On such occasions, the fuse usually blows off. Otherwise if no fuse was filted in
the circuit, the ‘sparking’ produced by the large current might burn the cable and
there are risks of fire being produced.
In a socket for 3-pin plug, the holes for the live and the neutral pins are usually closed
by an insulating material called a ‘blind’ (Fig. 10.11). This is a safety measure,
especially to children who may like to play with the circuit and might cause short
circuiting by putting wires in the socket. The ‘blinds’ are opened by the longer
earth pin of the 3-pin plug. The moment the earth pin touches and opens the
socket, any leakage current through the metal case will straightaway be earthed
hence making the appliance safe.
Insulating
material (blind)
Fig. 10.11: A socket
289
10.4 Household wiring

Activity 10.6 To describe domestic wiring system
Material:
• Chart showing domestic wiring system
Steps
1. Take a walk around the school premises and identify where the following
electrical components are situated:
(a) Main switch (b) Meter box
(c) Sockets (d) Circuit beakers
2. Discuss with your classmate how wiring is done in the school premises.
3. Now look at the chart provided to you. Discuss how domestic wiring is done.
How accurate were you before looking at the chart?
4. With the assistance of your teacher, make notes on your findings in your
notebook.
Fig. 10.12 shows the drawing for the domestic wiring system consisting of the
following: the main fuse, electricity meter, consumer unit or the fuse box, lighting
circuit and the ring main circuit.
N Electricity meter
Main fuse
L
Main switch
Consumer unit Lighting
5A circuit
fuse
30 A
30 A
15 A
L N
Cooker Heater
E
Staircase switch
N
L
Ring main circuit
E
S
N L
S S
N L N L
Fig. 10.12: Domestic wiring system
290
Electricity is supplied from a transformer to the house via two wire ( L and N)
cables. Earthing for one of the wires is done at the transformer. It then goes
through a fuse which usually differ depending on the amount of the current
required. It is then wired to the meter box which contains all the fuses and circuit
breakers.
The circuit breakers are normally labelled clearly to show to which each circuit
breaker belongs. Wiring for each part of the house is done starting at this unit
box also referred to as consumer unit.
Every circuit is connected in parallel with the power supply, i.e. across the live and the
neutral wires. Every circuit receives 240 V ac. There is no connection between the
live and the neutral wires except through an electrical appliance.
The electricity meter records the electric energy consumed in the whole house.
The consumer unit is a junction box which distributes current to several separate
circuits. The consumer unit also houses the main switch which can switch off all
the circuits in the house, if required.
The lighting circuit contains lights for different places and the 2-way switches for
places like the staircases. Each lamp is connected in parallel at a suitable point
along the cable. The lighting circuit does not require the earth connection, as the
current is normally quite low.
The ring main circuit provides parallel circuit connections to each electrical
appliance plugged into the sockets. Since the current drawn is high, the ring main
circuit incorporates the earth wire connection.
Activity 10.7 Replacing a single flash socket with a double socket

Materials
• A test light • Wire strippers • Wire cutters
• Double cutlet sockets • Screw drivers
Procedure
1. Switch off the power at the fuse box. Test the circuit with the test light to
ensure that there is no current flowing.
2. Unscrew the fuse plate and disconnect the cables from the terminals of the
single socket. Using the wire cutters cut off the extensions of the wires you
have unscrewed.
3. Using the wire strippers, strip off the wire casing and prepare the wires.
4. After preparing the wires of the junction box, screw the black wire to the
socket terminal with a neutral sign, the red wire to the terminal with live sign,
the green wire to the terminal with earthing sign.
5. Fit the new double socket on the wall and use the test light to conform its
correctly wired.
291
Installation of lightning arrestor
Activity 10.8 To discuss the importance of lightning arrestor and

how to install it.
Material
• A house or picture of a house filted with a lightning arrestor
• Reference books or Internet
Steps
1. Identity any house in your school compound or the neighbourhood with a
lightning arrestor.
2. Take your group members to the house and observe how the lightning
protection system has been fitted. Discuss how it works.
3. On going back to class, confirm the facts by conducting a research from the
Internet or reference book.
4. Give a summarised report of your findings to the whole class through the
secretary.
A lightning protection system performs a simple task. It provides a specific path

through which lightning can travel, see Fig 10. 13.
Lighting
Lighting rod
Lighting rod mounting base
Copper cable
Copper cable strap
Ground rod clamp

GEM
Ground enhancement
Ground rod
material
Fig. 10. 13: Lightning arrestor
292
When a house is filted with a lightning arrestor, the destructive power of a

lightning strike is directed safely to ground leaving the home, family members
and property unharmed.
Copper and aluminium cables are the most appropriate for use as the lightning
rods (Fig.10.13). Copper is the material of choices in the conductor cable since
it is a better conductor of electricity. Copper wire offers a higher fusing current
to discharge lightning and fault current into the ground.
Note: Electric grounding cable should not be used.
A lightning protection system should include the following primary component

which work together to prevent lightning damage.
1. Lightning rods (Air terminals)
2. Braided conductors (cables)
3. Bonds or metallic bodies
4. Ground rods or ground plates
5. Surge arrestor
The arrestor is the first line of defence against harmful electrical surges that can
enter a structure through power lines. On the other hands, lightning rods protect
the structure from a direct lightning strike.
To ensure the highest level of protection of buildings, the following general design
rule should be followed:
1. Lightning protection system shall be applied to metal covered buildings in
like manner as on building without metal coverings.
2. All buildings must have two groundings as widely separated as possible,
preferably at diagonally opposite corners if perimeter distance around the
buildings at ground level is 250 feet or less (1 ft = 2.54 cm).
3. Cables should be free of sharp turns. They should remain horizontal or in a
downward path towards the ground.
4. If building perimeter is between 250 and 350 feet , then three groundings
are required. If it is between 350 feet and 450 feet, then four grounding is
required etc.
293
Caution
• Copper equipment should not be used on aluminium roofs, aluminium

siding or other aluminium surfaces including bare galvanisied steel.
• Aluminium equipment should not be used underground.
• Aluminium equipment should not be used on copper roofing or other
copper surface.
• Copper and aluminium conductor should not be interconnected except
with acceptable bimetallic connectors.
10.5 Dangers of electricity
Activity 10.9 To describe the dangers of electricity and safety

measures
Materials
• Reference books • Internet enabled computer
Steps
1. Discuss the possible electrical hazards in our home and safety measures
needed to be taken.
2. Confirm your faults in step 1 above by conducting a research from the
internet or reference books on the dangers of electricity.
3. Present a brief report on your findings to the whole class through your group
secretary.
Working with electricity can be dangerous. Engineers, electricians and other

workers deal with electricity directly including on overhead lines, electrical
installation and circuit assembling. Users like office workers, farmers and
construction workers deal with electricity indirectly.
Some of us may have experienced some form of shock where electricity causes
our body to experience pain or trauma.
Those of us who were fortunate, the extend of that experience was limited to
tingles or jolts of pain. But electrical shocks from electrical circuits are capable
of delivering high power to loads, which cause much more serious damage.
It is therefore important to be aware of hazards brought about by electricity and
how to avoid them in order to be safe.
294
Electrical hazards
A hazard is a situation that poses a threat to life, health, property or environment.
The following are some common electrical hazards in our homes, offices and
factories.
• Poor wiring and defective electric wires can lead to electric shock and fires.
• Water outlets being close to electric outlets.
• Pouring water on electrical fire. This can lead to electric shock.
• Covering electrical cords and wires with heavy cover can lead to overheating.
• Overloading the outlets leading to overheating and electrical fire.
• Use of long extension cords which can cause tripping or accident.
• Touching electrical appliances with wet hands leading to shocks.
• Broken sockets and electrical appliances leading to electric shock.
10.6 Electrical safety

Every electricity user should observe safety measures when using electricity and
electrical appliances. The following are some of the electrical safety measures.
• Do not touch naked electric cables with bare hands to avoid electric shock.
• Always pay attention to the warning signals given out by your appliances e.g.
if a circuit breaker repeatedly trips, you should confirm the problem.
Be safe
Whenever you see this electricity sign it warns

you to keep off. You may be electrocuted.
• Use the right size circuit breakers and fuses to avoid overloading.
• Ensure that potentially dangerous electrical devices or naked wires are out of
reach of children.
• You should avoid cube taps and other outlet-stretching devices.
• Always replace broken plugs and naked wires.
• Use the correct appliances in a socket to avoid overload.
295
Exercise 10.2
1. State the international standard colours for the live (line), neutral and earth
leads of a 3-core flex.
2. Define 'fuse' and state its function in an electrical circuit.
3. Sketch and label a three pin plug.
4. Explain why the earth connection is important.
5. (a) Explain how to install lightining arrestor in a house.
(b) Highlight the general design rule that must be followed to ensure highest
level of electrical safety of morden houses.
6. A laboratory building in a of Voluntary Counselling and Testing Centre
(VCT) is to be supplied with electricity. Briefly explain how wiring would be
done in the laboratory building for effective supply of electricity.

• An electronic symbol is a pictogram used to represents various electrical
and electronic devices in a schematic diagram of an electrical or electronic
circuit.
• A fuse is a short thin wire of low melting point. It melts when a large current
flows through it hence breaking the circuit.
• The earth wire is connected to the ground and prevent an appliance or house
from becoming live. This is important incase there is an electrical fault.
• Electric Lamps are of many types. They are categorised into three major
groups namely: Incandescent, LED and gas-discharge lamps.
• A circuit breaker cuts off the flow of current when there is an electric fault
within the circuit thus keeping the premises and appliances safe from electric
fire.
• When wiring a house, there are many types of wire to choose from, some
copper, others aluminium, some rated for outdoors, others indoors.
• Every electricity user should be aware of electrical hazards and practice
safety measures.
296
Unit Test 10
1. (a) Name five electrical components used in house wiring.
(b) Draw the standard electrical symbols used for each of the component
named in (a)
(c) Briefly explain the functions of each of the component you named in (a)
above.
2. Explain why the earth connection is so important to appliance at home.
3. Fig.10.14 shows an electrical cable. Name the earth, live and neutral wire.
Fig. 10.14: Electrical cable
4. (a) Name three types of electric lamps.

(b) List any two gases used in lamps.
(c) What is the purpose of the presence of the gases inside the bulb?
5. Brief explain the function of the following electrical component in an electric
circuit
(a) a circuit breaker
(b) a fuse
6. Describe briefly how you can do electrical wiring in a house.
7. Mukantagara a student in Senior 3 saw an electric post with cables collapse
and block the path way. Describe some of the safety measures she and the
people near the cable should observe for them to be safe from any harm.
8. Explain why a fuse is always fitted to the live wire.
9. State three necessary precautions to be taken when connecting a metal-
framed electrical appliance to the mains power supply?
10. An electric iron rated 240 V, 750 W is to be connected to a 240 V mains supply
through a 3 A fuse. Determine whether the fuse is suitable or not.
11. Find the maximum number of 75 W bulbs that can be connected to a 3A fuse on a
mains power supply of 240 V.
12. In the circuit shown in Figure 10.15, each bulb is rated 6 V, 3 W.
297
Fig. 10.15
(a) Calculate the current through each bulb when the bulbs are working
normally.
(b) Is a 3 A fuse suitable for the circuit when all the switches are closed?
(c) Calculate the power delivered by the power supply
(d) What is the advantage of connecting all the bulbs in parallel rather than in
series?
14. Figure 10.16 shows a staircase double switch.
Fixed point
A
1 4
2 3
B
Fixed point
Fig. 10.16
In Table 10.4, write down whether the lamp will be ON or OFF for the various
combinations of switch positions.
Table 10.4
Position of Position of Lamp

switch A switch B ON/OFF
1 3
1 2
4 3
4 2
298
Basic alternating current circuits
UNIT 11 Basic alternating current circuits
Key Unit Competence

By the end of the unit, the learner should be able to design and analyse simple
alternating current circuits.
Learning objectives

• Identify circuit symbols representing electrical components.
• Describe design of electrical circuits using different electrical components.
• Design an electric circuit consisting of AC voltage and inductor, resistor and
capacitor.
• Describe the function of inductor in an electric circuit.
Skills
• Identify and apply electrical components in AC circuits

• Differentiate between an alternating current and a direct current.
• Explain function of inductor, resistor and capacitor in electric circuits.
• Manipulate apparatus and equipment, given procedures, to obtain data.
• Evaluate and draw conclusion using experimental results.
Attitude and value

• Appreciate applications of electric circuits.
• Be aware of the safety precaution to be observed while handling electrical circuits.
• Appreciate advantage of a.c. over d.c.
299
Introduction
Unit Focus Activity

Fig. 11.1 shows three different electronic components

(a) (b) (c)
Fig. 11.1: Pictures of electronic components
1. Identify each component.
2. Describe the function of each component in a circuit.
3. Sketch a symbol circuit diagram showing the three components connected
in series with an a.c supply.
4. Sketch the graph of the voltage signal output from the circuit in step 3 above.
In unit 7, we discussed and defined what alternative current (a.c) is. In this unit,
we are going to design and analyse simple alternating circuits.
300
11.1 Standard symbols used in electric circuit and function
Activity 11.1 To draw circuits symbols for electronic components

Materials
• Cell • Connecting wires • Inductor
• Bulb • Resistors • Switch
• Capacitor • A.C source • D.C source
Steps
1. Your teacher will provide you with the above circuit components.
2. Briefly describe the function of each in a circuit. Summarise your report in a
table.
3. Suggest three precautions that one should observe when designing and
connecting circuits.
4. Compare your findings with those of other groups.
We have already learnt some common electrical symbols used in d.c circuit. Let
us now consider some components used in simple a.c circuits namely inductor,
resistor, a.c power source, capacitor, wires.
The table 11.1 summarizes these symbols and their function.
Table 11.1: Summary of electrical components, symbol and function.
Circuit symbol Component name Function

Used to connect one
Electrical wire
component to another.
To show connected
Connected wires
crossing
To show wires are not

Not connected wires
connected
Disconnects current
Switch
when open
Used for zero potential
Earth ground and electrical shock
protection
Used to restrict the
Resistor (R) amount of current flow
through a device.
301
Adjustable resistor has

Potentiometer 3 terminals – used to
divide electric potential
Adjustable resistor has

Variable resistor/
2 terminals – used to
rheostat
set the resistance
Used to measure
Photoresitor/light
resistance that changes
dependent resistor
with the change in light
LDR
intensity
It acts as short circuit
Capacitor (C) with a.c and open with
d.c
Used to measure the
V Voltmeter
voltage.
Used to measure the
A Ameter
current.
Used to measure very
Galvanometer
small current.
Coil/solenoid that
Inductor (L)
generates magnetic field
Used in the tuned
Variable inductor circuit of radio
transmitter
Indicates current flow
Indicator Lamp
by lighting.
a.c supply Supplies alter nating
voltage.
11.2 Differences between alternating current (a.c) and direct

current (d.c)
Activity 11.2 To differentiate between d.c and a.c sources
Materials
• Dry cells • Galvanometer
• Bicycle dynamo
302
Steps
1. Connect a dry cell across a centre-zero milliammeter and a resistor,
making sure the positive polarity is connected to the positive polarity of the
milliammeter. Note and explain what happens to the pointer.
2. Connect a bicycle dynamo in series with a centre-zero galvanometer and
resistor. Make the dynamo turn by use of the wheel. Observe and explain
what happens to the galvanometer pointer.
3. Compare and discuss your observation in steps 1 and 2 with your classmates.
Fig. 11.2 shows an electric circuit of d.c and a.c sources respectively.
M.A M.A
(a) d.c Source (b) a.c Source

Fig. 11.2: The d.c and a.c source
The polarity for d.c source remains constant making the current to flow in only
one direction while that of a.c source switches polarity from positive to negative
and back again over time. This makes the current with respect to the voltage
oscillates back and forth.
The oscillating shape of a.c supply follows that of mathematical form of a sine
wave. For d.c, voltage is the same as that across the resistor. Vdc = VR. For the a.c,
the Vac = Vpeak Sinωt.
When working with a.c voltages and currents, we use the r.m.s. values which is the
equivalent steady d.c (constant) value which gives the same effects. For example,
a lamp connected to a 6 V r.m.s a.c supply will shine with the same brightness
when connected to a steady supply by d.c supply. Table 11.2 summarises the main
differences between ac and dc.
Table 11.2 difference between d.c and a.c power source
d.c a.c
1. Flows in one direction in the circuit Reverses its direction periodically while
and has of constant magnitude continuously varies in magnitude.
current
current
current currentcurrent
current
constant
constant or vary
vary in
in vary in
or
or
or constant oror
voltage magnitude
magnitude
magnitude voltagevoltage
voltage magni
magnitude magnitude
voltage
voltage time
time time
tude
tud. tud.
time
time
time
303 mA
centre zero mA
centre zero
A
2. d.c voltage cannot travel very long Easier, safe and cheaper to transfer
distance due to energy losses. over long distances.
3. Frequency = 0 Hz Frequency = 50 H3 or 60 H3 depending
on the country.
4. Obtained from a cell or battery Obtained from a.c generator and the
mains
5. Hindrance to flow of current in d.c Hinderance to flow of current in a.c
circuit called resistance circuit is called impendence
6. Power factor is always 1 Power factor lies between 0 and 1
7. Can be stored in batteries Cannot be stored.
11.3 The circuit for analysing resistors, capacitors and inductors

in a.c circuits
In order to compare the behavior of a resistor, inductor and capacitor in an
a.c circuit, we need to design a circuit diagram that can be used by the three
components.The source of a.c should be such that its voltage supply and frequency
can be varied. Hence a low frequency generator is recommended.
Activity 11.3 To design a circuit for analysing a.c circuits

involving resistors, capacitors and inductors
Materials
• Low frequency generator • Carbon resistor
• d.c voltmeter (centre – zero) • Frodile clips
• d.c ammeter (milliammeter and centre – zero)
• Inductor
current •
or constant
current
or
Capacitor vary in
magnitude voltage magnitude
voltage time
• Stop watch • Connecting wires tud.
time
Steps
Connect the circuit as shown in Fig. 11.3
mA
mA
centre zero
centre-zero
A
centre-zero Terminals for
V centre-zero
B connecting
B
min
min max
max components
switch
switch
low
Lowfrequency generator
frequency generator
(A Csource)
(A.C source)
Fig 11.3 circuit diagram to analyse resistors, capacitor and inductors in a.c circuits.
The circuit is now ready for use

power
Sin2 wt
(W)
304
11.4 A single resistor connected in series to an a.c source.
11.4.1 Variation of current and voltage with time
Activity 11.4 To analyse the behaviour of a single resistor

connected to an a.c source
Materials
• A carbon reister (2 kΩ) • Stop watch • Crodile clips
• Connecting wires
Steps
1. Set the low frequency generator to the minimum value.
2. Connect a carbon resistor between points A and B in the circuit made in
Activity 11.3.
3. Connect the low frequency generator to the main (make sure the generator is
“off”). Switch on the generator.
4. Set the generator at a frequency where both the milliammeter and voltmeter
show a reading. What happens to this reading with time.
5. Record the variation of current and voltage with time Table 11.3.
Table 11.3.
Time (S) current (mA) Voltage (V) V power (VI)

I
1
2
3
4
5
6. Repeat the experiment by increasing the frequency of a.c signal.
7. On the same axes, plot the graph of variation of current and voltage with
time.
V
8. Calculate the ratio of I
at various time.
9. Calculate the power in the resistor.
10. Show on the graph the variation of current and voltage with time. Also show
how the power varies with time.
305
When a single pure resistor is connected in series to an a.c source, the current
through it and voltage across it each vary sinusoidaly with time as shown in Fig.
11.4.
+V0 ε
voltage (v) I
current +I0
Time(S)
-I0
-V0
1 cycle
Fig. 11.4: Variation of current and also voltage with time for a resistor in a.c circuit
1 1
During 2 cycle, the current flows in one direction and in the other 2 cycle
the current flows in the opposite direction. The e.m.f and the current reach the
maximum and minimum at the same time. When this happens, we say that the
current and voltage are in phase. In such a case, we say the e.m.f. and the current
are directly proportional. This is the same case with a d.c. circuit, except that the
values of e.m.f. and current do not change.
Hence, Ohm’s law can be used in both d.c and a.c circuit to calculate resistance,
voltages, current and power. The ratio of voltage to the current gives the resistance
of the resistor to the flow of a.c current. In a.c, this resistive nature is called
impedance (Z). The units of impedance are ohms (Ω).
V(r.m.s)
Impendance Z = I
(r.m.s)
11.4.2 Electrical resistance and frequency
Activity 11.5 To show the relationship between resistance and

frequency
Materials
• Ammeter • Voltmeter • Bulb
• Connecting wires • Low frequency generator
Steps
1. Connect the circuit shown in Figure 11.5
306
Low frequency
generator
Fig. 11.5 relationship between resistance and the frequency signal
2. Note the brightness of the bulb.

3. Change the frequency of a.c. supply and again note the brightness of bulb.
What do you observe on the brightness of the bulb? Explain.
The frequency of the a.c source (at constant voltage) does not affect the brightness
of the bulb. Since the brightness is a measure of the resistance, we can conclude
that the frequency of an a.c source does not affect the resistance (impendance of
the circuit). This can be shown by the graph Fig. 11.6
+
impendance (Ω)
Frequency (Hz)
Fig. 11.6: A graph of resistance against frequency
11.4.3 Power in a.c circuit

The voltage and current reach their maximum and minimum value at the same
time. The instantaneous current and voltage are.
V = Vo sin ωt and I = Io sin ωt
Power = V0 sin ωt × I0 sin ωt = V0 T0 sin2 ωt
Fig. 11.7 shows the varitiom of current, voltage, power with time plotted on the
same axes.
P Pmax
max
Power
Power Sin
Sin2 wt
2
wt
Averagepower
Average power
V Vmax
max Y
V Sin
Sin wt
wt
I max 1I Sin wt
wt
1 max
Time (s)
time (S)
pure resistor
Fig. 11.7 Variation of current, voltage and power with time
centre-zero galvanometre
307
wire
bulb
The power dissipated in a pure resistor connected to a.c supply is given a

V2r.m.s
P = V (r.ms) × I r.ms = 1
2
r.m.s
R = R
Example 11.1
A 1000 W heater is connected to a 250 V a.c. supply voltage. Calculate:
(a) the current taken from the mains,
(b) the impedance (a.c resistance) of the heater when it is hot.
Solution
P 1000 W
I = V = 250 V = 4 A
V 250
Z= I = 4 = 62.5 Ω
Example 11.2
Find the power being consumed by a 100 Ω resistive element connected a
240 V a.c supply.
Solution
P max
Vresistor = Vsupply (only one component is connected)
Power Sin 2
wt
Average power
VR V max
240 V Sin wt
I = R =I max
100 = 2.4 A I Sin wt
P = I2R = 2.42 × 100 = 576 W time (S)
Exercise 11.1
1. Differentiate between resistance (R) and Impendance (Z)

2. Match the following symbols with their component.
pure
pureresistor
resistor
centre-zero galvanometer
galvanometre
wire
wire
bulb
n AA.C
C power supply
power supply
I V P
X 308
(A) (r) (W) Y
Z
wire
bulb
: 3. Using the standard symbols, draw a simple circuit showing a resistor, bulb,
connected to A.C power supply.
A.C power supply
n (v) and power (P) through
4. The variation of current (I), voltage a resistor
connected to a low frequency generator.
X
II V
V P
(A) (r) (W) Y
time
time(s)
(S)
Fig. 11.8
Identify which curve represents:

(a) Current (b) Voltage (c) Power
5. The current through a resistor in an a. c circuit in given by I = Io sin ωt.
(a) Explain the various symbols and terms in this expression.
(b) Sketch a graph representing this expression.
(c) Write down the expression for instantenous voltage across the resistor.
6. A direct current of 2.0A passes through a resistor of value R Ω. Calculate the
peak value of an a.c which produces three times the heat per second as the
2.0A:
2.0 A
RΩ
7. A low frequency a.c source gives a signal output of 4V peak value when
connected to a 100 Ω resister R.
(a) Draw a phasor diagram for this connection.
(b) Calculate the current through R in milliamperes.
8. The Fig 11.9 shows a resistor R connected to a signal generator while Fig.11.10
shows how the voltage (V) varies with time.
309
Voltage V
voltage
Voltage v
V
2.0 2.0
1.0
1.0
0
0 time (S)
R=Ω –1.0
time
R=Ω –2.0 –1.0
–2.00.01
0.01 0.02
0.02 0.03
0.03 0.04
0.04 0.05
0.05 0.06
0.06 0.07
0.07 0.08
0.08
Fig. 11.9 Fig. 11.10

(a) Find the frequency of the signal. 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08
(b) Calculate the value Vr.m.s. IL

MA
(c) Sketch the variation of current in R with time t for the same time interval.
Low
(d) Use one frequency
or both of theV graph to determine the power dissipated in R when
a bgenerator c d e
t = 0.0150.
IL
MA time (S)
11.5 A single capacitor connected in series to an a.c source

Low
frequency V
To describe inductor connected in series with a.c
Activity 11.6
generator
supply source
Materials.
• Capacitor • Crocodile clips • Milliameter
• Connecting wires • Voltmeter • A.C. power supply
Procedure a b c d e
90º 180º 270º 360º 360º time (S)
• Repeat activity 11.3 but use a capacitor across A and B as shown in Fig 11.11
MA
one cycle
Low
Low
frequency
frequency V
generator
generator
a b c d e
90º 180º
Fig.270º
11.11 360º 360º time (S)
When the capacitor is connected directly across an a.c supply, it will alternately
charge and discharge at a rate determined by the frequency of the supply. The
capacitance in a.c circuits varies with frequency,
one cycle as the capacitor is being alternately
310
I
charged and discharged. The bulb in the circuit will light due to the charging and
discharging current that passes through it.
The charging current is directly proportional to the rate of change of the voltage
across the plates, with the rate of charge at its greatest as the supply voltage crosses
over from its positive half cycle to its negative half cycle or vice versa at points 0°
and 180° along the sine wave.
Consequently, the least voltage change occurs when the a.c sine wave crosses over
at its maximum or minimum peak voltage level. At these positions in the cycle,
the maximum or minimum currents are flowing through the capacitor circuit.
See Fig. 11.12.
Vmax V
Imax I
90° 180° 360°

0°
ωt
1f 1f 1f 1f
4 4 4 4
Charge Discharge Charge Discharge
Fig. 11.12: A graph showing charging and discharging at a capacitor in a.c circuit
At 0°, the rate of change of the supply voltage is increasing in the positive direction
resulting in a peak value at 90°. For a very brief instant in time, the supply voltage
is neither increasing nor decreasing so there is no current flowing through the
circuit.
As the applied voltage begins to decrease to zero at 180°, the slope of the voltage
is negative so the capacitor discharges in the negative direction. The bulb's lighting
increases slowly. At the 180° point, the rate of change of the voltage is at its
maximum again so maximum current flows at that instant and so on.
We can therefore say that for capacitors in an a.c circuits, the instantaneous current
is at its minimum or zero whenever the applied voltage is at its maximum and
likewise the instantaneous value of the current is at its maximum or peak value
when the applied voltage is at its minimum or zero.
From the waveform above, we can see that the current is leading the voltage by
1
cycle.
4
311
Like resistors, capacitors also offer some form of resistance against the flow of
current through the circuit. The resistance to the flow of a.c current by a capacitor
is known as reactance xc or capacitive reactance, its SI unit is ohms (Ω). Capacitive
Vr.m.s
Reactance in an a.c is given by: Xc = I (ohms).
r.m.s
In a graph of Vr.m.s against Ir.m.s (Fig. 11.13) the slope gives the reactance.
Vr.m.s
Slope = Xc
Ir.m.s
Fig. 11.13: Graph of Vr.m.s against Ir.m.s for capacitor.
Capacitive reactance, current, capacitance and frequency

The capacitive reactance of the capacitor decreases as the frequency across it
increases. Therefore, capacitance is inversely proportional to frequency, that is
1
XC α f
The graphs in Fig. 11.14(a), (b) and (c) show the variation of capacitive reactance,
against current, frequency of a.c source and capacitance.
XC XC XC
(Ω) (Ω) (Ω)
( c is constant) (f is constant)
XC
Current (A) Frequency Hz C (µf)
(a) (b) (c)
Fig: 11.14: A graph of capacitive reactance against current and frequency and capacitance.
It has been shown that.

1 1
XC = 2πf C ⇒ XC = Cω (since 2πf = ω)
Where, f is in Herts (Hz) and C is in Farads (F), and angular frequency ω is in
radians per second.
Looking at the graph in Fig. 11.14 (b), we see that as the frequency increases,
312
the current flowing through the capacitor increases in value because the rate of
voltage change across its plates increases. At very high frequencies, a capacitor
has zero reactance (short-circuit).
From the graph in fig. 11.14(c), we see tha for a d.c supply, a capacitor has infinite
reactance (open-circuit).
Fig. 11.15 shows the variation of power in a capacitor in a.c circuit.
P(W)
P (W)
A B
0
time
time (s)
(S)
charging
charging discharging
discharging
Fig. 11.15:Variation of power in a capacitor in a.c.

In the first ¼ cycle, OA, power is drawn from the source and energy is stored in
the capacitor. In theVsecond quarter cycle
R
V AB, the energyV is being returned back
L L
to the source. The mean power is therefore zero over a cycle.
Example 11.3
resistor
resistor capacitor
When a parallel plate capacitor was connected to a 60 Hz a.c supply, it was found to
have a reactance of 390 ohms. Calculate the value of the capacitance in microfarads.
1 1
V
XC = 2πfC ⇒ C = 2πfX
C
MA 1
C = 2π × 60 × 390 = 6.8 µF
Example 11.4
(a) Draw a circuit with 100 nF connected to 2 Vr.m.s a.c source of 1 kHz supply.
(b) Calculate the:
(i) XC
(ii) Ir.m.s (Give your answer in appropriate form)
313
Solution
(a) The circuit is as shown in Fig. 11.16
C = 100 nF
V = 2 Vr.m.s
f = 1 kHz
Fig. 11.16: Electric circuit

1 1
(b) (i) XC = 2πfC = 2π × 1 × 103 × 100 × 10–9 = 1591.5 Ω
Vr.m.s 0.707 × 2V
(ii) Ir.m.s = = 1591.5 Ω = 4.44 × 10–4 A
XC
To give your answer in appropriate form, use mA, µA etc e.g.
Ir.m.s = 444 µA
Exercise 11.2
For questions 1 - 9. Select the most appropriate answer for the choices
given.
1. As the size of the plates in a capacitor increases, all other factors being
constant,
A. The value of XC increases negatively
B. The value of XC decreases negatively.
C. The value of XC does not change.
D. We cannot say what happens to XC without more data.
2. If the dielectric material between the plates of a capacitor is changed, all
other things being equal,
A. The value of XC increases negatively.
B. The value of XC decreases negatively
C. The value of XC does not change.
D. We cannot say what happens to XC without more data.
3. As the frequency of a wave gets lower, all other things being equal, the value
of XC for capacitor
A. Increases negatively. B. Decreases negatively.
C. Does not change. D. Depends on the current.
314
4. What is the reactance of a 330-pF capacitor at 800 kHz?

A. −1.66 Ω B. −0.00166 Ω
C. −603 Ω D. −603 kΩ
5. A capacitor has a reactance of −4.50 Ω at 377 Hz. What is its capacitance?
A. 9.39 µF B. 93.9 µF
C. 7.42 µF D. 74.2 µF
6. A 47-µF capacitor has a reactance of −47 Ω. What is the frequency?
A. 72 Hz B. 7.2 MHz
C. 0.000072 Hz D. 7.2 Hz
7. A capacitor has XC =−8800 Ω at f = 830 kHz. What is C?
A. 2.18 µF B. 21.8 pF
C. 0.00218 µF D. 2.18 pF
8. A capacitor has C = 166 pF at f = 400 kHz. What is XC ?
A. −2.4 kΩ B. −2.4 Ω
C. −2.4 × 10−6 Ω D. −2.4 MΩ
9. A capacitor has C = 4700 µF and XC =−33 Ω. What is f?
A. 1 Hz B. 10 Hz
C. 1 kHz D. 10 kHz
10. Briefly describe what happens when a capacitor is connected to an a.c source.
11. A sinusoidal p.d of r.m.s value of 10V is applied across a 20µF capacitor.
(a) What is the peak capacitance of the capacitor.
(b) Draw a graph of current on capacitor with time.
(c) 50Hz Calculate the r.ms. current through the capacitor.
12. (a) What do you understand by the term reactance.
(b) (i) Describe an experiment to investigate the dependence on frequency
of the reactance of a capacitor.
(ii) Sketch a graph to illustrate this dependence.
11.6 A single inductor connected in series to an a.c source

An inductor is basically a coil or loop of wire that is either wound around a hollow
tube former or wound around a ferromagnetic material.
315
a b c d e
Voltage V
time (S)
2.0
1.0 Basic alternating current circuits
11.6.1 Variation of voltage and current through an inductor in a.c

0
circuit
Activity 11.7 R To
= Ωanalyse the behaviour–1.0
of a single inductor in an
a.c circuit
Materials –2.0
• Inductor ` • Crocodile clips • Voltmeter

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0
• Connecting wires • Low a.c frequency generator • Ammeter
MA
Produre
Low
• Repeat activity 11.6 but use an inductor
frequencyacross A and
V
B instead of capacitor
as shown in Fig. 11.17 generator
IL
L2
mA
MA
Low
Low
frequency
frequency V
generator
generator
Fig. 11.17
When a single inductor is connected in series with an a.c source, the voltage (VL)
and current (IL) through the inductor vary with time as shown in Fig. 11.18.
Voltage
Voltage (VL) and Current (A)
I
current
V
a b c d e
90º0
90 180º0
180 270º0
270 360º0
360 360º0
360 time
time (s)
(S)
a b c d e
one cycle
one cycle
90º 180º 270º 360º 360º time (S)
Fig. 11.18:Variation of voltage and current with time for an inductor.
From the graph, we see that when the voltage become negative just after point
a, the current starts to decrease and become onezero
cycle at point b. The current then
becomes negative, following the voltage. The voltage becomes positive at point c
where it begins to make the current less negative. At point d the current becomes
zero again just as the voltage reaches its positive peak value to start another cycle.
Hence, when a sinusoidal voltage is applied to an inductor, the voltage leads the
316
current by ¼ of a cycle or by a 90o phase angle as shown. In other words the

current lags behind voltage.
When an inductor is connected across an a.c supply, the sinusoidal voltage will
cause a current to flow and rise from zero to its peak value. This rise or change in
the current induces a magnetic field within the coil which in turn will oppose or
resists this change in current in accordance with lenzs law. Hence, like a capacitor,
an inductor offers opposition to the flow of a.c.
This opposition to the flow of current in inductor called inductive reactance XL.
The unit symbol for inductive resistance (XL) is ohms (Ω)
Like in a resistor, the effective opposition to the flow of current in the inductor
to a.c in given by a version of ohms law Fig. 11.19.
Vr.m.s
Vr.m.s
(V) Slope = XL = I
r.m.s
Ir.m.s (A)
Fig. 11.19
11.6.2 Variation of inductive reactance XL, with frequency and

inductor
Activity 11.8 To investigate relationship between inductive

reactance and frequency.
• Repeat activity 11.3 but change the frequency of the low frequency generator,
keeping the inductance L constant.
• Repeat activity 11.3 but change the inductance L keeping frequency constant.
• Describe the relationship between XL and f, XL and L.
Fig. 11.20 and 11.21 shows variation of inductive reactance with inductance and
frequency respectively.
XL
XL(Ω)
L(H) f(HZ)

Fig. 11.20:Variation of XL with L Fig. 11.21:Variation of XL with f
317
From the graphs, we observe that increasing frequency (f) and inductance (L )
increase inductive reactance XL.
XL ∝ fL ⇒ XL = 2πfL
Since 2πf = ω then XL = ωL
Where ω is angular frequency in radians per second.
11.6.3 Variation of power with time in an inductor

Since P = VI, the variation of V and I shows give how the power in and an inductor
varies (Fig.11.22).
energy from sources
P(w) V + stored in inductor
0 A
B time seconds
energy returned to
source from inductor
Fig. 11.22:Variation of power with time in an inductor.

In the first ¼ of cycle OA, power is drawn from the source and energy is stored
in the magnetic field of the inductor. In the second quarter cycle OB, the current
and magnetic field decrease and the e.m.f induced in the inductor causes it to
act as a generator i.e turning the energy stored in its magnetic field back to the
source. The average power is therefore zero over a cycle.
Example 11.5
A 4.0 H iron core inductor is connected in series with a 300 Ω resistor to a 240V,
50H power supply.
(a) Using standard symbols draw the circuit used.
(b) Find the reactance of the inductor.
318
Solution
(a)
Low
frequency V
generator
Fig. 11.23
(b) (i) Reactance XL = 2πfL = 2 × 50 s-1 × 4 H × π = 1256 Ω

Example 11.6
In a purely inductive AC circuit, L = 25.0 mH
I and the r.m.s voltage is 150 V.
90º 180º 270º 360º 360º
Fig. 11.24: An inductor connected to an a.c supply

one cycle
(a) Calculate the inductive reactance and rms current in the circuit if the
frequency is 60.0 Hz.
(b) Suppose the frequency increases to 6.00 kHz? What happens to the rms
current in the circuit?
Solution
(a) inductive reactance XL = 2πfL = 25.0 × 10-3 × 2π × 60 = 9.42 Ω
Vr.m.s 150 V
The r.m.s current is Ir.m.s = X = 9.42 Ω = 15.9 A
L
(b) If the frequency increases, the inductive reactance increases because the
current is changing at a higher rate. The increase in inductive reactance
results in a lower current. We can find the new inductive reactance:
XL = Lω = 25.0 × 10-3 × 2π × 60 × 103= 9.42 Ω
Vr.m.s 150
The new current is Ir.m.s = X = 942 = 0.159 A
L
319
Exercise 11.3
1. A pure inductance of 1 henry is connected across a 110 V, 70Hz source.

A. reactance of the circuit is 440 B. current of the circuit is 0.25 A
C. reactance of the circuit is 880 D. current of the circuit is 0.5 A
2. A coil of inductance 5.0 mH and negligible resistance is connected to an
oscillator giving an output voltage E = (10V) sin θ. Which of the following is
correct
A. for f = 1000 s–1 current is 225 A B. for f = 500 s–1 current is 4 A
C. for f = 100 s–1 current is 2.25 A D. for f = 1000 s–1 current is 4 A
3. With increase in frequency of an a.c supply, the inductive reactance
A. decreases.
B. increases directly proportional to frequency.
C. increases as square of frequency.
D. decreases inversely with frequency.
4. An a.c voltage of 50 Hz is connected to an inductor of 2 H. A current of Ir.m.s.
frequency of the voltage V is changed to 400 Hz keeping the magnitude of
V the same. In terms of I, give the magnitude and direction of current now
flowing in the coil.
5. An inductor of 2 H is connected to a 12 Vr.m.s mains supply f = 50 Hz
(a) Find the current flowing through it.
(b) What current flows when the inductance is increased to 6 H.
6. For the circuit shown in Fig. 11.25 below, find the current in the circuit.
10 V L 10 mH
60 Hz
Fig. 11.25: An inductor connected to an a.c supply
320
11.7 Resistor, inductor and capacitor (RLC) in series with an a.c

power supply.
P (W)
Activity 11.9 To demonstrate the behavior of a resistor (R)

Inductor (L) and capacitor when they are connected
in series with an A.C power supply.
Materials 0 A B
time (S)
• A carbon resistor • Capacitor • Resistor
• Inductor • Low range (mv) centre – zero voltmeter
• Low range (mp) centre zero ammeter
• Low frequency generator • Connecting wires
Steps charging discharging
• Set up the circuit as shown in Fig. 11.26
VRR VL
VL
V
VLC
resistor
resistor capacitor
capacitor
inductor
resistor
MA
Low frequency generator
Fig. 11.26 RLC in series
• Make sure the low frequency generator is set to its minimum value.
• Switch on the mains and then the generator.
• Adjust the generator until the four voltmeters and the milliameter are active.
• Record the value of VR, VL and VC and fill Table 11.4.
• Record what happen to the current on the frequency is increased.
• Find from the internet how the RLC series may be applied.
321
Table 11.4
F H3 I MA VR (V) VL(V) VC (V) (VL – VC)2 + V2R V2 (V2)

1
2
3
4
5
When a resistor, capacitor and inductor are connected in series with an a.c voltage
supply, they form circuit called series RLC circuit (Fig 11.27)
VR VC VL
R C L
I = Io sinωt
VS
Fig. 11.27: Series RLC circuit
Since these three components are in series, the current through each of them is
the same
IR = IL = IC = I = Io sinωt
Let VR be the voltage across resistor, R.
VL be the voltage across inductor, L.
VC be the voltage across capacitor, C.
XL be the inductive reactance.
XC be the capacitive reactance.
R be the resistance of the resistor.
Since for a resistor the voltage is in-phase with the current, for inductor the voltage
leads the current by 90o and for capacitor, the voltage lags behind the current by
90o; the total voltage in the RLC circuit is not equal to algebraic sum of voltages
across the resistor (VR), inductor (VL) and capacitor (VL). In other words, these
voltages are not in phase with each other; hence cannot be added arithmetically.
The total voltage is actually the vector sum of these three voltages.
Fig. 11.28 shows the phasor diagram for the three voltages with the current as
the reference (because circuit the current is the same in all components). It shows
the vector addition of the three voltages
322
VL
V L – VC ω
V V
VR
)+
–VC
(V L
90o θ θ
90o VR I VR
VC
Fig.11.28: Phasor diagram for vector addition of VR,VL, and VC.

The total voltage V is given by:
V2 = VR2 + (VL – VC )2
Where VR = IR, VL = IXL, VC = IXC,
The Impedance for a Series RLC Circuit
The impedance Z of a series RLC circuit is the opposition to the flow of current
due the resistance R, inductive reactance, XL and capacitive reactance, XC.
We know that :
V2 = VR2 + + (VL – VC )2
Substituting for VR = IR, VL = IXL, VC = IXC, we get
V2 = (IR)2 + (IXL– IXC )2
This equations simplifies to
V2 = I2 [R2 + (XL – XC )2]
⇒ V = R2 + (XL – XC )2
2
(Dividing by I2(both sides)
I2
V
⇒ = √(R2+(XL2 – XC2) (Finding square roots both sides)
I
From this equation we get other equations like

V
I=
√(R + (XL2 – XC2)
2
V
And, Z=√(R2 +(XL – XC)2 (since = Z)
I
Fig. 11.29 shows the phasor diagram for the vector addition of R, XL and XC to
get Z
323
XL
Z2 = R2 + (XL – XC)2
XT = XL – XC
Z
Φ
R
XC
Fig.11.29: Phasor diagram for vector addition of R, XL and XC.
1
Since XL=Lω and XC= , then the equation =√(R2 +(XL – XC)2 is also expressed as
Cω
√
1
Z= (R2 +(Lω – )2
Cω
Activity 11.10 To show the variation of current with frequency in an

RLC series circuit
Materials
• Milliammeter • Variable capacitor
• Inductor • Signal generator (0-100 Hz: 5V)
Steps
1. Connect the circuit shown in fig. 11.30.
30 mA a.c
MA
Signal generating 1100 turns

(0 -100 HZ : 5V)
0 - 100 µF
Fig. 11.30:
2. Starting with 0 Hz increase the frequency.

3. Observe and explain what happens to the milliameter.
324
The graph of current against frequency for an RLC is as shown in Fig. 11.31
I
(mA)
f0 f (Hz)
Fig. 11.31: resonant frequency fo
The frequency at which the current is maximum is called resonant frequency.

This occurs when XL = XC
1
∴ 2πfoL = 2πfoC

1 1
fo = 2π √ LC ωo = 2πfo = LC
√
A good example of application of RLC is the tuning circuit of

a radio.
Example 11.7
A series RLC circuit with L =160 mH, C = 100 μF, and R = 40.0 Ω is
connected to a sinusoidal voltage V(t) = 40sinωt , with ω = 200 rad/s.
Fig. 11.32: An RLC circuit

(a) What is the impedance of the circuit?
(b) Let the current at any instant in the circuit be I(t) = I0sinωt . Find I0.
Solution
(a) The impedance of a series RLC circuit is given by
√
√ ( )
Z = (R2 +(Lω – 1 )2 = 402 + 0.16 × 200 –
1 2
= 43.9 Ω
Cω 100 × 200
325
40
(b) The amplitude of the current is given by I0= V0 = = 0.911 A
Z 43.9
Example 11.8
An a.c generator with V(t) = 150sinωt is connected to a series RLC circuit with
R = 40.0 Ω, L = 80.0 mH, C = 50.0 μF, and ω = 100 rad/s.
(a) Calculate VR, VL and VC , the maximum of the voltage drops across each circuit
element.
(b) Calculate the total potential difference across the three components between
points a and d.
Solution
1
(a) The inductive reactance is given by XC= 1 = = 200 Ω,
Lω 50 × 100
The capacitive reactance by given is by XL = Lω 0.08 × 100 = 8.0 Ω
The impedance of the circuit is given by Z = √R +(X – X L C

)2
= √40 +(8 – 200) = 196 Ω

2 2
Therefore, the corresponding maximum current amplitude is

150
I0= V0 = = 0.765 A,
Z 196
The maximum voltage across the resistance would be the product of maximum
current and the resistance:
VR = IR = 0.765 × 40 = 30.6 Ω
Similarly, the maximum voltage across the inductor is
VL = IXL = 0.765 × 8.0 = 6.12 V
326
And the maximum voltage across the capacitor is

VL = IXC = 0.765 × 200 = 153 V
Note that the total voltage V0 is is given by

V2 = VR2 + (VL – VC)2 =30.62 + (6.12 – 153)2
= 22510.09
V = √ 2 2 5 1 0 . 0 9 =150 V
Exercise 11.4
1. How much current will flow in a 100 Hz RLC series circuit if VS = 20 V,
R = 66 ohms, and XL = 47 ohms?
A. 1.05 A B. 303 mA
C. 247 mA D. 107 mA
2. Draw an electric circuit with the following components; resistors, capacitors,
and inductor in series. Hence, sketch a graph to show how their values change
with change in a.c frequency.
3. Draw an electric circuit showing an a.c source connected to a single inductor.
4. A series RC circuit with R = 4.0 × 103 Ω and C = 0.40 μF is connected to an
a.c voltage source V (t) = (100 V) sin ωt , with ω = 200 rad/s .
(a) What is the rms current in the circuit?
(b) Find the voltage drop both across the resistor and the capacitor.
5. A series RLC circuit with R =10.0 Ω, L = 400 mH and C = 2.0 μF is connected
to an a.c voltage source which has a maximum amplitude V0 = 100 V .
(a) What is the resonant frequency ω0?
(b) Find the rms current at resonance.
(c) Let the driving frequency be ω = 4000 rad/s . Compute X X and Z.
C, L,
6. A series LCR circuit has a supply current I(t) = I0 sin 2πf where I0 = 0.1 A
and the supply frequency f = 50 Hz . If the components have values R = 100 Ω,
C = 50 μF and, L = 50 H, calculate the impedance of the circuit and hence
obtain an expression for the supply voltage.
7. A resistance of 100 Ω is connected in series with a capacitor of 25 μF and an inductor.
The r.m.s voltage across the terminals of the source is 240 V and its frequency is
800. π-1 Hz. Given that VR = 80 V, find (a) Z (b) L.
327

• An electric symbol is a pictogram used to represent various electrical and
electronic devices (such as wires, batteries, resistors, and inductor) in a
schematic diagram of an electrical or electronic circuit.
• In direct current (dc), the electric charge (current) only flows in one direction.
Electric charge in alternating current (ac), on the other hand, changes direction
periodically. The voltage in ac circuits also periodically reverses because the
current changes direction.
• A resistor is an electrical component that limits or regulates the flow of
electrical current in an electronic circuit. It is also used adjust signal levels, to
divide voltages, bias active elements, and terminate transmission lines, among
other uses.
• A capacitor is a device used to store an electric charge, consisting of one or
more pairs of conductors separated by an insulator.
• An inductor is a passive two-terminal electrical component, which resists
changes in electric current passing through it. It consists of a conductor such
as a wire, usually wound into a coil. Energy is stored in a magnetic field in the
coil as long as current flows.
• An RLC circuit is an electrical circuit consisting of a resistor (R), an inductor
(L), and a capacitor (C), connected in series or in parallel.
Unit Test 11
For questions 1 to 3, choose the most appropriate answer.
1. What is the applied voltage for a series RLC circuit when IT = 3 mA,
VL = 30 V, VC = 18 V, and R = 1000 ohms?
A. 3.00 V B. 12.37 V
C. 34.98 V D. 48.00 V
2. Which of the following is true about both alternating current (a.c) and
direct current (d.c)?
(i) Causes heating
(ii) Can be stepped up or down with transformer.
(iii)
Can be used to charge a battery
A. (i), (ii), (iii) B. (i) and (ii)
C. (i) and (iii) D. (ii) and (iii)
328
3. What is the impedance of the circuit shown in Fig. 11.34 if ω = 200rad/s?
+ 1 kΩ
V1
–
L1 5H
C1 1 µF
A. 4123 KΩ B. 4.123 KΩ
C. 17 mΩ D. 16 KΩ
4. (a) Briefly explain the meaning of:

(i) A resistor (ii) A capacitor
(iii) An inductor
(b) State two uses of each electric component in 1 (a).
For questions 5 and 6 calculate and indicate the most correct answer.
5. Consider the LC circuit in Fig. 11.35. If one needs to tune this circuit to a
frequency of 84 kHz, and the capacitor has a capacitance C = 3.0 µF, what
inductance (L) is needed?
C I
I
Fig. 11.35:An LC circuit
6. Consider the LC circuit shown in Fig 11.36.
329
Fig. 11.36: An LC circuit
(a) What is the impedance of the circuit if: f = 60 Hz, L = 20 mH, R = 4.0
W?
(b) If the r.m.s. voltage of the source is Vr.m.s = 110 V, what is the r.m.s.
current?
(c) What is the peak current?
(d) What is the power dissipated in the resistor?
7. Consider the RLC circuit shown in Fig 11. 37.
R
C L
(a) What is the capacitance such that the current through the circuit is maximum?
(b) What is the r.m.s. current through the circuit?

(c) Find the capacitance that would make the impedance equal to 8 Ohms.
330
UNIT 12 Refraction of light
Key Unit Competence

By the end of this unit learner should be able to explain refraction of light
phenomenon
Learning objectives

• Recall the propagation of light and reflection.
• Explain phenomenon of refraction of light.
• State the laws of refraction of light.
• Explain total internal reflection of light and its application.
• Explain how spherical thin lenses form images.
• Explain defects of lenses and how they occur.
• Describe the spectrum of light by glass prism.
• Outline different applications of refraction of light.
Skills
• Analyse graphical construction of images formed by converging or diverging thin
lens.
• Evaluate the correction of defects in lenses.
• Describe types of light refraction.
• Analyse the dispersion of light by glass prism.
• Measure refractive index.
Attitude and value

• Appreciate the formation of real and virtual image by lenses.
• Appreciate the applications thin of lenses.
• Appreciate different colours in white light.
• Internalize the formation of a rainbow.
• Appreciate the bending of light when it moves from one medium to another.
331
Introduction
Unit focus activity

To explain the phenomenon of refraction of light
Materials
• Microscope • Optical fibre cable
Steps
Part 1
1. The fig 12.1 shows light signals being transmitted through an optical fibre.
Light Signal A
Light Signal B
Fig. 12.1 : Optical fibre transmitting light signals
332
Refraction of light
2. Name the phenomenon of light used in it.

3. Briefly describe the working principle of optical fibres.
Part 2
Your teacher will provide you with a microscope as the one shown in fig. 12.2.
Fig. 12.2: Microscope
Precaution!
Handle the microscope with a lot of care! It is costly to replace once damaged.
4. Make a specimen from a plant and observe the tissues with unaided eyes
then using a microscope. Do you notice any difference? Comment.
5. With the help of a simple light rays diagram, describe how the microscope
magnifies the image of very tiny objects.
We have learnt that light travels in a straight line. We also looked at reflection
of light at plane surfaces and characteristics of images formed under the plane
mirror. In this unit, we will introduce another property of light. In general life,
we have observed that:
1. A thin rod dipped obliquely into water appears to be bent at the water
surface.
333
Refraction of light
2. A pool of water appears to be shallower than it actually is.

3. A colourful rainbow is formed in the atmosphere usually after some rainfall.
4. A ‘shimmering’ pool of water seems to be ahead of a traveller on tarmac road
or desert sand on a hot day.
Explain these phenomena to your classmates.
These and many other similar effects are caused by a property of light called
refraction of light.
12.1 Phenomena of refraction of light

The following activities will help us illustrate the refraction of light.
12.1.1 Demonstrations of refraction of light
Activity 12.1
To describe the refraction of light
Materials
• Plastic ruler • Water in a transparent container
Steps
1. Dip a plastic ruler into a transparent container of clean water
2. View the ruler from the top and the side of the container (Fig. 12.3). What
do you observe on the shape of the ruler? Explain.
10
10
9
Ruler appear
9
8
bent
8
7
7
6
6
5
5
4
4
3
3
2
1
1
m
0c
0
cm
Fig. 12.3: Appearance of a ruler in water.
334
Refraction of light
The ruler appears to be bent at the point where it enters into water.This is because
light rays change direction (bend) when traveling from air to water. Therefore, a
ruler appears bend due to refraction of light.
Activity 12.2 To demonstrate how light rays travels using a

rectangular glass block
Materials
• A rectangular glass block. • A ray box.
Steps
Pass a narrow beam of light through a rectangular glass block in a semi-dark room
and observe the path of light (Fig. 12.4).
A weak reflected ray
Incident ray
Air
Refracted ray Glass block
Emergent ray
Fig. 12.4: A ray of light passing through a rectangular glass block.
The direction of the ray of light inside the glass changes. Some of the light is also
reflected from the surface of glass. The emergent ray is parallel to the incident ray.
In all the above experiments, when light travels from air to another medium like
water or glass and vice versa, there is a change in the direction of the path of light
at the boundary of the two media. This property of light is called refraction.
When light travels from one medium to another of different optical density, it
bends. The bending of light is called refraction.
Refraction of light is the bending of light rays when they travel from one medium to
another of different optical density. Also, refraction is the change of direction when
light rays travel from one medium to another.
Refraction is caused by the change of velocity of light as it travels from one

medium to another. Experiments show that the velocity of light in air (vacuum) is
3 × 108 m/s. The velocity of light is less in all the other media. Hence air is
considered as an optically rarer medium. All the other media, are considered as
optically denser media than air.
335
Refraction of light
12.1.2 Terms associated with refraction of light

Consider a rectangular glass block ABCD (Fig. 12.5). AB is a boundary that
separates the two media i.e. air and glass. Ray PQ travelling in air is incident at
the point Q at the boundary. On entering the glass, the ray travels along a path
QR (Activity 12.1). NQM is the normal drawn at Q to the line AB.
P N
Incident ray Normal

i
Air B
A
Q Glass
r
Refracted ray
Glass block
M
Glass
D R Air C
T
S Emergent ray
Fig. 12.5: Terminologies used in refraction of light.
The ray PQ is the incident ray and the ray QR is the refracted ray. The angle PQN,
between the incident ray and the normal, is the angle of incidence, i. The angle RQM,
between the refracted ray and the normal, is the angle of refraction, r. The ray RS is
the emergent ray. As seen in Fig. 12.5 the emergent ray RS is parallel to the incident
ray PQ, shown by the dotted line QT.
A ray passing from a rarer medium to a denser medium bends towards the normal.
On the other hand, a ray passing from a denser medium to a rarer medium bends
away from the normal (Fig. 12.5).
At the boundary or the surface that separates the two media, there is a change in
velocity of light that causes the change of direction. However, if light travels at right
angles to the boundary as shown in Fig. 12.6 (c) there is no change in direction.
Light continues to travel in a straight line but the speed of light is reduced in the
glass. This is, sometimes, referred to as the normal refraction.
N N
N i Air
i 90º 90º Air
Air Air
i 90º Air
(rarer medium)
(rarer medium) Glass Glass
Air
(rarer medium)Glass Glass Glass
r (denser medium) N N
Glass r (denser medium)
r (denser medium) N
M i Glass Glass
M i
(denser medium) Glass Glass
Glass (denser medium)
M i
(denser medium) Glass
r Air Air
r
r Air
M M
M
Air Air
(rarer medium)
(rarer medium)
Air
(rarer medium)
(a) Refracted ray moves (b) Refracted ray moves (c) No refraction
towards the normal away from the normal at normal
incidence
Fig. 12.6: Refraction of light in different media
336
Refraction of light
Exercise 12.1
1. Define the term:
(a) Refraction of light
(b) Angle of incidence
2. Explain why light bends when it travels from one medium to another.
3. Draw diagrams to illustrate refraction for a ray of light on:
(a) Glass – air boundary
(b) Water – air-glass boundaries
(c) Water – glass boundary
4. Complete the ray in Fig. 12.7 to show refraction.
incident
ray
Air
Water
Fig. 12.7: Refraction of light
12.1.3 Laws of refraction of light

Many scientists have contributed in the study of refraction of light. One such
scientist who discovered the relationship between the angle of incidence and the
angle of refraction was Willebrord Snell.
The following experiments will help in establishment of the relationship between
angle of incidence and the angle of refraction.
Activity 12.3 To investigate the relationship between the angle of

incidence and the angle of refraction
Materials
• A thick glass block • A paper protractor
• A ray box
Steps
1. Fix on the top of a table a thin circular “paper protractor” graduated in
degrees as shown in Fig. 12.8.
337
Refraction of light
2. Place a rectangular glass block ABCD such that the edge AB coincides with
the 90º–90º mark of the protractor. The line along 0º–0º mark represents the
normal NQM at Q.
Ray box
N
P 0o
10 0 170
20 160
180 10 15
30 170 20 0
160 14
50 30
40 1 0
0 40
i
14
Paper protractor
13
50
0
0
50
13
graduated in degrees
12
60
0
0
60
12
110
70
110
70
100
80
100
80
A B
Q
90
90
100
80
100
80
110
70
110
70
12
60
0
0
60
12
13
50
Glass block
0
30
50
14
1
40 0
0 150 40
14 30
20 160
30
150 10 180 170
160 20
170 0 10
0o
M
D c
R
C
Fig. 12.8: Relationship between the angle of incidence and the angle of refraction
3. Using a ray box, in a semi-dark room, pass a ray of light through the glass
block at an angle of incidence, i, say 20º. Measure the angle of refraction, r.
4. Repeat the experiment for different angles of incidence and measure the
corresponding angles of refraction.
5. Record the results in a table similar to Table 12.1. Complete the other
columns in the table. What do you notice about the ratio of sin i to sin r, i.e.
sin i
?
sin r
Table 12.1
sin i
iº rº sin i sin r
sin r
338
Refraction of light
In summary
sin i
• The ratio sin r
is practically a constant.
• The incident ray, the refracted ray and the normal all, lie in the same plane at
the point of incidence.
The two observations constitute the laws of refraction.
Laws of refraction
1. The incident ray, the refracted ray and the normal, at the point of incidence,
all lie in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of the angle of
refraction is a constant for a given pair of media (Snell’s law) i.e
sin i
sin r = constant
sin i
The expression sin r = constant is the mathematical expression of snell's law.
Activity 12.4
To verify Snell’s law using pins and a glass block
Materials
• White plain paper • Four pins
• A ruler • Softboard
• A glass block
Steps
1. Fix a white plain paper on a softboard. Draw a line XY and mark its midpoint
Q. At Q draw a normal NQM perpendicular to XY and a line PQ such that
the angle of incidence, i, (∠ PQN) is 20º.
2. Place a rectangular glass block such that the midpoint of the edge AB
coincides with the midpoint Q of the line XY (Fig. 12.9). Draw the outline
of the glass block ABCD.
3. Stick two pins O1 and O2, called the object pins on the straight line PQ
vertically into the softboard about 5.0 cm apart. View from the side CD and
look for the images of the first two pins.
4. Keeping the eye along the plane of the paper, move the head to and fro
slowly until in one particular position the images of the two pins lie on a
straight line.
339
Refraction of light
5. Fix a third pin S1 called the search pin, such that this pin and the images of
the first two pins as seen through the glass block lie along the same straight
line.
6. Repeat the above procedure with the fourth pin S2, so that the images of the
pins O1 and O2 and the search pins S1 and S2 lie along the same straight line.
7. Using a pencil, mark the positions of the four pins with a small circle and
remove the pins and the glass block.
8. Join the points S2 and S1 to meet the line DC at R. Join points O2 and O1 to
meet at Q. Join Q to R to make line QR. QR is the refracted ray in the glass
for the incident ray PQ in air. Measure the angle of refraction, r, (∠MQR).
P
O1
N
O2
Image of O1
Image of O2 i
Q
X A B Y
M R
D C
S1
S2
Fig. 12.9: Verifying Snell’s law.
9. Repeat the experiment for different angles of incidence and record the
readings in a table similar to Table 12.1.
10. What do you observe on the column sin i ? What does it represent?
sin r
Once again it is seen that the ratio sin i is a constant for the two given media.
sin r
Draw a graph of sin i (y-axis) against sin r (x-axis).
The graph drawn is a straight line passing through the origin as shown in Fig.
12.10. The gradient of the graph gives sin i which is a constant.
sin r
Snell's law states that for two refracting media, the ratio of the sine of the angle
of incident to the sine of the angle of refraction is a constant.
340
Refraction of light
Sin i
sin r
Gradient =
sin i
O
Sin r
Fig. 12.10: Graph of sin i against sin r
12.1.4 Refractive index

The refractive index (η) is the measure of bending of light i.e is the ratio of sine
of angle of incident to the sine of angle of refraction (hence Snell’s law).
sin i
Snell’s law can also be stated as = a constant.
sin r
This constant is known as refractive index, η or index of refraction. Therefore,

sin i
Refractive index (η) = It has no units
sin r

Since refractive index is the property of two media, it is necessary to indicate
sin i
them. If, for example, light travels from air to glass the ratio is found to be
sin r
1.50. This is the refractive index of glass with respect to air. We can rewrite this
using the following symbols.
airηglass = 1.50 i.e. aηg = 1.50
The absolute refractive index of a medium is the value of η when the first medium
is a vacuum.
sin i
The absolute refractive index of a medium η = sin r
Where the angle of incidence is in vacuum and the angle of refraction is in the medium
341
Refraction of light
The difference between the refractive indices of air and vacuum is very small and
hence the refractive index of a medium with respect to air is taken as the absolute
refractive index (unless extreme accuracy is required).
sin i
Refractive index of a medium (η) = .
sin r
Where the angle of incidence is in air and the angle of refraction is in the medium
The refractive index of a medium can also be expressed in terms of velocity of

light in the media.
Refractive index is the ratio of velocity of light in a vacuum to velocity of light in
a medium.
velocity of light in vacuum
Refractive index of a medium =
velocity of light in the medium
η = cv
Table 12.2 gives the refractive indices of some substances with respect to air (taking
the refractive index of air as 1.00). Materials with higher refractive indices bend
light more than those with lower refractive indices.
Table 12.2
Solid refractive index (η) Liquid refractive index (η)

Ice 1.31 Water 1.33
Glass (crown) 1.50 Alcohol 1.36
Glass (flint) 1.65 Paraffin 1.44
Ruby 1.76 Glycerine 1.47
Diamond 2.40 Turpentine 1.47
Example 12.1
A ray of light passing from air to glass is incident at an angle of 30°. Calculate the
angle of refraction in the glass, if the refractive index of glass is 1.50.
Solution
sin i
Refractive index of glass aηg =
sin r
sin i sin 30° 0.50
∴ sin r = = = = 0.33
ηg 1.50 1.50
∴ r = 19.5°
The angle of refraction in glass is 19.5°
342
Refraction of light
Example 12.2
In Fig. 12.11 calculate the refractive index of glass.
60°
35°
Fig. 12.11: Glass block
Solution
sin i sin 60°
Refractive index of glass (aηg) = =
sin r sin 35°
0.866
η = = 1.51
a g 0.574
The refractive index of glass is 1.51.
Example 12.3
The angle of incidence for a ray of light passing from air to water is 30° and the
angle of refraction is 22°. Calculate the refractive index of water.
Solution
sin i sin 30° 0.500
ηwater = = = = 1.33
sin r sin 22° 0.375
∴ The refractive index of water is 1.33 (hence aηw = 1.33)
Example 12.4
Calculate the refractive index of water, given that the velocity of light in air is
3 × 108 m/s and velocity of light in water is 2.25 × 108 m/s.
Solution
velocity of light in air 3 × 108 m/s
ηw = velocity of light in water = 2.25 × 108 m/s = 1.33
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Refraction of light
Example 12.5
The velocity of light in glass is 2.0 × 108 m/s. Calculate (a) the refractive index
of glass and (b) the angle of refraction in the glass for a ray of light passing from
air to glass at an angle of incidence of 40°.
Solution
C air (b) aɳg = sin i = 1.50
(a) aηg = = 3 × 108 m/s sin r
V glass 2 × 108 m/s
sin 40°
η = 1.50 sin r = sin i = 1.50 = 0.428
a g 1.50
r = sin–1 0.428
= 25.4°
12.1.5 The principle of reversibility of light

Just like light rays travel from medium 1 to a medium 2, it also travels in the
reverse direction i.e travel from the medium 2 to medium 1. This is known as
the principle of reversibility of light. It states that light will follow exactly the same
path if its direction of travel is reversed. The following experiment will help us
to establish that light rays can travel from a medium to air or vacuum and back
through the same path.
Activity 12.5 To verify the principle of reversibility of light
Materials
• A glass-air boundary • Plane mirror
Steps
1. Place a plane mirror M perpendicular to the refracted ray of light QR as
shown in Fig. 12.12.
P
i
Medium 1
Q Medium 2
r
M
R Mirror
Fig. 12.12: The principle of reversibility of light
2. Observe what happens to the reflected ray. Explain your observation.
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Refraction of light
In this activity, the reflected ray is reversed along RQ and is refracted through
QP in medium 1.
This means that the ray retraces its entire path. This is the principle of reversibility
of light.
sin i
From Snell’s law, 1η2 =
sin r
sin r
For the reversed path, 2η1 =
sin i
By multiplying the respective sides of the above two equations we get
η × η = sin i × sin r = 1
1 2 2 1
sin r sin i
η × 2η1 = 1.
1 2
Therefore, 1η2 = 1η
2 1
The refractive index for a ray moving from air to water is 1.33 but when the ray
1
moves from water to air the refractive index is 1.33 = 0.75.
Exercise 12.2
1. Define the term refractive index.
2. (a) State the laws of refraction.
(b) Describe an experiment to determine the refractive index of a rectangular
glass block.
3. A ray of light is passing from air into water along PQ. The ray strikes the
bottom surface at T instead of R as shown in Fig. 12.13. Calculate:
P
Q Air
Water
55° 40°
T R
Fig. 12.13: A ray passing air-water interface
(a) The angle of incidence

(b) The angle of refraction
(c) The refractive index of water.
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Refraction of light
4. Copy and complete Fig. 12.14 to show the path of light through and out of
the glass block of refractive index 1.50.
Incident ray Normal
40°
Air
A B
Glass
D Air C
5. Light travels through glass of refractive index 1.60 with a speed of v m/s.
Calculate the value of v, if the speed of light in air is 3.0 × 108 m/s.
6. In a semicircular glass block for the incident ray PQ inside glass, QR is the
refracted ray in air as shown in Fig. 12.15.
P T S
30°
Q r
R
Fig. 12.15: Semi circular glass block
(a) What do the dotted lines QS and QT represent?

(b) Calculate the angle of refraction, r, in air, if the refractive index of glass
is 1.50.
7. State Snell’s law of refraction. Describe an experiment to verify it.
8. Table 12.3 shows the angles of incidence i and the angles of refraction, r,
when light passes from air to glass. Complete the table and draw a graph of
sin r (y-axis) against sin i (x-axis). From the graph determine the refractive
index of glass.
Table 12.3
iº rº sin i sin r
15 10
30 19
45 28
60 35
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Refraction of light
12.1.6 Real and apparent depth
Activity 12.6 To describe real and apparent depth
Materials
• Water in a transparent container
• A coin
Steps
1. Place the coin at the bottom of the container with water.
2. From the top position your eyes perpendicularly to the coin and look at it.
Do you see the coin? Where is it?
3. Now shift eye to an angle and look again at the coin. Where do you see it?
4. Explain your observations in steps 2 and 3.
5. Discuss with your classmates what real and apparent depth are. Deduce the
relationship between the two.
Fig. 12.16 shows how to locate the image of an object in a denser medium (glass)
using a ray diagram.
Eye
C D
r
Air
A B
B
y i
x Glass
I
O Air
Fig. 12.16: Image formation in a glass block
A ray OA is incident along the normal. The ray goes undeviated as AC at the
surface of the two media. Another ray OB, incident obliquely at B and close to
A bends away from the normal and proceeds along BD. When DB is produced
backwards, it meets OC at I. This is the position of the virtual image of the object
O. OA (x) is the real depth of the object below the surface of separation. IA (y) is
the apparent depth of the image below the surface of separation.
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Refraction of light
The relationship between refractive index, real depth and apparent depth
Fig. 12.17 shows the image I of an object O inside water. Point B is very close
to point A.
Eye
C D
r
A B
i
y r
x
Fig. 12.17: Real and apparent depth
sin i
From the Snell’s law wɳa =
sin r
By the principle of reversibility of light,
sin r
ɳ =
a w
sin i
∠ AOB = i (alternate angles); ∠ AIB = r (corresponding angles)
AB AB
sin r = and sin i =
IB OB
AB
sin r IB AB OB OB
ɳ = = = × =
a w
sin i AB IB AB IB
OB
OB OA
Since B is very close to A then =
IB IA
OA x (real depth)
Therefore, ɳ = =
a w
IA y (apparent depth)
real depth
hence ɳ =
a w apparent depth
This relation can be used to determine the refractive index of a transparent solid
or a liquid. Since the refractive index of a denser medium is greater than 1, y
is always less than x. Thus the image of an object situated in a denser medium
348
Refraction of light
appears to be raised towards the surface. For example, if a pool of water is 4 m

deep, the apparent depth of the pool is given by
x 4.00
y = = = 3.0 m
ɳ 1.33
Activity 12.7 To determine refractive index of water using two pins

Materials
• A beaker with water • A clamp and stand
• 2 search pins • A ruler
Steps
1. Take a beaker containing clean water and measure the real depth x from the
base of the beaker to the water surface. Place an object pin O inside the beaker
and adjust its position so that the tip of the pin just touches the edge of the
beaker.
2. Locate the image of the object pin by keeping the eye vertically above water.
Adjust the position of the search pin S by moving it upwards or downwards
slowly till its tip and the tip of the image pin as seen through water move
together when the eye is moved across the beaker to and fro.
3. Fix the search pin S at this position (Fig. 12.18). Measure the apparent depth.
Eye
Clamp
and stand
y
x
Image I
S
Object
Fig. 12.18: Refractive index of water using two pins
4. Use the values of real and apparent depth to determine the refractive index
of water.
5. Present four findings to the class.
349
Refraction of light
Example 12.6
The real depth of a pool of water is 6 m and the refractive index of water is 1.33.
Calculate the apparent depth of the pool of water.
Solution
real depth
ɳwater =
apparent depth
6.0
∴ apparent depth = real depth = = 4.5 m.
ɳwater 1.33
Example 12.7
The real thickness of a glass block is 12 cm and apparent thickness is 8 cm.
Calculate the refractive index of glass.
Solution
real depth 12
ɳglass = apparent depth = = 1.5
8
Example 12.8
The graph in Fig. 12.19 shows the results obtained when a pin was viewed through
different sizes of glass of same material.
15
Apparent depth (cm)
10
0 10 20 30
Real depth (cm)
Fig. 12.19: A graph of apparent depth against real depth
Calculate the:
(a) gradient of the graph.
(b) refractive index of the glass.
350
Refraction of light
Solution
apparent depth 18 – 6 2
(a) The gradient of the line = real depth = =
27 – 9 3
real depth
(b) Refractive index, ɳ = apparent depth
1 1 3
ɳ= = 2 = = 1.5
gradient 2
3
Exercise 12.3
1. In Fig. 12.20, the eye can see point P inside an empty cup, but not the coin
inside. Suggest a simple method by which the observer can see the coin
without moving the position of the eye, the coin or the cup.
Eye
P Coin
Fig. 12.20: Empty cup
2. The length of a glass block is 6 cm (Fig. 12.21). Using a ray diagram, show
how the eye can see the virtual image of object O, if the refractive index of
glass is 1.50.
O
6 cm
Eye
3. Describe an experiment to determine the refractive index of a glass block

using two pins.
4. In a transparent liquid container, an air bubble appears to be 12 cm when
viewed from one side and 18 cm when viewed from the other side (Fig.
12.22). Where exactly is the air bubble, if the length of the tank is 40 cm?
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Refraction of light
Eye Eye
12 cm 18 cm
40 cm
Fig. 12.22: Transparent liquid container
5. The graph in Fig. 12.23 shows the real depth against the apparent depth of
a swimming pool as water is being filled.
6
Real depth (m)
0 1 2 3 4 5
Apparent depth (m)
Fig. 12.23: A graph of real depth (m) against apparent depth (m)
(a) Use the graph to calculate the refractive index of water.

(b) Which physical property of light changes as light leaves the pool of
water?
6. Describe an experiment to determine the refractive index of water.
7. Copy and complete a ray diagram to show how the eye sees the image of the
dipped part of the pencil (Fig. 12.24). (Refractive index of water is 1.33).
Eye
Air
Water
4 cm
Fig. 12.24: Pencil dipped in water
8. A pool of water seems to be shallower than the real depth whereas the
apparent height of a star in the sky is more than the real height. Explain this
observation.
352
Refraction of light
12.1.7 Critical angle and total internal reflection

For light to refract from a more optical dense to a less optical dense medium,
the light ray must be incident at an angle less than a certain angle. The following
activity will help us to determine this angle.
Activity 12.8 To investigate the critical angle and total internal

reflection
Materials
• Table • Protractor • Semi circular glass block
• Ray box • Paper protractor
Steps
1. Fix a thin circular ‘paper protractor’ calibrated in degrees on the table top.
2. Place a semicircular glass block such that the edge AB coincides with the
90º–90º mark of the protractor ensure the midpoint of the line AB is at Q
(Fig. 12.25 (a)). The line along 0º–0º mark represents the normal NQM at Q.
3. In a dark room, use a ray box to pass a ray of light through the glass block
at an angle of incidence (i) say 20º (Fig. 12.25 (a)). Observe the path of the
ray CQ in glass and ray QR in air, for the incident ray PC.
N R N Paper protractor
(a) Paper protractor (b) 0º graduated in degrees
0º
0 170
graduated in degrees 0 170
10 10
20 160 20 160
180 10 15 180 10 15
30 170 20 0 30 170 20 0
160 14 160 14
0 30 0 30
40 15 0 40 15 0
0 40 0 40
14 14
13
13
50
50
0
0
0
0
50
50
13
13
12
12
60
60
0
0
0
0
60
60
12
12
110
110
70
70
110
110
70
70
r Air Air B
100
100
80
80
100
100
80
80
A B A R
Q i
90
9
90
9
0
90º 90º 90º Q i 90º
Weak
0º P C 0º reflection
M M
Ray box Semicircular
Semicircular
C glass block
glass block
P Weak reflection
Ray box
Fig. 12.25: Critical angle.
4. Increase the angle of incidence, i, gradually and observe the relative change in
the path of the refracted ray in air. Explain your observation.
At C the ray strikes the glass block at an angle of 90˚ and is therefore not deviated
inside the glass along QC. At Q, the refracted ray in air moves away from the
normal and the angle of refraction, r, in air is greater than the angle of incidence,
i, inside the glass.
353
Refraction of light
As the angle of incidence increases, the ray is refracted further away from the
normal until, at one angle, it falls along the edge AB of the semicircular glass
block. At this angle, the refracted ray cannot be located on the air. This particular
angle of incidence is called critical angle, c, (Fig. 12.25 (b)).
The critical angle is the angle of incidence in a denser medium for which the angle of
refraction is 90º in the rarer medium.
When the angle of incidence is greater than the critical angle, there is no refraction
and all the light is reflected back inside the denser medium. This phenomenon
is known as total internal reflection (Fig. 12.26). The angle of reflection, r, is equal
to the angle of incidence, i.
Normal
Rarer
medium Q
Denser
medium i r
i>c
C
P R
Fig. 12.26: Total internal reflection
The following conditions must be satisfied for total internal reflection to occur:
1. Light must travel from a denser medium to a rarer medium.

2. The angle of incidence in the denser medium must be greater than the
critical angle.
Refractive index using critical angle

Using Snell’s law, the refractive index of glass-air boundary
sin i
ɳ =
g a
sin r
But if i = c, r = 90˚ and principle of reversibility (see Fig. 12.25(b)).
sin 90° 1
∴ aɳg = =
sin c sin c
1
Hence, refractive index of a medium aɳg = where c is the critical angle.
sin c
Example 12.9
Calculate the critical angle for glass-air interface, if the refractive index of glass
is 1.50.
354
Refraction of light
Solution
1
ɳglass =

sin c
1 1
sin c = = = 0.667
ɳg 1.50
c = sin–1 0.667= 41.8°
Example 12.10
Calculate the critical angle at the water-air interface if the refractive index of
water is 1.33
Solution
1
ɳw =
sin c
1 1
sin c = = = 0.752
ɳw 1.33
c = sin–1 0.667 = 48.8°

Example 12.11
Calculate the refractive index of diamond, if the critical angle for the diamond
is 24°.
Solution
1 1 1
ɳD = = = = 2.46
sin c sin 24º 0.407
∴ Refractive index of diamond is 2.46
12.2 Refraction of light through a prism
12.2.1 Demonstration of refraction of light through a prism
Activity 12.9 To show refraction of light through a prism
Materials
• A glass prism • White plain paper
355
Refraction of light
Steps
1. Place a glass prism on a white paper as shown.
A B
Fig. 12.27
2. Incident a ray of light on the edge CA.

3. What happens to the ray of light as it enters and leaves the glass prism?
4. Compare its behaviour between when it enters and leaves a rectangular glass
block.
In real glass block the 1st refraction cancels the 2nd refraction. Hence, the incident
ray and emergent ray are parallel. In the case of a glass prism, the 2nd refraction
adds to the 1st refraction. This is the importance of a prism. It can be used to
analyse the components of white light.
A prism has a refracting medium bound by two plane surfaces inclined to each
other at an angle. The two planes are the refracting faces(ADEC and ADFB in Fig.
12.28 (a)) and the angle between the faces is called the angle of the prism (∠CAB
in Fig. 2.28 (a)). The line along which the two faces meet is the refracting edge
of the prism. The face opposite to the angle of the prism is called the base of the
prism (Fig. 12.28 (a) ). The section of the prism cut by a plane perpendicular to
the edge of the prism is the principal section of the prism Fig. 12.28 (b) )
D Angle of prism
Angle of prism
Refracting edge A
A
E F
Refracting face Refracting face
C B
C B Base
Base
(a) CB is the base (b) Principal section of a prism
Fig. 12.28: Glass prism
When light passes from air into the triangular glass prism (ray PQ), it undergoes
refraction. The refracted ray QR inside the glass bends towards the normal N1N2.
The emergent ray RS bends away from the normal N3N4 (Fig. 12.29). Ray PQ
produced meets ray RS produced backwards at T. Notice that, the incident ray
has deviated from its original direction. Angle VTR is called the angle of deviation.
Hence the action of a prism is to deviate light rays.
356
Refraction of light
A
V
T
N1
i N3
Q r
N2 N R
4
P
S
C B
Base
Fig. 12.29: Refraction of light through a prism
12.2.2 Dispersion of white light by a prism
Activity 12.10 To illustrate dispersion of white light
Materials
• Triangular prism • White screen
• Ray of white light
Steps
1. A narrow beam of white light (such as sunlight, light from carbon arc lamp
or a mercury vapour lamp) from a narrow slit, in a semi-dark room, to an
equilateral glass prism.
2. Adjust the angle of incidence until a distinct band of colours is obtained on a
white screen placed on the other side of the prism as shown in Fig. 12.30.
Ray of white light
Screen
Red
Orange
Yellow
S Green
Blue
Indigo
Violet
Fig. 12.30: Dispersion of white light forming a spectrum of colours
3. What colours are obtained on the white screen? How many of the colours
can you identify? Is the angle of deviation the same for each colour?
A monochromatic light is one that has a single colour and a single frequency or single
wavelength. White light is not monochromatic because it is made up of seven
different colours. Non-monochromatic light is also called composite light.
357
Refraction of light
When white light is passed through a triangular prism, it is split up into a series
of colours as it enters the glass prism. Different colours are deviated to different
angles. The colours are red, orange, yellow, green, blue, indigo and violet. These colours
gradually blend into one another.
The above experiment was first carried out by Sir Isaac Newton. He noticed that
violet light is the most deviated colour while red light is the least deviated colour. The
splitting of white light into its constituent colours is called dispersion. The coloured
band produced is called visible spectrum (Fig. 12.30).
For the same angle of incidence, each colour inside the glass prism has its own
angle of refraction and angle of deviation. Since refractive index is given by
sin i
n = sin r , it follows that each colour has its own refractive index for glass. But
speed of light in air
refractive index is also given by n = speed of light in glass .
Therefore each colour travels with its own speed inside glass. For example, violet
light having the least angle of refraction has the greatest refractive index for glass.
This means that the speed of violet light is the least in glass.
If two identical prisms are placed as shown in the Fig. 12.31, the final spectrum
produced is more spread out. This means that the angle of deviation of each
colour is increased.
Fig. 12.31: More spread out spectrum.
12.2.3 Combination of spectrum colours

The dispersion created by one prism can be reversed by a second prism
(Fig. 12.32). When reversed, a white light parallel to the incident white light
emerges from the second prism.
White light parallel to

White light
the incident ray
Coloured beam
Fig. 12.32: Separation and combination of colours by prisms
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Refraction of light
Exercise 12.4
1. (a) Define critical angle.

(b) Write down the relationship between the critical angle and the refractive index
of a medium.
2. State the two conditions under which total internal reflection occurs.
3. Calculate the value of critical angle for a liquid-air interface, if the refractive index
of the liquid is 1.40.
4. Copy and complete Fig. 12.33 to show
the path of a ray of light through the glass
prism. Mark the angle of incidence and
the angle of deviation in your diagram
5. Distinguish between monochromatic and
composite light. Give an example of each. Fig. 12.33: Separation and combination
of colours by prisms
6. What do you understand by the terms
deviation and dispersion.
7. Write down the colours in a visible spectrum in the decreasing order of angle
of deviation.
8. Fig. 12.34 shows the path of a ray of white
light incident on a glass prism.
(a) What name is given to the spreading out
of white light inside the prism?
(b) What name is given to the band of colours
seen on the screen?
(c) Name two physical properties which Fig. 12.34
change for all the colours when they enter
the prism.
(d) Mark the position of violet, blue and red light as seen on the screen.
9. Describe an experiment to illustrate the dispersion of white light.
10. A beam of light made of a mixture of red and green light is shone through a
prism (Fig. 12.35).
Mixture of red
and green light
White screen
Fig. 12.35
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Refraction of light
(a) What name is given to the ‘bending’ of light as it enters the glass prism?
(b) Why does light ‘bend’ as it enters glass prism?
(c) Why do the two colours of light bend by different amount inside the prism?
(d) Which colour ‘bends’ the least?
12.2.4 Total internal reflection of light by a prism

Under certain circumstance total internal reflection can occur in a prism. The
following activity will demonstrate this.
Activity 12.11 To show total internal reflection by a prism

Materials
• Right-angled prism, white plain paper
• Mathematical set
Steps
1. Place a right-angled prism on a white paper.
2. Direct a ray (ray 1) towards side AB at a large angle of incident say 60°. Trace
the outline of the prism and record the angle of incidence on the side AB.
3. Repeat with (ray 2) less than 60° this time say 40°.
4. Continue decreasing the angle until the angle of incidence is zero (ray 3).
What happens to the angle of incidence on AB?
The angle is seen to increase up to the critical angles of glass (i.e. 42°). Refraction
stops and reflection occurs. (See Fig. 12.36)
A
1
2
3 45°
45°
C B
Fig. 12.36:
360
Refraction of light
For ray 3, the angle of incidence (45°) on the side AB is greater than the critical
angle hence total internal reflection occurs and the ray emerges perpendicular to
the other smaller side CB.
Example 12.12 B
Copy and complete Fig. 12.37 to show 45°

the path of light PQ as it enters
the right angled isosceles prism
P Q
Air
of refractive index 1.50.
45°
A C
Fig. 12.37: Isosceles prism
Solution
Critical angle c for glass-air interface is given by
Critical angle c for glass-air interface.
1 1
sin c = = = 0.667
ɳglass 1.5
c = 41.8°
At the surface AB, the ray passes undeviated B
along QR in glass. At the surface BC, 45°

the angle of incidence, i = 45°, Normal
since ∠BRQ = BCA = 45° R
P Q i
(corresponding angles), r = 90° – 45° = 45°. Air r
45°
Since i is greater than the critical angle c, A S C
total internal reflection occurs (ray RS).
T
Ray RS proceeds undeviated as ST in air
(Fig. 12.38). Fig. 12.38: Total internal reflection in
isosceles prism
Example 12.13
A ray of light passes from a liquid to air. Calculate the critical angle for the liquid-
air interface, if the velocity of light in the liquid is 2.4 × 108 m/s, while in air is
3.0 × 108 m/s.
361
Refraction of light
Solution
velocity of light in air
Refractive index of a liquid = velocity of light in liquid
3 × 108
= = 1.25
2.4 × 108
1 1 1
ηmedium = sin c = ɳ = = 0.80
sin c medium 1.25
c = 53.1°
The critical angle for the liquid is 53.1º.
12.2.5 Application of total internal reflection of light
Activity 12.12 To describe the application of total internal

reflection of light by a prism
Materials
• Right-angled prism • white plain paper
• Mathematical set • Blue coloured plate (filter)
Steps
1. Incident two rays of red light as shown in Fig 12.39(a). The two rays emerge
on side CD.
B
45°
Ray 1
Ray 2
45°
C D
Fig. 12.39 (a)
2. Hold a blue glass plate (filter) in front of ray 2. The ray 1 is coloured blue
(see Fig. 12.39(b)).
3. Turn the prism as shown in Fig 12.39(b) The blue ray enters as upper ray,
but emerges on lower ray.
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Refraction of light
45o
90o
45o
Fig. 12.39 (b)
This shows that a right-angled prism can therefore be used to reverse an image
laterally. The following are some of the applications of total internal reflection of
light in prism.
1. Totally reflecting prism

Right angled isosceles prisms deviate light rays through an angle of 90º (Fig.
12.40).
45°
45°
Fig. 12.40: Action of isosceles prisms
Reflecting prisms have more advantages than plane mirrors. Since plane mirrors
are silvered at the back surface of a thick glass plate, they give rise to a number
of faint images besides the main image. They also lose light intensity due to the
refraction in glass. In the case of reflecting prisms, the light is totally reflected
without much loss of intensity. They do not have invisible faint images found
in mirrors. The silvering of the mirror may wear off with time whereas a prism
has a tough structure. Due to these advantages, reflecting prisms are preferred in
optical instrument such as periscopes, binoculars and camera.
2. Prism periscope
A periscope is a device which enables us to see over the top of an obstacle. Two
right angled isosceles prism are used in prism periscope instead of the two plane
mirrors used in a simple periscope. This periscope produces brighter images than
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Refraction of light
those formed by plane mirrors.

A parallel beam of light normally incident on the first prism is turned through
90º and proceeds to the second prism and is again turned through 90º to reach
the eye of a person. The final image produced is virtual and upright (Fig. 12.41).
B
O
Object
Virtual
image
M
Eye
Fig. 12.41: Action of a prism periscope
The periscope of this type is normally used in submarines to sight enemy ships
over the surface of the sea.
3. Optical fibres
An optical fibre is a long clear glass rod which can be of thickness of a fraction of
a millimeter. When a ray of light enters the fibre, the angle of incidence, i, inside
the glass is always greater than the critical angle c for glass-air interface and hence
undergoes total internal reflection repeatedly on the boundary of glass fibre and
air (Fig. 12.42). Light travels along the length of an optical fibre without much
loss of light intensity.
Optical
fibre
Fig. 12.42: Passage of light through an optical fibre
Optical fibres are used in transmitting signals in communication. In medicine,

bundles of very fine flexible fibres are used to view the internal parts of a human
body using an instrument called the endoscope.
364
Refraction of light
4. Rainbow
White light from the sun undergoes dispersion as it enters into the raindrops of
water in the sky. Total internal reflection takes place at the opposite side of the
raindrop and different colours emerge from the raindrop after refraction (Fig.
12.43).
White light
from the sun
Raindrop
Violet
Red
Fig. 12.43: A single raindrop produces a spectrum
5. Mirage
The mirage is an optical illusion that takes place in a hot desert or a hot road due
to total internal reflection. A traveller sees in the distance a shimmering pool of
water in which the surrounding objects, like a tree, appear inverted (Fig. 12.44).
Light travelling from a denser medium (warm air) towards the earth, enters
regions of rarer medium (very hot air of low density in contact with the earth) and
undergoes total internal reflection at a certain point when the angle of incidence is
greater than the critical angle. An observer on a distance sees the inverted image
of the tree. Further, as hot air in contact with the earth rises up due to convection
currents, the image appears shimmering.
denser medium
rarer medium
Fig. 12.44: Mirage
Exercise 12.5
1. Certain prism may be used in such a way that refraction takes place when light
passes through it. Draw a diagram of a prism acting in this way and deviating a ray
of light through 90º.
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Refraction of light
2. A ray of light strikes a prism as shown in Fig. 12.45. Copy and complete the
diagram to show the path followed by the ray as it passes through and out of the
prism.
Fig. 12.45: Triangular prism
3. (a) Draw a diagram to show how two right angled isosceles prisms can be
used in a periscope.
(b) State a reason why glass prisms are preferred to plane mirrors for use in
periscopes.
4. The critical angle for a glass-air interface is 42º. Copy and complete Fig. 12.46 to
show the path of PQ and RQ, when they leave the glass block.
Glass block
R P
42°
Q Air

5. A narrow beam of light is incident on the face PQ of a glass prism of refractive
index 1.50, as shown in Fig. 12.47.
Q
30°
P R
Fig. 12.47: Glass prism
(a) What is the angle of incidence at the face QR of the prism?

(b) Copy and complete the diagram to show the path of light.
(c) Why does the beam not change direction as it passes through the face PQ?
(d) Give one change in property that occurs to the light when it passes through the
face PQ.
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Refraction of light
6. Fig. 12.48 shows the path of a ray of light passing through a length of an optical
fibre. Explain why the ray of light (a) does not change direction at B or at E (b) is
totally reflected at C and D.
C D
E
B F
A
Fig. 12.48: Optical fibre
12.3 Refraction of light through a thin lens
12.3.1 Definition of a lens
Activity 12.13 To find out what a lens is
Materials:
• Water • Round bottom flask • Plain paper
• Retort stand • Sun
Steps
1. Fill a volumetric flask with clear water.
2. Cork the flask.
3. Tilt the flask such that the neck of the cork is horizontal.
4. Place a source of light (sun, bulbs) above the flask.
5. Place a white paper under the flask preferably on the ground.
6. Move the flask to or away the white paper.
7. What happens on the plain paper?
8. Record and discuss the results.
Sun
Round bottomed
Clamp flask
Water
Stand Cork
Fig. 12.49
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Refraction of light
A lens is a transparent medium bounded by two spherical surface or a planed

curved surface.
12.3.2 Types of lenses
Activity 12.14 To describe types and shapes of lenses
Materials:
• Charts showing converging and diverging beam through lens.
• Plane lenses.
• Convex lenses
• Spherical lenses
Steps
1. Place some lenses available in your school on a labeled white plane paper.
Trace their outlines. Describe the shapes of the lenses?
2. Identify the lenses in Fig.12.5 b by name.
(a) (b) (c)
Fig. 12.50
3. Identify the lenses in Fig. 12.51
(a) (b) (c)
Fig. 12.51
There are two main groups of lenses. A type that is thick in the middle and thin
at the edges, causing rays of light to converge. This is called converging or convex
lenses. The other type is thin in middle and thick at the edges causing the rays of
light to diverge. This lens is called diverging or concave lens.
Concave lenses are of different shapes as shown in Fig. 12.52
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Refraction of light
A bi-convex or double convex lens has both its surfaces ‘curving out’. (Fig. 12.52).
(a) (b) (c)
Bi-convex or double convex Plano - convex Concavo - convex

Fig. 12.52: Types of convex lenses
Concave lenses are also of different shapes (See Fig. 12.53)
A bi-concave or double concave lens has both its surfaces ‘curving in’. Other concave
lenses are plano-concave and convexo-concave or diverging meniscus (Fig. 12.50).
(a) (b) (c)
Bi-concave or Plano-concave Convexo-concave

double concave (diverging meniscus
Fig. 12.53: Types of concave lenses
12.3.3 Terms used in thin lenses
Activity 12.15
To find out the terms used in thin lenses
Materials
• Double convex lenses • Sunshine • Piece of paper
Steps
1. Take a double convex lens outside on a very sunny day.
2. Place a dry tissue paper (or leaves) on a flat open ground.
3. Place the lens on the tissue paper.
4. Slowly lift lens vertically away from the paper until a spot of light is formed
on the paper.
5. Hold the lens at this position for some time.
6. Observe what happens to the paper. Explain
7. Repeat the activity with a concave lens. What do you observe? What happens
to the beam of light?
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Refraction of light
The paper starts to burn. This shows that a convex lens brings to a focus point light
energy from the sun and since light is in form of energy, a lot of it is concentrated
at a point. This point where the rays are brought together after passing through the
convex lens is called principal focus. This point is real.
When the activity was repeated with a concave lens, nothing happens.The principle
focus of a concave lens is virtual.
The following are other common terms used in thin lenses:
(a) The centre of curvature (C)

The centre of curvature of the surface of a lens is the centre of the sphere of which
the surface forms a part (Fig. 12.54 (a) and (b)). For each spherical lens there are
two centres of curvature (C1, C2) due to the two curved surface.
(b) The radius of curvature (r)

The radius of curvature of the surface of a lens is the radius of the sphere of which
the surface forms a part (Fig. 12.54 (a) and (b)). Each surface has its own radius
of curvature (r1 or r2).
(c) Principal axis

The principle axis of a lens is a line passing through the two centres of curvature
(c1 and c2) as shown in Fig. 12.54.
r1 r1
c2 c2
c1 c1
r2 r2
(a) Convex lens (b) Concave lens

Fig. 12.54: Principal axis
(d) The principal focus
A prism always deviates the light passing through it towards its base. A convex
lens may be regarded as being made up of large portions of triangular prisms as
shown below. The emergent beam, therefore, becomes convergent in a convex
lens (Fig. 12.55 (a)). The reverse is the effect in a concave lens (Fig. 12.55(b)).
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Refraction of light
(a) (b)
Fig. 12.55: Action of lenses on incident rays of light.
(i) Principal focus of a convex lens
Consider a set of incident rays parallel and close to the principal axis of a convex
lens (Fig. 12.56). These rays, after refraction through the lens, pass through
point F on the principal axis. Since all the rays converge at this point, it is called
principal focus. Since this point can be projected on a screen, it is said to be a real
principal focus.
Incident rays Refracted rays
P F C
Principal axis
2F
Fig. 12.56: Principal focus on a convex lens
(ii) Principal focus of a concave lens

For a set of incident rays parallel and close to the principal axis of a concave lens,
the refracted rays appear to diverge from a fixed point on the principal axis. This
point is called the principal focus F, of a concave lens (Fig. 12.57). This principal
focus is virtual since it cannot be projected on a screen.
Lens
Incident rays Refracted rays
C F P Principal axis
2F
f
Fig. 12.57: Principal focus on a concave lens
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Refraction of light
(e) The focal plane

When a set of parallel rays are incident on a convex lens at an angle to the principal
axis, as shown in Fig. 12.58, the refracted rays converge to a point, on a line passing
through F and perpendicular to the principal axis. The plane passing through F
is the focal plane.
Lens axis Focal plane
90º
1
Fig. 12.58: Focal plane of a convex lens
3
F F
(f) The optical centre (P) F 2 P
The optical centre of a lens is a point which lies exactly in the middle of the lens
(PA = PB) as shown in Fig. 12.59(a) and 12.59(b). Light rays going through this
point go straight through without any deviation or displacement.
(a) (b)
A P B
Fig. 12.59: Optical centre of a convex and concave lenses
(g) The focal length of a lens (f)

This is the distance from the optical centre to the principal focus of the lens (see
Fig. 12.51(a) and 12.51(b). Biconvex and biconcave lenses have a focal length
on each side of the lens.
The concept of centres of curvature of the surfaces is required only in drawing the
principal axis. Otherwise these points are referred to as 2F, as they are situated at
a distance twice the focal length from the centre of the lens (PC = 2PF).
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Refraction of light
Exercise 12.6
1. Distinguish between converging and diverging lenses.

2. Define the following:
(a) Principal axis
(b) Optical centre
3. Differentiate between the principal focus of the concave and convex lens.
4. How many principal foci does a biconcave lens have?
12.3.4 Image formation by converging lenses
Activity 12.16 To analyse the image formed by converging lenses

Materials
• Convex lenses
• Tree, Screen (white wall can act as screen)
Steps
1. Place a convex lens between a wall of a lab and a far away object e.g. a tree
outside the lab. (See fig 12.60)
White screen
Lens
Object
(tree)
Real image
Fig. 12.60: Object at infinity
2. Adjust the distance between the lens and the wall until the image of the tree
is observed on the wall.
3. What are the characteristics of the image formed?
4. In groups, discuss how you can explain the formation of such an image using
ray diagrams.
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Refraction of light
Ray diagrams are used to illustrate how and where the image is formed. The
following are the important incident rays and their corresponding refracted rays
used in the construction of ray diagrams.
Ray 1: A ray of light parallel and close to the principal axis, passes through the
principal focus F (Fig. 12.61).
P
F F P
Fig. 12.61: Ray 1
Ray 2: A ray of light through the principal focus F emerges parallel to the
principal axis after refraction (Fig. 12.62).
F P P F
Fig. 12.62: Ray 2
Ray 3: A ray through the optical centre, P is undeviated after refraction through
the lens (Fig. 12.63).
P P
Fig. 12.63: Ray 3
12.3.5 Locating images by simple ray diagrams and describing

their character
To locate the image of an object, we need a minimum of two incident rays from
the object. From the three standard rays discussed above, any two incident rays
and their corresponding refracted rays can be drawn to locate the image. If the
refracted rays converge, a real image is obtained. If the refracted rays diverge, then
a virtual image is obtained.
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Refraction of light
12.3.5.1 Convex lens
Activity 12.17 To describe the characteristics of images formed by

convex lenses when the object is at infinity
Materials
• White screen • Lens
• Tree • Metre rule
Steps
1. Position a convex lens vertically in front of a tree far away from the lens (through
a window).
2. Place a white screen on the other side of the lens as shown in Fig. 12.64. Adjust the
screen until you see a sharp image of the tree.
3. Compare the size of the image with that of the object (tree). Is the image bigger or
smaller? Is it upright or inverted?
4. Measure the distance from the lens to the screen. What does this distance represent
in this set up?
White screen
Lens
Object
(tree)
Real image
Fig. 12.64: Object at infinity
Note: The distance from the centre of the lens to the screen is nearly
equal to the focal length, f, of the lens.
(a) Object far away from the lens (at infinity)

Since the object is at infinity, all the rays from the object, incident on the lens
are almost parallel. The refracted rays converge at a point on the focal plane, as
shown in Fig. 12.65.
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Refraction of light
Image characteristics
A diminished, real, inverted image is formed at F.
F
F P I
Fig. 12.65: Object OB at infinity
(b) Object OB just beyond C (2F)
Activity 12.18 To describe images formed by convex lens when the

object is beyond 2F and at 2F
Materials
• Screen • Lens • Candle
Steps
1. Mark the positions of the principal focus F and 2F on both the sides of the lens
with a piece of chalk.
2. Place a lit candle on the table along the principal axis of the lens, slightly away
from 2F.
3. Place a white screen, on the other side of the lens, perpendicular to the principal
axis of the lens and adjust its position to and fro to the screen and observe what
happens. What are the characteristics of the image formed?
Real image
Lens Object
(candle)
Screen
F 2F
2F F P
v u
Fig. 12.66: Object beyond 2F
4. Repeat step 3 by placing the candle at 2F and observe what happens. What
are the characteristics of the images formed?
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Refraction of light
Fig 12.67 shows the ray diagram to locate the images when the object is beyond C.
F 2F
0 2F F P I
M
Fig. 12.67: Object OB just beyond 2F
A diminished, real, inverted image is formed between F and 2F.
(c) Object OB at 2F
The ray diagram when the object (candle) was placed at 2F is as shown in Fig. 12.68
0 F 2F
2F F P I
Fig. 12.68: Object OB at 2F
A real, inverted image of the same size as the object is formed at 2F
(d) Object OB between 2F and F
Activity 12.19 To describe the images formed by convex lens when

the object is between F and 2F and at F
Materials
• Candle • Lens
• Screen
Steps
1. Repeat activity 12.16 by placing the candle between 2F and F.
2. What can you say about the characteristics of the image formed on the screen
(Fig. 12.69)?
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Refraction of light
Screen
Lens
Real
image
Object
2F F F 2F
P
v u
Fig. 12.69: Object between 2F and F
3. Now move the candle to point F. What do you observe? describe the
characteristics of the image formed.
The simple ray diagram when the object is between F and 2F is as shown in Fig.
12.70.
0 F 2F I
2F F P
Fig. 12.70: Object OB between 2F and F
A real, inverted and magnified image is formed beyond 2F
(e) Object OB at F
When the object was at F, the refracted rays are nearly parallel and converge at
infinity as shown in Fig. 12.71.
O F
F P par
alle
l ra
ys
Fig. 12.71: Object OB at F
A real, inverted, magnified image is formed far away from the lens i.e. at infinity.
(cannot be described)
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Refraction of light
(f) Object OB between F and P
Activity 12.20 To describe the image formed by convex lens when

the object is between F and P
Materials
• Candle • Lens
• Screen
Steps
Repeat activity 12.66 keeping the candle close to the lens, between F and P. Can
you get an image on the screen? Describe its characteristics.
The image formed is virtual and cannot be projected on the screen. An enlarged,
upright image can be seen through the lens on the same side with the object
(Fig. 12.72).
Lens
Eye
Upright,
Object virtual and
enlarged
F F image
Fig. 12.72: Object between F and P
A simple ray diagram to locate image when the object is placed between F and
P is as shown in Fig. 12.73.
M
F
I F O P
Fig. 12.73: Object OB between F and the lens.
A magnified, upright and virtual image is formed on the same side as object.
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Refraction of light
12.3.5.2 Concave lens

When the object is at infinity, an upright, diminished and virtual image is formed at
principal focus F. For all other positions of the object OB, an upright, diminished,
virtual image is always formed between F and P (Fig. 12.74).
B
M
O F I P
Fig. 12.74: Image formation by a concave lens.
Example 12.14
A convex lens has a focal length of 12 cm and a real object 6 cm tall is placed
18 cm from the centre of the lens. By means of an accurate scale diagram, find
the position, size and nature of the image formed.
Solution
Using rays 1 and 3 of the image construction (Fig. 12.61 and 12.63), two incident
rays are drawn from B and the corresponding refracted rays through the lens. The
refracted rays converge at M where the image of B is formed.
Scale chosen: 1 cm = 6 cm.
P F 2F
F I
2F O F
M
u v
Fig. 12.75: Graphical construction of images formed by convex lens
The image of O is formed at I. IM is the real image formed at 6 cm from the

lens. The height of the image is 2 cm. Since the scale is 1 cm represents 6 cm, the
image is 36 cm from the lens and the height of the image is 12 cm (Fig. 12.75).
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Refraction of light
Example 12.15
A concave lens has a focal length of 2 cm and real object 1.0 cm tall is placed at
3 cm from the centre of the lens. By means of an accurate scale diagram, find the
position, size and the nature of the image formed.
Solution
Scale chosen: 1 cm to represent 1 cm
Similar to Example 12.14, draw minimum two incident rays from B and the
corresponding refracted rays. Since the refracted rays diverge, a virtual image is formed.
The image is 1.2 cm from the lens and the height of the image is 0.4 cm (Fig. 12.76).
B
M
O F I P F
Fig. 12.76:
Exercise 12.7
1. Name two features of the image formed by a convex lens when:
(a) The object is between F and optical centre (b) The object is at F.
(c) The object is at infinity.
2. Sketch a ray diagram to show image formation for an object placed between 2 F
and F of a converging lens. State four characteristics of the image.
3. (a) If a convex lens picks up rays from a very distant object, where is the
image formed?
(b) If the object is moved towards the lens, what happens to the position and size
of the image?
4. An object 2 cm high is placed 12 cm away from a convex lens of focal length 6 cm. By
using an accurate drawing on graph paper, find the position, height and type of image.
12.3.6 The lens formula
Activity 12.21 Lens formula
Work in groups, review the relationship between similar triangles and make a
report.
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Refraction of light
The lens formula is a formula relating the focal length, image and object distance.
Consider a convex lens of focal length, f, which forms a real image IM of an object
OB as shown in Fig. 12.77.
B D
2F F I
O F P 2F
f M
u v
Fig. 12.77: Lens formula
Triangles OBP and IMP are similar (3 angles are equal)

OB OP
∴ = ……………………………………… (1)
IM IP
Draw a line DP perpendicular to the principal axis where DP = BO. Triangles
PDF and IMF are similar (3 angles are equal)
DP PF
∴ = …………………………………… (2)
IM IF
Since DP = OB, from equations (1) and (2),
OP PF
=
IP IF
u f
v = v–f
Cross multiplying, uv – uf = vf
Dividing both sides by uvf
uv uf vf 1 1 1
– = ⇒ – = .
uvf uvf uvf f v u
1 1 1
Hence = + . This is the Lens formula, where
f u v
u stands for the distance of the object from the optical centre.
v stands for the distance of the image from the optical centre.
f stands for the focal length of the lens.
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Refraction of light
12.3.7 Sign Convention (Real is positive)
Activity 12.22 To show how sign conventions are used in converging

lenses
Material
• Football
Steps
1. Kick a football vertically upward.
2. Observe what happens to the ball’s motion.
3. How would you differentiate the two motions of the ball?
We can adopt a method or a convention to describe the upward motion and

downward motion. For example let the distances up be negative and down positive
or vice versa.
∴ 3 m up = -3 m
3 m down = +3 m
There are several sign conventions used when the distances of the object and the
image are measured from the lens. In this book, we shall adopt the real is positive
in which:
1. All the distances are measured from the optical centre.
2. The distances of the real objects and the real images measured from the optical
centre are taken as positive, while those of virtual objects and virtual images are
taken as negative. From this convention, the focal length of a convex lens is positive
and that of a concave lens is negative. See Fig. 12.78 (a) and (b).
P F F
P
+f –f
(a) (b)
Fig. 12.78: Real and virtual focal lengths of lenses
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Refraction of light
Example 12.16
An object is placed 24 cm from the centre of a convex lens of focal length 20 cm.
Calculate the distance of the image from the lens.
Solution
1 1 1 1 1 1
From = + ⇒ = –
f v u v f u
1 1
= –
20 24
6–5 1
= =
120 120
The image distance (v) = 120 cm
Example 12.17
An object is placed 12 cm from the centre of a concave lens of focal length
20 cm. Calculate the distance of the image from the lens.
Solution
1 1 1 1 1 1
From lens formula; = + ⇒ = –
f v u v f u
1 1
+ =
–20 12
–3 – 5 –8
= =
60 60
–60
v = = –7.5 cm
8
v is negative because the image is virtual.
12.3.8 Magnification formula of the lens

The term magnification refers to how many times an image is bigger than the
object. Linear magnification (m) is defined as the ratio of the height of the image
to the height of the object.
Activity 12.23
To derive magnification formula
Material
• Graph papers
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Refraction of light
Steps
1. Draw three vertical lines on a graph paper.
18 cm
6 cm
2 cm
A B C
Fig. 12.79
2. How many times in line B bigger than A.

3. How many times in line B bigger than C.
4. What are the units of these comparisons?
Earlier in this unit we have done activities where we saw that the size of images
formed by lenses are either bigger or smaller than the object. The increase or
decrease in size of an object is called magnification. That is
height of the image h1
Linear magnification (m) = =
height of the object h0
image distance (v) v
= =
object distance (u) u
Note: Since magnification is a ratio, it does not have units
Sometimes it becomes difficult to measure the height of the image or the object
accurately. In such cases, magnification can be calculated in terms of distances u
and v. For example, consider a convex lens where a magnified image is formed
(Fig. 12.80).
B
P F 2F I
2F O F R
u v M
Fig. 12.80: Magnification
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Refraction of light
Since triangles OBP and IMP are similar (3 angles are equal), the ratios of
corresponding sides are equal i.e,
IM OB
=
IP OP
IM IP v
∴
= =
OB OP u
IM v
Hence magnification, m= =
OB u
image distance (v) v

Magnification (m) = or m =
object distance (u) u
h1 v hi is
Therefore m = = , therefore, the ratio of image to object sizes ––
h0 u ho
also equal to the ratio of image to object distances v– measured from the optical
u
centre.
Example 12.18
An object of height 2 cm is placed 20 cm infront of a convex lens. A real image
is formed 80 cm from the lens. Calculate the height of the image.
Solution
hi v h 80
m = = ⇒ i =
h0 u 2 20
∴ hi = 80 ×12 = 8 cm
20
1
Example 12.19
An object placed 30 cm from a convex lens produces an image of magnification
1. What is the focal length of the lens?
Solution
OB OP
Magnification, m = = = 1. (Fig. 12.81)
IM IP
Since m = 1; then v = u
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Refraction of light
This occurs when object is at 2f.

Hence 2f = 30
∴ f = 15 cm
B
I
O P
30 cm
M
u v
Fig. 12.81: Image formed by convex lens
Example 12.20
An object of height 1.2 cm is placed 12 cm from a convex lens and real image is
formed at 36 cm from the lens. Calculate
(a) the focal length of the lens
(b) magnification produced by the lens
(c) the size of the image.
Solution
1 1 1
(a) From lens formula, + =
u v f
1 1 1
+ =
12 36 f
3+1 4 1
= =
36 36 f
1 1
= ⇒ f = 9 cm
9 f
Focal length of the lens = 9 cm
v 36
(b) m = = = 3
u 12
hi
(c) m =
h0
∴ hi = 3 × 1.2 = 3.6
Size of the image = 3.6 cm
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Refraction of light
Example 12.21
An object of height 2 cm is placed 8 cm from a convex lens and a virtual image is
formed on the same side as the object at 24 cm from the lens. Calculate (a) the
focal length of the lens (b) the height of the image formed.
Solution
1 1 1
(a) From lens formula, + =
u v f
1 1 1
+ – = (v = –24 cm because the image is virtual)
8 24 f
3–1 1 1 1
= ⇒ =
24 f 12 f
∴ focal length, f = 12 cm
v h –24 h
(b) Magnification m = = 1 ⇒ = 1
u h0 8 2
–24 × 2
∴ hi = = –6 cm (negative sign indicate image is virtual)
8
Example 12.22
A convex lens produces a real image of an object and the image is 3 times the size
of the object. The distance between the object and the image is 80 cm. Calculate
the focal length of the lens.
Solution
v
Magnification m = =3
u
∴ v = 3u ………… (1)
u + v = 80 ………… (2)
Solving equations (1) and (2)
u + 3u = 80 ⇒ 4u = 80
∴ u = 20 cm
Hence v = 3u = 60 cm
1 1 1 1 1
From lens formula
+ = + =
20 60 f u v
3+1 4 1
= = =
60 60 15
∴ focal length, f = 15 cm
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Refraction of light
12.3.9 Power of a lens
Activity 12.24 To explain what is the power of a lens
Material
• A lens
Steps
1. Discuss with your classmates what the power of the lens is.
2. It is possible to increase the power of a lens? Discuss.
3. Share your findings with other classmates.
The ability to collect rays of light and focus them at a point in the case of a
converging, or to diverge them so that they appear to come from a point in the
case of diverging lens is called the power of a lens. It is calculated from its focal
length using the formula
Power = 1f
The unit for power is the dioptrie represented by the symbol D. The f must be
in S.I units of length.
Example 12.23
A lens has a focal length of 25 cm. Find the power of the lens.
f = 25 cm = 0.25 m. The focal length of convex lens = +ve (It forms real image)
1
∴ Power = +0.25 = +4 m–1
NB: For a concave lens f = -ve (because a concave lens forms a virtual image)
1
∴ Power = -0.25 = -4 m–1
Exercise 12.8
1. Define the terms: principal axis, optical centre and focal length of a convex
lens.
2. With the help of a diagram, show the action of a convex lens as a converging
lens.
3. The focal length of a diverging lens is 15 cm. With the help of a diagram
explain the meaning of this statement.
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Refraction of light
4. Fig. 12.82 shows a convex lens of focal length 15 cm and two rays of light
parallel to the principal axis.
Copy and complete the diagram to show the path of these rays as they pass
through the lens. Label the position of the principal focus as F.
P axis
Fig. 12.82
5. Draw ray diagrams showing how a convex lens could be used to produce
(a) a real and diminished image
(b) a virtual and magnified image of a real object.
6. Fig. 12.83 is drawn to scale. One incident ray from the object is parallel to
the principal axis and other ray passes through the principal focus of a convex
lens. Copy and complete the diagram to show the path of the ray through
the lens. Hence determine (i) position of the image (ii) the magnification
produced by the lens.
Convex lens
Object
Fig. 12.83
7. Copy the table and put a tick () in three of the boxes to describe the image
formed by a diverging lens.
Table 12.4
Magnified
Diminished
Upright
Inverted
Virtual
Real
8. Draw a diagram to show how a convex lens produce a virtual image.
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Refraction of light
9. Fig. 12.84 shows two rays of light approaching a thin diverging lens. Copy
and complete the diagram and show the path of the rays as they pass through
and emerge out of the lens. Label the position of the principal focus F.
Fig. 12.84
10. Fig. 12.85 is drawn to scale. An object OB placed in front of a convex lens
of focal length 5.0 cm. Copy and complete the diagram and (a) show the
position of the image (b) find the size of the image
Convex lens
F O F
Fig. 12.85
11. A convex lens is used to form an upright, magnified image of an object placed
6 cm from the lens. Calculate the focal length of the lens, if the magnification
produced is 4.
12. An object 3 cm high is placed 20 cm from a lens of focal length –25 cm. Find
the position, size and the nature of the image formed.
13. At what distance must an object be placed from a convex lens of focal length
20 cm so as to get real image 4 times the size of the object?
14. An object 3 cm high is placed 30 cm from a convex lens of focal length 20 cm
(a) Find the position, size and the nature of the image formed (b) If the same
object is now moved by 20 cm towards the lens, calculate the magnification
produced by the lens.
15. A convex lens forms a focused image on a screen when the distance between
an illuminated object and the screen is 1 m. The image is 0.25 times the size
of the object. Calculate (a) the object distance from the lens (b) the focal
length of the lens used.
16. An object 3 cm high is placed 150 cm from a screen. Calculate the focal
length of the lens that has to be placed between the object and the screen, so
as to produce a real image 6 cm high on the screen.
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Refraction of light
17. An object 6 cm high is placed 30 cm from a diverging lens of focal length 15

cm. With the help of a scale diagram determine
(a) the position of the image.
(b) the magnification produced by the lens.
18. A real object placed 8 cm in front of a converging lens produces an image
at a distance of 12 cm from the lens and on the same side as the object.
Calculate the focal length of the lens.
19. A diverging lens of focal length 24 cm forms an image at 18 cm from the
lens. Calculate the distance of the object from the lens.
20. In an experiment to determine the focal length of a converging lens, a student
obtains the results shown in Table 12.5.
Table 12.5
u (cm) 21.0 24.0 33.0 36.0 45.0 60.0

v (cm) 50.0 40.0 22.5 25.0 22.0 20.0
(i) Plot a graph of u (x-axis) against v (y-axis)
(ii) Using the graph, determine the focal length of the lens.
12.3.10 Defects of lens
Activity 12.25 To analyse the defects of a lens
Materials
• Converging lens • Internet reference books
Steps
1. Take a keen look at the lens given to you. What do you understand by the
terms ‘lens defect’? Do you think the lens has any defect?
2. With the help of the internet and reference books conduct a research on
defects of lens.
3. Discuss you findings with your classmate.
4. With the assistance of your teacher, make short note on defects of lenses
from your findings.
In the derivations of lens formulae, we consider rays that are close to and parallel
to the principal axis of the lens (paraxial rays). The assumption made is that the
lens has a small aperture and that object points are on or close to the principal axis.
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Refraction of light
However, when we consider extended objects, the rays are non-paraxial. In this
case the image and its object appears different from its object in colour, shape or
sharpness of definition. This phenomenon is called image defects or aberrations.
There are two important aberrations in lens namely; spherical aberration and
chromatic aberration.
(a) Spherical aberration

Spherical aberration is associated with lenses with a large aperture. As a result
the image of an object point is not a point. The focal length of non-paraxial rays
is less than for paraxial rays which is a characteristic of spherical surfaces.
Non-paraxial ray or
marginal rays
Longitudinal
Spherical
Aperture aberration
Paraxial ray
F1 Principal axis
Fig. 12.86
Paraxial rays – rays parallel to and close to the principal axis

Non-paraxial rays – rays parallel to but far from the principle axis.
The image formed is a circular blurred and not a point. The distance F1 is the
longitudinal spherical aberration for a particular object distance.
Correction
1. The effect can be corrected for a given object distance by grinding the lens
surface to make them spherical.
2. The effect can be reduced by cutting off the marginal rays and placing a
stopper in front of a lens. This is called stopping down the lens aperture. This
has the disadvantage of making the image lens bright.
3. The effect can be minimized by keeping the angles of incidence at each
refracting surface small. This ensures that the deviation of light is shared or
divided equally between surfaces. In the plano-convex this is achieved when
curved surface receives or emits those rays which are most clearly parallel to
the axis.
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Refraction of light
F2 F2
F1 F1
(a) (b)
Fig. 12.87: F1F2 in (a) is less than F1F2 in (b)
(b) Chromatic aberration
Activity 12.26 To demonstrate chromatic aberration
Materials
• Converging lens • White screen
• Red light • White light
Steps
1. Place a converging lens near a white screen.
2. Shine red light on the lens.
3. Locate the principal focus using red light (Fig. 12.88).
Red light F
Fig. 12.88
4. Repeat steps 2 and 3 with white light. What do you observe?
5. Compare the positions of the principle focus located using the two different
types of lights. Do the principal focus coincide if not, explain why.
If a white object is viewed through a converging lens, its image is blurred with
coloured edges. This is because as seen above, a lens has a greater focal length for
violet light than for red light. This is the case since ηviolet>ηred. The effects resulting
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Refraction of light
from dispersion in which there is failure of lens to focus all colours to the same
convergence point is called chromatic aberration (Fig.12.89).
Fig. 12.89: Chromatic aberration
Correction
This defect of a lens can be eliminated by a combination of two lens cemented
together. This combination is called a chromatic doublet. A chromatic doublet
consists of a converging lens of crown glass combined with a diverging lens of
flint glass.
Fig. 12.90: Correction of chromatic aberration
One surface of each lens has the same radius of curvature to allow them to be
cemented together using Canada balsam which reduces the loss of light by
reflection.
The flint glass of diverging lens produces the same refraction as the crown glass of
the converging lens but in opposite direction. The end result is a converging effect.
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Refraction of light
Converging crown
glass lens
(low dispersive power) Diverging flint glass
lens
(high dispersive power)
Canada balsam
Fig. 12.91: A chromatic doublet
If the power of the lenses of the doublet give a focus point infront of the doublet,
it is said to be a positive achromatic.
The idea of combining lenses to bring the image of an object into focus is used in
compound lenses. How can one determine the effective focal length the compound
lens? The following discussion will help us to answer this question.
Activity 12.27 To demonstrate and explain lens combination
Materials
• Two converging lenses • Candle
• Screen
Steps
1. Revisit activity 12.66. Follow the steps in the activity to locate the image of
an object placed between F and 2F using a converging lens.
2. Now once you locate the image move the screen and place the second
converging lens on that point.
3. Move the screen to and from till you locate the image again. Where is the
image located? Describe the characteristic of the image.
4. Using a simple ray diagram, draw a diagram to illustrate this lens combinations.
Consider a case of a real object placed between F and 2F
2F F F 2F
v u
Fig. 12.92: Object between F and 2F
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Refraction of light
If a second converging lens intercepts the rays converging to a point, that point
is the virtual object for any subsequent image.
F Final image
2F F
v
u Virtual image
Fig. 12.93: Intercepting rays with a converging lens.
1 1 1
v1
+ u1 = f1
f1 is the effective focal length.

• Refraction of light is the bending of light as it passes from one medium to
another.
• Refraction of light takes place as the velocity of light is different in different
media.
• Laws of refraction of light states that:
1. The incident ray, the refracted ray and the normal to the surface at the
point of the incidence are all in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of the angle of
refraction is a constant for a given pair of media (this law is also known as
Snell’s law).
sin i
• According to Snell’s law, = a constant known as the refractive index of
sin r
the medium with respect to air, when the incident ray is in air.
• Refractive index of a medium (η) is defined as
η =
velocity of light in a medium
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Refraction of light
• A pool of water appears to be shallower than the real depth, due to refraction
of light.
real depth
Refractive index of a medium = .
apparent depth
• As the angle of incidence increases in a denser medium, the angle of refraction
also increases in the rarer medium. Critical angle is the angle of incidence
in the denser medium for which the angle of refraction is 90º in the rarer
medium.
1
ηmedium=
sin c
• Total internal reflection occurs when the angle of incidence in the denser
medium is greater than the critical angle.
• Prism periscopes, prism binoculars, optic fibres etc., apply the properties of
total internal reflection in their construction. Formation of a rainbow is due
to dispersion and total internal reflection.
• A glass prism deviates a monochromatic light from its original path and the
light bends towards the base of the prism.
• There is deviation as well as dispersion when white light is incident on a glass
prism.
• Dispersion is the separation of white light into its component colours.
A lens is a transparent medium bound between two surfaces of definite
geometrical shape.
• Thin lenses may either be converging or diverging.
• A convex lens is thicker at its centre than its edges and converges the light
incident on it.
• A concave lens is thicker at its edges than at the centre and diverges the light
incident on it.
• The following are some of the important terms used in spherical lenses:
principal axis, optical centre, principal focus, focal length.
• The focal length of a convex lens is positive, while that of a concave lens is
negative.
• The characteristics of the image formed by a converging lens depends on the
position of the object (see Table 12.6).
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Refraction of light
Table 12.6
Position of Position of Nature of Size of image formed

object image image formed compared to object
At infinity (far F real and inverted diminished
away)
Beyond 2F Between F and real and inverted diminished
2F
At 2F At 2F real and inverted Same size
Between 2F Beyond 2F real and inverted Magnified
and F
At F At infinity (far real and inverted Magnified
away)
Between F and Same side as Virtual and Magnified
P object upright
• A diverging lens always forms a virtual, upright, diminished image between F

and P (except when the object is at infinity).
• Magnification (m) is defined as the ratio of the height of the image to the
height of the object
magnification = height of image = image distance

height of object object distance
1 1 1
• Lens formula is given by u + v = f where u is the object distance, v the image
distance and f the focal length of the lens.
• Short-signtedness and long-sightedness are two most common defects of a
human eye
• A lens camera, simple microscope, compound microscope are some examples
of optical instruments.
Unit Test 12
For questions 1 - 6, select the correct response from the choices given.
1. Bending of light is called
A Reflection B Dispersion
C Refraction D Incidence ray
2. Give the reason why the speed of light is slower in diamond than in air.
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Refraction of light
A Diamond is optical dense than air.

B Diamond is optical less dense than air.
C Diamond have a hard structure.
D Diamond have soft structure.
3. The relationships between, velocity of light in air and velocity of light in a
medium is given by
velocity of light in air 1
A velocity of light in the medium =
η

B =η
C velocity of light in medium = η

velocity of light in air × η

D
4. A ray of light from air to glass is incident at an angle 35°. Calculate the angle
of refraction in the glass if the refractive index of glass is 1.50
A 22.5° B 59.4°
C 20° D 21.9°
5. The critical angle of a certain refracting material is 30°. Calculate the
refractive index of the material.
A 1.5 B 2
C 0.5 D 1
6. Which diagram in Fig. 12.94 correctly shows a ray of light passing through
a rectangular glass block?
A B

Fig. 12.94 (a) and (b): Rectangular glass block
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Refraction of light
C D
Fig. 12.94 (c) and (d): Rectangular glass block
7. Use the words provided to fill in the blank spaces.

spectrum, deviation, refraction, dispersion, velocity, composite, monochromatic,
seven
____ is the bending of light when it travels from one medium to another. It
takes place because the ____ of light is different in different media.
9. A beam of light is incident on a glass block as shown in Fig. 12.95. Copy
and complete the diagram to show the path of the beam of light in glass after
refraction.
Normal
Glass
Air
Normal
Fig. 12.95: Rectangular glass block
10. (a) Table 12.7 below gives the values of angle of incidence and angle of
refraction when light passes from air into a plane glass surface. Calculate
the mean refractive index of glass.
Table 12.7
i° r°
25 17
40 26
70 40
(b) The incomplete Table 12.8 shows the values of angle of incidence, i, and
of angle of refraction, r, when light passes from glass to air.
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Refraction of light
Table 12.8
i° r°
10
30 46.5
40
Calculate r, when i = 10º and 40º and complete the table.

11. Calculate the refractive index of diamond if the velocity of light in air is
3.0 × 108 m/s and that in diamond is 1.25 × 108 m/s.
12. Fig. 12.96 shows a ray of light passing from air into oil. Calculate the angle
of refraction in oil if the velocity of light in air is 3.0 × 108 m/s and that in a
transparent oil is 2.2 × 108 m/s.
30°
Air
Oil
Fig. 12.96: Air-oil interface
13. In an attempt to determine the refractive index of a glass block, a learner

finds the displacement produced due to refraction by glass as d and apparent
thickness of the block as y (Fig. 12.97). Show that the refractive index of
d
glass may be expressed as η = ( 1 + )
y
Eye
Image I
d
O object
Fig. 12.97: Rectangular glass block
14. A learner attempted to find the refractive index of a liquid using a concave
mirror and a pin. The radius of curvature of the mirror is 20 cm. When a small
quantity of a liquid was placed in the mirror, the pin had to be moved down
by 4 cm. Calculate the refractive index of the liquid.
15. The light ray passing from glass to air is monochromatic and has a frequency
of 4 × 1014 Hz and a wavelength of 5 × 10–7 m in glass (Fig. 12.106).
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Refraction of light
N1 R
Q Air
Glass
P
N2
(a) What is meant by monochromatic?

(b) Calculate the velocity of light in glass.
(c) Calculate the velocity of light in air. (Refractive index of glass is 1.50).
16. A single ray of light is incident on an equilateral glass prism as shown in
the Fig. 12.99. Copy and complete the diagram to show the path of light
through and out of the prism (refractive index of glass is 12.107).
A
60°
60°
C B
Fig. 12.99: Equilateral glass prism
17. Table 12.9 shows the reading of the angle of incidence and angle of refraction
for a rectangular glass block.
Table 12.9
iº 10 25 40 55 70 80
rº 8 17 25 32 38 41
(a) Draw a graph of sin r (y-axis) against sin i (x-axis).

(b) What can you conclude from the graph about the relationship between
sin i and sin r ?
(c) Sketch another graph on the same axes in part (a), if a transparent
substance of higher refractive index had been used instead of the glass
block. Explain how you arrive at your answer.
18. Copy and complete the diagram in Fig. 12.100 to show the path of the ray
through the prisms arranged as shown below.
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Refraction of light
(a) (b) (c)

Red light Violet light
Red light
Fig. 12.100
(a) a ray of red light is incident on the prism.
(b) a ray of red light is incident on 2 identical prisms placed as shown.
(c) the red light in (b) is replaced by violet light.
19. Describe an experiment to illustrate that white light is composite in nature.
20. Fig. 12.101, drawn to scale, shows two rays starting from the top of an object
OB incident on a converging lens of focal length 2 cm.
Fig. 12.101 Equilateral glass prism
(a) Copy and complete the diagram to determine where the image is formed.
(b) Add one more incident ray from B through the principal focus and draw
the corresponding refracted ray through the lens.
(c) Calculate the magnification produced by the lens.
21. Fig. 12.102 shows an object placed at right angles to the principal axis of a
thin converging lens.
Lens
B
F F
O
8 cm 6 cm
(a) Calculate the position of the image formed.

(b) Give an application of this arrangement of a lens.
(c) Describe the nature of the image formed.
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Refraction of light
22. Describe with the aid of a ray diagram, how an image is formed in a (i)
simple microscope (ii) lens camera.
23. A converging lens is used to form an upright image, magnified 5 times of an
object placed 6 cm from the lens. Determine the focal length of the lens
24. Fig. 12.103 shows two converging lenses L1 and L2 placed 8 cm from each
other. The focal length of the lens L1 is 2 cm and that of L2 is 2.8 cm. An
object 1.0 cm high is placed 3 cm from lens L1.
L1 L2
Object Eye
3 cm 8 cm
(a) Construct a ray diagram to scale, on a graph paper to show the position
of the final image as seen by the eye of a person.
(b) Determine the magnification obtained by this arrangement.
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Telecommunication Channels
UNIT 13 Telecommunication Channels
Key Unit Competence

By the end of this unit the learner should be able to differentiate telecommunication
channels.
Learning objectives

• Recall refraction of light and total internal reflection.
• Define and explain the term communication.
• Outline different channels of communication.
• Distinguish between digital and analogue signals.
• Explain difficulties related to signal transmission.
• Outline difficulties experienced in communication system.
• Describe simple block diagram of communication system.
Skills
• Apply knowledge acquired to characterize quality of a communication system.

• Use computer simulations to demonstrate channels of communication.
• Suggest different channels of communication applied in telecommunication.
• Evaluate difficulties experienced in communication system.
Attitude and value

• Appreciate applications of communication
• Adapt scientific techniques in analysing and modelling simple block diagram of
communication systems.
• Acquire ability to deal with difficulties related to signal transmission.
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Introduction
Unit Focus Activity

1. Identify some of the means of communication that were being used by our fore
fathers in historic times and their disadvantages in terms of efficiency, timeliness,
effectiveness over long distances and security/confidentiality of the message.
2. The following are some of the modern communication channels being used
to transmit messages in form of data signals:
Twisted pair cables, coaxial cables, fiber optical cables, radio waves and
satellite communication. Describe the types of communication where each
of them is most suited.
Humans are social beings and hence they have an inherent need to communicate
with one another in one way or another. The need to keep on communicating is so
strong that humans have from time immemorial invented better ways of passing
messages to each other. In this unit, we will start by defining communication and
then discuss modern communication channels.
13.1 Definition of terms used in communication
Activity 13.1 To identify and discuss the term used in

communication
Figure 13.1 shows two children talking via a simple string telephone.
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Fig. 13.1: Children talking via a string telephone
1. What do you understand by the term communication?

2. With the help of Fig. 13.1, identify five elements of a communication system.
3. Compare your elements with those identified by your classmates.
Communication is the process of using sounds, words, symbols, signs, pictures or

signals to pass a message or information from one person to another. The message
origin is called the source or sender while the target recipient is the receiver. The
message is usually targeted for sending to the receiver.
A communication system has five basic elements
1. Message source
2. Transmitter
3. Communication channel
4. Receivers
5. Message user
13.1.1 Message source

This is the person who wants to send the message across the communication
system. The message source may want to make a telephone call, send an email,
chat etc.
13.1.2 Transmitter
Transmitter refers to a terminal equipment that receives a message from the source
and converts it to a format that can be transmitted on the channel. A transmitter
could be a computer, radio or TV transmitter station in broadcasting houses.
13.1.3 Communication channel

A communication channel is a transmission media through which data and
information flows. A channel carries the coded message from the transmitter to
the receiver using signals that can flow through it. The channel could be made of
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one or more transmission media especially if the message is travelling over long
distances. As mentioned in the introduction, the main focus in this unit is to
explore in details the various communication channels in use in the world today.
13.1.4 Receivers
A receiver is a terminal equipment that gets the transmitted message from the
channel and decodes it before presenting it to the user. It could be a computer,
mobile phone, radio receiver etc.
13.1.5 Message user

This is the person who is the target recipient of the message. The message is
important to the user. Figure 13.2 is a block diagram depicting the elements of
a communication system.
Noise
Fig. 13.2: Simple block diagram of a communication system
13.2 Types of data Signals
Activity 13.2 To analyse types of data signals
In the year 2015, Rwanda completed what is referred to as migration from analogue
to digital broadcasting.
From this understanding,
1. Point out the differences between digital and analogue signals
2. Highlight the advantages of digital data transmission over analogue
transmission.
3. Present your findings to the rest of your class.
In modern telecommunication channels, the message is sent in the form of

electrical signals known as data signals.
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A data signal refers to a voltage level in the circuit which represents the flow of
data. In data communication, data signals can either be analog or digital in nature.
Analog data is made up of a continuous waveform while digital data is made up
of a non-continuous discrete signal.
Figure 13.3(a) shows a digital signal represented graphically as a square wave,
while Figure 13.3(b) shows an analog signal represented as a continuous sine wave.
+1
–1
(a) Digital signal
+1
–1
(b) Analog signal
Fig. 13.3: Digital and analog data representation
Looking at Figure 13.3(a), you will observe that a digital signal rises suddenly to
a peak amplitude of +1, holds for sometime then suddenly drops to –1 level. On
the other hand an analog signal rises to +1 and falls to –1 in a continuous version
(Figure 13.3(b)).
Although the two graphs look different in their appearance, notice that they repeat
themselves at equal time intervals. Electrical signals or waveforms of this nature
are said to be periodic. Generally, a periodic wave representing a signal can be
described using the following parameters.
1. Amplitude (A)
2. Frequency (f)
3. Periodic time (T)
Amplitude (A): Amplitude is the maximum displacement that the waveform of
an electrical signal can attain. For example, the amplitude of the electrical signals
in Figure 13.3 is 1.
Frequency (f): Frequency of an electrical signal is the number of cycles made
by the signal in one second. It is measured in units called Hertz (Hz). 1Hz is
equivalent to 1 cycle/second.
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Periodic time (T): The time taken by a signal to complete one cycle is called
periodic time.
1
Periodic time, T, is given by the formula T = f , where f is
the frequency of the wave.
Exercise 13.1
For questions 1– 7, select the most appropriate answer
1. Three of the following are basic elements in a communication system. Which
one is not?
A. transmitter B. channel
C. amplifier D. receiver
2. If the maximum amplitude of a sine wave is 1.5 V, the minimum amplitude
is ________ V.
A. 1.5 V B. 1.0 V
C. -1.5 V D. 0 V
3. A ___________ signal is continuous and takes continuous values.
A. digital B. analog
C. (a) or (b) D.
none of the these
4. A ___________ signal has discrete states and takes discrete values.
A. analog B. digital
C. (a) or (b) D.
none of these
5. What is the bandwidth of a signal that ranges from 2 MHz to 5 MHz?
A. 2 KHz B. 3 MHz
C. 5 MHz D. none of these
6. A periodic signal completes 4 cycles in 0.002 s. What is the frequency?
A. 2 Hz B. 200 Hz
C. 2 KHz D. 4 KHz
7. The period of a signal is 0.005 s. What is its frequency?
A. 5 Hz B. 20 Hz
C. 0.2 KHz D. 2.5 KHz
13.3 Data transmission media

Data signal cannot be sent from one place to another without a medium of
communication. A communication medium is a physical or wireless channel
used for transmitting data and information from one point to another.
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The communication medium will more often than not determine the type of signal
used to transmit a message. In networking, data communication media can be
divided into two broad categories:
1. Communication using cable (physical media)
2. Wireless communication (wireless media)
13.3.1 Communication using cables (physical media)
Activity 13.3 To analyse the use of cables in communication
1. Identify 4 types of data transmission cables in use today.

2. Evaluate the strengths (advantages) and weakness (disadvantages) of each.
3. Identity areas or conditions under which each type is most suitable.
4. Present your findings to the rest of the class.
The main distinguishing characteristic of physical media is that data is transmitted

from the source to destination through a physical channel such as copper cables.
For example, if the cable is a copper conductor, the data signal which may be
in form of an electrical signals is propagated through the cable from the source
to the destination. Degradation of the signal as it travels through the medium is
referred to as signal loss. There are several types of bound transmission media
but the most common ones are:
1. Two-wire open line cables.
2. Twisted pair cables.
3. Coaxial cables.
4. Fibre optic cables.
(a) Two-wire open line cable

Two-wire open line cable is made up of two parallel copper wires separated
by a plastic insulator as shown in Figure 13.4. They are mostly used in
telecommunication network to transmit voice signal.
Insulator Wire
conductor
Fig. 13.4: Two wire open lines cables
Although the plastic insulator is meant to reduce interference called crosstalks

their linear nature allows an electromagnetic field to build around them during
heavy data transmission which may cause interference to the signal. The wires
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also capture environmental frequencies e.g. radiowaves hence causing noise in the
transmission channel. In data communications, the word noise refers to unwanted
signals picked up by the channel.
(b)Twisted pair cables
Activity 13.4 To describe the use of twisted pairs cables in

communication
Your teacher will provide you with pieces of two different kinds of twisted power
cables.
1. Observe them and discuss their structural differences.
2. Highlight the main advantage of STP cable over UTP cable.
3. Present your findings to the rest of the class.
A twisted pair cable is made up of two solid copper wires twisted around each
other in a double helix manner as shown in Fig. 13.5 (a) and (b). The winding of
the wires is meant to reduce the build-up of an electromagnetic field around the
two wires as they transmit data. Twisted pair cables are mostly used to transmit
both voice and data signals. The two common types of twisted pair cables are the
unshielded twisted pair (UTP) and shielded twisted pair (STP) shown in Figures
13.5 (a) and (b).
Shield
twisted
pairs
Outer cover
cover stripper
(a) Unshielded Twisted
twisted pair cables pair
(b) Shielded twisted pair cables
Fig. 13.5: Twisted pair cables
Notice that unlike STP, UTP cable pairs cables do not have a shield that prevents
electromagnetic interference (EMI) from the environment. The cable is therefore
susceptible to noise and signal interference. Noise may come from lightening
sparks, radio frequency or radiations from spark plugs in motor vehicles. Unshielded
twisted pair is therefore not suitable for environments that are electrically “noisy”.
The alternative is to use STP that has cable pairs. Shielded twisted pair (STP)
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is similar to unshielded twisted pair except that a shield is wrapped around the
wires to protect them from noise.
Although twisted pair cables support high data rates of upto 100 Mbps, they suffer
from attenuation. For every cable length of 90 metres, a device for amplifying the
signal called a “repeater” must be installed.
Example 13.1
A student typed an e-mail to send over the internet at a speed of 100 Mbps.
Calculate the maximum number of characters that can be sent per second if each
character consists of 8 bits.
Solution
100 × 1 × 106 100 × 1 000 000
Characters per second = =
8 8
= 12 500 000 characters per second
The advantages of twisted pair cabling include:

1. It is easier to set up network media because UTP cables are widely available.
2. Devices used to setup UTP connection are cheap and readily available.
3. UTPs are cheaper because of mass production for telephone use.
The disadvantages of twisted pair cabling include:
1. UTP connection suffers high attenuation rate.
2. It is sensitive to electromagnetic interference and eavesdropping.
3. It has low data transmission rate as compared to fibre optic cables.
(c) Coaxial cables
Activity 13.5 To analyse the use of coaxial cables in

communication
Your teacher will provide you with a piece of TV cable that you use at home to
connect the antennae to your television set.
With the help of research from books or the internet:
1. Draw its structure and label all its parts
2. Discuss the function of each part.
A coaxial cable resembles the cable that is used to connect television antenna
to a television set. It is called coaxial cable because it has a copper core (coax)
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which may be of solid copper wire surrounded by a dielectric material as shown

in Fig 13.6 (a). The dielectric material is then surrounded by mesh conductor
which is covered by a shield making the cable more resistant to electromagnetic
interference than the twisted pair cable.
The mesh conductor is made of copper or aluminium and serves as the earthing for
the carrier copper core. Together with the insulation and any foil shield, the shield
protects the core from radio frequency interference (RFI) and electromagnetic
interference (EMI). Although the cable has better protection against electrical
interference than the twisted pair cables, it has a moderate protection against
magnetic interference. The diameter of the centre core determines the attenuation
rate. The thinner the core, the higher the attenuation rate. Data is carried on coax
in the form of direct current (d.c).
Coaxial cables have bandwidths of upto 1 Gbps (Gigabits per second) hence,
they are used as network backbone. A good example where these cables are used
is connecting different networks between buildings and routing trunk calls in
telecommunication. There are two types of coaxial cables:
1. Thin coaxial cable also known as thinnet that has one dielectric insulator as
shown in (Figure 13.6 (a).
2. Thick coaxial cable also known as thicknet that has two dielectric insulators
around the core and is thicker than the thinnet (Figure 13.6 (b)).
Aluminium foil Copper core
Core insulation
(a) Thinnet
Braided shielding
Outer covering
(b) Thicknet
Fig. 13.6: Thinnet and Thicknet coaxial cables
Fig. 13.6: Thinnet and Thicknet coaxial cables
Advantages of coaxial cables include:

1. They are stable even under high transmission loads.
2. They have high bandwidth compared to twisted pair cables.
3. They are capable of carrying voice, data and video signal simultaneously.
4. They are more resistant to radio, and electromagnetic interference than
twisted pair cables.
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Disadvantages of coaxial cables include:

1. Thick coaxial cable is hard to work with.
2. Coaxial cables are relatively expensive to buy and to install as compared to
twisted pair.
3. Local area network established using coaxial cable is difficult to troubleshoot
and maintain. This has made coaxial cabling unpopular in LANS hence the
wide usage of twisted pair cables.
(d) Fibre optic cables
Activity 13.6 To analyse the use of fibre optic cables in

communication
“Optical cables have become the most reliable and efficient channels of data
transmission”.
1. Discuss three justifications for this proposition.
2. Draw and label the structure of an optical cable on a manila paper.
3. Discuss the function of each part.
4. Illustrate with a diagram the process of data transmission in the cable.
Fibre optic is one of the latest physical transmission media to be used in local
and wide area networks. Instead of transmitting data signals using electronic
signals, fibre optic cable uses light to transmit data from one point to another on
the network. The electrical signals from the source are converted to light signals,
then propagated along the cable. To convert electrical signal to light, the source
has a transmitter system made of light emitting diode (LED). At the receiving
end, a photosensitive device is used to convert the light back to electric signals.
The fibre optic cable is made up of the core, cladding, buffer, strength members
and the jacket.
• Core: The core is the central part of the cable and is made of a hollow
transparent plastic or glass.
• Cladding: This is a single protective layer surrounding the core. It has some
light bending characteristics in that, when the light tries to travel from the
core to the cladding, it is redirected back to the core. This is why even if a fibre
optic cable is bent into coils and a light signal is inserted at one end it will still
be seen coming out from the other end.
• Buffer:The buffer surrounds the cladding and its main function is to strengthen
the cable.
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• Jacket: It is the outer covering of the cable.

Fibre optic cables can be classified into two categories namely: Single mode and
multimode fibre optic cables.
The single mode fibre cable has a very narrow centre core (Figure 13.7 (a)). The
light in the cable can therefore take only one path through it. Because of this, it
has a very low attenuation rate and is preferred for long distance transmission.
It has a bandwidth of 50 Gbps which is higher than that of the twisted pair’s
100 Mbps. Single mode fibre is very expensive and requires very careful handling
during installation.
Multimode fibre cable has a thicker core than the single mode (Figure 13.7 (b)). It
allows several light rays to be fed in the cable at an angle. Because of multiple light
signals navigating the cable at the same time, distortion of the signal is possible.
Multimode cables have a high attenuation rate and are usually used for shorter
distances than single mode.
Cladding Thin core
Cladding
Jacket
Thick core
(b)
(a)
Fig. 13.7 (a) Single mode fibre cable (b) Multimode fibre cable
In fibre optic cables, the data signal travels as light through the core due to
total internal reflection. Total internal reflection occurs when light travels from
optically dense medium such as glass to less optically dense medium such as air.
The phenomenon that causes total internal reflection is called refraction. When
light travels from optically dense medium, it is refracted away from the normal to
a point that the ray deviates so far away from normal making it reflected rather
than being refracted. When light signal is fed into fibre optic cable, it rises to
cross from the core to the cladding. The light is bent back into the core and hence
propagated along the length of the cable as shown in Figure 13.8.
Light rays
Core Cladding
Fig. 13.8: Multimode fibre
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Figure 13.9 below shows how a fibre based network transmits data from source
to destination.
Sending
computer
1010011
Fibre optic cable Photodetector transforms
LED changes light signal to electric signals.
electrical signals
to light signals
1010011
Receiving
computer
Fig. 13.9: Fibre network
The advantages of fibre optic cabling include:

1. Fibre optic cable is immune to electromagnetic interference and
eavesdropping.
2. Fibre optic cables supports high bandwidth.
3. It can be used as backbone in wide area networks because it has low
attenuation.
4. Can be used in highly flammable places because they do not generate
electrical signal.
5. Fibre optic cable is smaller and lighter than copper cables hence good for
space between ceiling and rooftop.
The disadvantages of fibre optic cabling include:
1. Installation and configuration of fibre optic network devices and the media
are expensive.
2. Installation is difficult because the cable is dedicated.
3. Fibre optic network is difficult to troubleshoot and complex to configure.
Exercise 13.2
Part 1: Select the most appropriate answer.
1. ___________ cable consists of an inner copper core and a second conducting
outer sheath.
A. Twisted-pair B. Shielded twisted-pair
C. Coaxial D. Fiber-optic
2. In an optical fibre, the inner core is ___________ the cladding.
A. less dense than B. denser than
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C. the same density as D. another name for

3. ___________ cable is used for voice and data communications.
A. Twisted-pair B. Coaxial
C. Fiber-optic C. All of these
4. ___________ cables carry data signals in the form of light.
A. Twisted-pair B. Coaxial
C. Fiber-optic D. one of these
5. In a fiber-optic cable, the signal is propagated along the inner core by
___________.
A. refraction B. total internal reflection
C. modulation D. none of these
6. ___________ cables are very cheap and easy to install, but they are badly
affected by the noise interference.
A. STP B. UTP
C. Co-axial D. Optical Fiber
7. Twisting of wires in twisted pair cable helps to
A. increase the data speed
B. reduce the effect or noise or external interface
C. make the cable stronger
D. make the cable attractive
8. Signals with a frequency above 30 MHz use ___________ propagation.
A. line-of-sight B. sky
C. ground D. none of these
9. Twisted pair cables offers data rates up to ___________
A. 100 Mbps B. 15 Mbps
C. 20 Mbps D. 25 Mbps
10. State whether the following statements are True for twisted pair cable.
(i) Signal attenuation is lower in STP than in UTP
(ii) The cost of UTP is higher than STP
(iii) The installation of UTP is fairly easy than STP
A. (i) and (iii) only B. (i) and (ii) only
C. (ii) and (iii) only D. All (i), (ii) and (iii)
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11. ___________ is the overlapping of frequency bands, which can distort/wipe-

out a signal.
A. Noise B. Attenuation
C. Interference D. Distortion
12. What is the major factor that makes coaxial cable less susceptible to noise
than twisted-pair cable?
A. insulating material B. inner conductor
C. diameter of cable D. outer conductor
13. Which of the following are the advantages of fiber optic communication
over other conventional means of communication.
(i) Small size and light weight
(ii) Easy availability and low cost
(iii) No electrical or electromagnetic interference
(iv) Large bandwidth
A. (i), (ii) and (iii) only B. (ii), (iii) and (iv) only
C. (i), (iii) and (iv) only D. All (i), (ii), (iii) and (iv)
Part 2
1. A _____ fibre has a thick core while a _____ fibre has a narrow one.
2. Multimode fibre has high attenuation because of _____.
3. _____ is caused by refraction of light at the critical angle of incidence.
4. State one factor that affect the attenuation rate of a fibre cable.
5. Highlight two advantages and disadvantages of the following cables:
(a) Coax (b) UTP (c) Fibre (d) Two wire open lines
6. Describe the fibre optic data transmission system. Draw a diagram to
illustrate its elements.
13.3.2 Wireless communication
Activity 13.7 To identify wireless communication channels

Match the cited communication to the wireless communication channels used.
TV remote device to TV set Blue tooth
Continent to continent Infrared
Between two distant mobile phones Satellite
From a phone to a computer Radio waves
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Wireless transmission medium is used to transmit data from one point to another
through non-physical channel. Examples of wireless transmission media include
microwaves, satellite, radiowaves, and infrared all of which use different frequencies
of the electromagnetic spectrum shown in Figure 13.10.
In this section, we discuss three types of wireless transmission media namely;
microwaves, radiowaves and infrared waves.
Wavelength decreases
Radiowaves
1016Hz Ultra-violet (UV)
Infra-red (IR)
Gamma rays
Microwaves
Visible light
X - Rays
UHF
HF
VHF
1013Hz
1022Hz
1020Hz
107Hz
106Hz
1015Hz
10 Hz
108Hz
10
Frequency increases
Fig. 13.10: The electromagnetic spectrum
(a) Microwave transmission

Microwave frequencies range from about 3GHz to 40GHz on the electromagnetic
spectrum. Due to small wavelength, microwaves easily release energy in water
as heat hence they are also used in making microwave ovens used as domestic
appliances. In networking, microwaves are suitable for point to point transmissions.
This means that, a signal is directed through a sharp beam from transmitter to
receiver station. Figure 13.11 shows an illustration of point-to-point transmission
in microwaves connecting two local area networks in different buildings.
Fig. 13.11: Point to point microwave transmission
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(b) Satellite communication
Activity 13.8 To analyse satelite communication
1. Draw and label a conceptual diagram depicting a satellite communication

system.
2. Highlight the advantages of satellite communication system.
A satellite is a special type of microwave relay station in space. The satellite earth
stations are microwave dishes with antenna used for relaying a narrow beam to
the satellite in space. A satellite transmission system has three main components:
1. Transmitter earthstation that has uplink frequency for transmitting data to
the satellite.
2. Satellite that is relay station in orbit that receives, amplifies and retransmits
the signal to a receiver earth station via a downlink frequency that is different
from the uplink frequency.
3. Receiver earthstation that receive data signals from the satellite. Figure 13.12
shows an illustration of a satellite system.
Satellite in space
D
ow
k nlin
p lin k
U
Transmitter Recieving
earthstation earthstation
Fig. 13.12: Satellite transmission system
Communication satellites used for data transmission are usually launched into
space about 36 000 km above the earth in such a manner that their rotational
speed is relatively equal to that of the earth. An observer on earth will therefore
see as if the satellite is stationary in space. These type of satellites are called
geostationary satellites. They are convenient because they eliminate the task of
moving parabolic dishes in a bid to track the line of sight. A geostationary satellite
offers a large constant line of sight to earthstations. The area where the line of
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sight can easily be located is called the satellites footprint. The satellite transmits
the signal to many earthstations to form a point to multipoint transmission. In
multipoint transmission the transmitted signal scatters in all directions forming
a cell of access radius.
New trends in microwave transmission systems has seen the use of very small
aperture terminal (VSAT) technology. Very small aperture terminal refers to a
small satellite dish used in data, radio and TV communication. In recent times,
most organizations have mounted USAT in order to access satellite communication
directly instead of having to go through state owned gateways. Figure 13.13 shows
how VSAT is used to connect two locations to a communication network set up
to enable two laptops in geographically disperate locations to communicate.
Satellite
VSAT dish
Broadcasting
station
Decoders
Home Uplink centre
Fig. 13.13: VSAT technology
In VSAT, a satellite produces strong signals that can be received by a dish of only
2 metres in diameter. The signals are decoded using a decoder which is plugged
directly to a television set or a computer.
(c) Radio communication
Activity 13.9 To analyse radio communication
1. Describe the acronym names HF, VHF and UHF in regard to radio waves.
2. Sketch a conceptual diagram showing the transmission of HF radio waves
from a broadcasting station the radio receiver.
3. Identify three uses of radio waves in our daily lives.
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Radio waves travel just like surface water waves, i.e. they are omnidirectional.
This means that radiowaves start from a central point and spread outwards in
all directions. As they travel outwards, the energy emitted by the waves spreads
outwards over the covered area. The waves are radiated into the atmosphere by
a radio frequency antenna at constant velocity. Radio waves are not visible by
human eye. Figure 13.14 shows how radio waves are propagated between the
transmitting station and the receiving station.
Fig. 13.14: Radio waves transmission
Radio waves are used in radio and television broadcasts. Data can also be
transmitted over radiowaves communication channels. For example, instead of
installing telephone cables between two distant towns, radiowaves can be used to
connect the two towns. Radiowaves can be of high frequency, very high frequency
or ultra-high frequency.
The high frequency (HF) radio waves signal is propagated by directing it to the
ionosphere of the earth as shown in Fig 13.15. The ionosphere reflects it back to
the earth’s surface and the receiver picks the signal. Before invention of satellite
communication, high frequency radio transmission was a preferred mode of
communication in marine transport to direct ships and other maritime objects.
However, the main drawback in use of high frequency communication is the
danger of signal interception by unauthorised parties.
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Earth’s ionosphere
Sender Receiver
Fig. 13.15: HF radio transmission
Very high frequency (VHF) radio waves are transmitted along the earth’s surface.
Due to shape of the earth, the signal attenuates mostly at the horizon. This means
that repeater stations have to be placed strategically to maintain a line of sight in
order to receive, amplify and propagate the signal. Very high frequency is mostly
used on hand held radio devices like “walkie-talkie” radios. The range of VHF
is limited but it is preferred to high frequency where no major obstructions are
encountered. This is because with very high frequency, it is possible to make the
radiowave follow narrow and more precise or focussed path to the receiver. To
overcome the obstructions by the earth surface like mountains and buildings,
repeater stations are built on high grounds like hills and mountains.
Ultra high frequency (UHF) radiowaves are very high frequency when it comes
to the line of sight principle. This means that there should be no barrier between
the sending and the receiving stations. Notice that the television aerial for very
high frequency is bigger than the one for ultra high frequency radiowaves. This
is because UHF radiowaves can be made to follow very narrow and precise path
to the receiver than VHF radiowaves. Therefore, ultra high frequency is popular
for horizon limited broadcasts.
(d) Bluetooth transmission

One of the latest short range radio transmission technologies is called bluetooth
technology. Bluetooth is short range radio transmission that enables devices located
within a limited geographical location to communicate and share resources such
as files. The main objective of bluetooth communication is to define a common
standard that allows most devices regardless of their architectural or software
differences to share data and other resources. The main component in bluetooth
is a small low power two-way radio transceiver, small enough to be inserted in
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small devices. A network of bluetooth-enabled devices is called a wireless personal

area network (WPAN) or piconet because bluetooth networks are best suited for
hand held devices. This has made radio transmission become popular in mobile
communication and Internet connectivity. Figure 13.16 shows a piconet between
a mobile phone and a portable computer.
Fig. 13.16: A piconet of two devices
(e) Infrared transmission

Infrared waves fall below visible light on the electromagnetic spectrum. Like the
radiowaves, infrared waves are not visible to the human eye. Communication
through infrared is achieved by having transmitters and receivers (transceivers).
Transceivers of infrared signals must be within a line of sight because unlike
radio signals, they cannot penetrate obstacles like walls. However, the signal can
be reflected by surfaces like walls and ceiling before they are received.
An example of an infrared device is the transceiver installed on some mobile
phones. Once activated, two people in the same room can send messages to each
other using infrared without going through the mobile service provider hence
saving on network charges.
In networking, infrared can be used to connect devices without need for cables
e.g. connecting a computer to a printer.
Advantages and disadvantages of wireless communications

Wireless communication offers numerous advantages which justify the cost of
installation. Some of the advantages are:
1. Wireless transmission is flexible compared to physical media i.e. devices can
be moved around without disconnecting them.
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2. Wireless networks can span large geographical areas.

3. Wireless communication can take place via satellite even in remote areas
that do not have physical infrastructure like telephone lines.
Some of the disadvantages of wireless communications include:

1. It is relatively difficult to establish or configure.
2. The initial set up cost may be high.
Exercise 13.3
Part 1: Select the most appropriate answer from the choices given.
1. ___________ are used for mobile phone, satellite, and wireless network
communications.
A. Radio waves B. Infrared waves
C. Microwaves D. none of these
2. ___________ are used for short-range communications such as those between
a PC and a peripheral device.
A. Radio waves B. Infrared waves
C. Microwaves D. none of these
3. Which of the following primarily uses guided media?
A. radio broadcasting B. satellite communications
C. local telephone system D. cellular telephone system
4. ___________ use line of sight propagation
A. Radio waves B. Microwaves waves
C. Infrared D. none of these
Part 2: Fill in the gaps.
1. The _____ depicts the various frequencies and wavelenghts of electromagnetic
waves.
2. The _____ waves have smaller wavelengths, require smaller aerials for
reception and are suited for metropolitan broadcasts.
3. In a satellite transmission system, the _____ frequency must be different
from the _____ frequency to avoid interference.
4. A _____ satellite is usually launched approximately _____ km above the
earths surface and set to circle the earth at a speed that makes it to appear
stationary to a passive observer on earth.
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5. A device can receive a satellites signal only if it is within the satellites.

6. _____ is a contemporary satellite technology for transmitting satellite
content to offices and homes without installing the large satellite dishes.

• Communication is the process of using sounds, words, symbols, signs, pictures
or signals to pass a message or information from one person to another.
• The five elements of communication system are: message source, transmitter,
communication channel, and receiver and message user.
• There four commonly used bounded transmission media are: two wire open
lines cables, twisted pair cables, coaxial cables and fibre optic cables.
• The two-wire open line cable is made up of two parallel copper wires separated
by a plastic insulator. The insulator reduces crosstalks. The wires are highly
susceptible to noise and electromagnetic field.
• A twisted pair cable is made up of two solid copper wires twisted around
each other in a double helix manner. The twisting reduces the build up
of electromagnetic field. There are two categories of coaxial cables: the
unshielded twisted pair (UTP) and shielded twisted pair (STP. The UTP
cable pairs has a shield that further prevents electromagnetic interference
(EMI) from the environment.
• Coaxial cable has a copper core (coax) which may be of solid copper wire
surrounded by a dielectric material The dielectric material is then surrounded
by mesh conductor which is covered by a shield making the cable more
resistant to electromagnetic interference. There are two categories of coaxial
cables: thinnet and thicknet
• The fibre optic cable is made up of the core, cladding, buffer, strength
members and the jacket. The data signal travels as light through the core, due
to total internal reflection.
• Examples of wireless communication are microwave, radio waves, infrared,
satellite and bluetooth communications
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Unit Test 13
Part A: Select the most appropriate answer from the choices given.
1. One of the following is relatively cheap and suffers from crosstalk. Which
one?
A. Two wire open lines B. Coaxial cable C. Fiber Optic cable
2. To avoid crosstalk, UTP cables are:
A. Insulated with a thick casing
B. Shielded with a braided wire
C. Pairs are twisted at a certain pitch
3. The following statements best describes a coaxial cable:

A. A pair of copper wires twisted together; cheap; readily available.
B. Have low attenuation; uses light signals to transmit messages.
C. Has single central core surrounded by dielectric material,
aluminium foil and braided shield.
4. One of the following cables has a dielectric insulators around the core
A. Coaxial B. Multimode fibre
5. Fibre optic cables transmit light using the following concept:
A. Total internal reflection caused by refraction at the critical angle.
B. Total internal refraction caused by reflection at the critical angle.
C. Total internal reflection caused by logical mirrors at the critical angle.
6.
Which one of the following show the correct sequence of sending a message
via a satellite transmission system:
A. Source, base station, downlink,satellite,uplink,base station,destination
B. Source, base station, uplink, satellite, downlink,base station,
destination
7.
What name is given to the surface area on earth where a satellites signal
can be received.
A. Satellites shadow B. Satellites footprint C. Satellites beam.
8. Radio waves travel in a _____ manner from source to destination while

microwaves do the same in a _____ manner respectively.
A. Unidirectional, omnidirectional
B. Singledirectional, multidirectional
C. Omnidirectional, unidirectional
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9. The following types of radio waves are suitable for over the horizon
transmission:
A. High frequency (HF)
B. Ultra-High frequency (UHF)
C. Very High frequency (VHF)
10. A student realised that a phone could be linked to a computer wirelessly
to transfer photo’s from the phones memory to the computer and vise
versa within a limited distance of a few meters. What technologies could
be used to achieve this phenomena?
A. Bluetooth and infrared transmission.
B. VHF and UHF transmission.
C. Microwave transmission.
11. Which cable would you recommend for use in the following networking
environments:
A. In a petrol handling terminal.
B. In a busy office LAN.
C. A campus backbone.
D. Setting up an undersea marine cable.
E. In an electrically noisy environment.
F. To carry video and sound for cable TV or CCTV system.
G. Telephone system trunching.
Part B: Fill in the gaps.

12. The _____ cable is made up of parallel insulated copper wires.
13. Twisting of UTP cables is meant to reduce _____ between cables that are
close to each other.
14. The _____ fiber is preferred for long distance backbones while the _____
is used for short distance backbones.
15. Repeater stations are constructed on raised ground in order to maintain
a _____ with one another.
16. _____ have a limited range of about 10m radius, cannot penetrate surfaces
and can be used to wirelessly connect computers to peripheral devices.
17. _____ have a limited range of about 300m radius, can penetrate surfaces
and can be used to wirelessly connect computers to peripheral devices or
mobile devices.
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18. The range within which computing devices can wirelessly connect to a
wired network for internet access is called _____..
19. Explain the following statement: “. . radio waves are omnidirectional while
microwaves are unidirectional . .”
20. Describe the factors that are driving more and more people to set up wireless
networks.
Part C: Write a brief correct response.
21. State two differences between single mode and multimode fibre optic cable.
22. If you wanted to link networks separated by long distance, what fibre optic
cable would you select and why?
23. Describe waves available in electromagnetic spectrum.
24. What type of radio communication relies on the earths ionosphere?
25. Differentiate between single mode and multimode fibre cables.
26. Explain the importance of the wire mesh in a coaxial cable.
27. Explain the line of sight principle in infrared transmission.
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Properties of physical processes
UNIT 14 affecting plant growth
Key Unit Competence

By the end of this unit the learner should be able to describe the physical properties
affecting plant growth.
Learning objectives

• Explain environmental factors.
• Explain composition of the atmosphere, soil aeration, soil structure and soil reaction,
• Explain Biotic factors, and mineral supply.
• Describe impact of environmental factors on range plant productivity, water,
temperature, light, atmosphere, nutrients, fire and grazers.
• Describe physical properties of soil (soil structure and texture) and their application
on plant nutrition and plant growth.
Skills
• Explain aspects of physics in physical processes affecting the environment.

• Apply knowledge of physics to explain factors affecting plant growth.
• Identify physical properties of soil (soil structure and texture) and describe their
relation to plant nutrition and plant growth.
Attitude and value

• Appreciate the need to think scientifically, and evaluate environmental factors that
affects plant growth.
• Appreciate the benefits of mineral elements (cobalt, vanadium, sodium, silicon,
selenium) to plants.
• Adapting scientific method of reasoning about the environment.
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es
Introduction
Unit Focus Activity

1. Discuss and make presentation on environmental factors: Temperature,
moisture supply, radiant energy, composition of the atmosphere, soil aeration
and soil structure, soil reaction, biotic factors, supply of mineral nutrients.
2. In your view, how do temperature, radiant energy and moisture supply affect
plant growth?
3. Name other examples of factors that can affect plant growth.
4. What are biotic factors? How do they affect plant growth?
5. From the above discussed factors, discuss and make debate on the impact
of selected environmental factors on range plant productivity: water,
temperature, light, atmosphere, nutrients, fire and grazers.
6. Discuss and make presentation on physical properties of soil (soil structure
and texture) and their role in plant nutrition and plant growth.
7. A research from Internet literature on environment protection and note
down your findings.
8. Read your findings to the class.
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Properties of physical process affecting plant growth
14.1 Environmental factors and their impact on plant growth
Activity 14.1 To find out environmental factors that affect plant

growth
Materials
Internet articles on environmental factors affecting plant growth.
Reference books
Steps
1. Do a research from the internet and reference materials meaning of
environmental factors and name some factors.
2. In your view, how do temperature, radiant energy and moisture supply affect
plant growth?
3. Name other examples of factors that can affect plant growth.
5. Share your findings with your friend and then with the whole class
Environmental (abiotic) factors are all external conditions and influences

affecting the life and development of an organism. It is important to understand
the environmental aspects that affect plant growth because plant growth and
distribution are limited by these environmental factors. The following are the
most important environmental factors;
• Radiant energy • Temperature
• Moisture supply • Composition of the atmosphere
• Soil aeration and soil structure • Soil reaction
• Supply of mineral nutrient • Absence of growth-restricting
substances
14.1.1 Light (Radiant Energy)
Activity 14.2 To discuss how light affects, plant growth
Materials
• Internet articles • reference books
Steps
1. Conduct a research on the definition of light and how it is useful to plant growth.
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3. Share your findings with your friend and then to the whole class
There are three aspects of light that affect plant growth. These are quantity,
quality, and duration.
Light quantity is the intensity or concentration of sunlight and varies with
the period of the year. We experience the maximum in the summer and the
minimum in winter. The more sunlight a plant receives (up to a point), the more
food it produces through photosynthesis. As the sunlight quantity decreases the
photosynthetic process decreases and vice versa. Light quantity can be decreased
in a garden or greenhouse by using shade-cloth or shading paint above the plants.
It can be increased by surrounding plants with white or reflective material or
supplemental lights.
Light quality is the color or wavelength reaching the plant surface. Sunlight can
be broken up by a prism into respective colors of red, orange, yellow, green, blue,
indigo, and violet. Red and blue light have the greatest effect on plant growth.
Green light is least effective to plants as most plants reflect green light and
absorb very little. It is this reflected light that makes them appear green. Blue
light is primarily responsible for vegetative growth or leaf growth. Red light when
combined with blue light promotes flowering. Fluorescent or cool-white light is
high in the blue range of light quality and is used to encourage leafy growth. These
lights are excellent for starting seedlings.
Light duration or photoperiod refers to the amount of time that a plant is
exposed to sunlight. The length of uninterrupted dark periods is critical to floral
development. The ability of many plants to flower is controlled by photoperiod.
Exercise 14.1
For the following questions, select the most appropriate answers.

1. The following characteristics of light that affect plant growth except
A. Light quality B. Light quantity
C. Light duration D. Refraction of light
2. Light duration in plant growth refers to
A. The amount of time that a plant is exposed to sunlight.
B. Intervals between light bright and dull lights
C. Color of light
D. The intensity of light
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3. Light quality refers to

A. The color or wavelength reaching the plants surface
B. The amount of time plants are exposed to light
C. Color of light
D. The intensity of light
4. Light quantity refers to
A. Wave length reaching the plant
B. The intensity or concentration of sunlight
C. Color of light
D. The amount of time plants receive light
5. Light affects plant growth except
A. Dominance B. Light quantity
C. Light duration D. Light quality
14.1.2 Temperature
Activity 14.3 To discuss the effects of temperature on plant growth
Materials
• Reference materials or resource persons
Steps
1. From the knowledge of physics, define temperature.
2. Research from biology, agriculture and geography the aspects of plant growth
that temperature affects directly.
Temperature is the degree of hotness or coldness of a body. It is measured using

a thermometer. It's also a measure of how fast the atoms and molecules of a
substance are moving. Temperature is measured in degrees on the Fahrenheit,
Celsius scales. On the Kelvin scale it is measured in kelvins
Temperature directly affects the following on plant growth;
• Photosynthesis
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• Flowering
• Transpiration
• Respiration
• Sugar storage; Low temperatures reduce energy use and increase sugar
storage
• Dormancy: there is breakage of dormancy if warmth comes after a period of
low temperature and it will make plants to resume active growth.
• Absorption of water and nutrients
The rate of these processes increases with an increase in temperature. Responses
are different with different crops.Temperature also affects soil organisms. Nitrifying
bacteria is inhibited by low temperature. pH may decrease in summer due to
activities of microorganisms.
High temperatures cause increased respiration sometimes above the rate of
photosynthesis. This means that the products of photosynthesis are being used
more rapidly than they are being produced. For growth to occur photosynthesis
must be greater than respiration
Low temperatures can result in poor growth. Photosynthesis slows at low
temperatures. Since photosynthesis is slowed, growth is slowed and this results
in lower yields. Not all plants grow best in the same temperature range.
As temperature is also the measure of the intensity of heat, soil micro-organisms
show maximum growth and activity at optimum soil temperature range. The
biological processes for nutrient transformations and nutrient availability are
controlled by soil temperature and soil moisture. Soil temperature has a profound
influence on seed germination, root and shoot growth, and nutrient uptake and
crop growth. Seeds do not germinate below or above a certain range of temperature
but micro-organisms functioning in the soil are very active while a certain range
of temperature, which is about 27°C to 32°C. It is necessary to know whether the
soil temperature is helpful to the activities of plants and micro-organisms so that
it could be suitably controlled and modified. The various factors that control the
soil temperature are soil moisture, soil color, slope of the land, vegetative cover
and general soil tilth. Aeration can be used to control soil temperature, regulate
soil moisture, improve drainage, stimulate microbial activity and improve overall
soil tilth.
Exercise 14.2
1. Explain how temperature can be controlled in the soil.
2. Explain what temperature effects on plant growth directly.
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14.1.3 Moisture Supply (Humidity)
Activity 14.4 To discuss how moisture supply affects plant growth
Materials
• Science encyclopedia or • Internet articles
Steps
1. Conduct a research from the internet and reference books on the meaning
of moisture.
2. Discuss how it affects plant growth as an environmental factor.
4. Share your findings with your friend and then with to the whole class
Soil moisture refers to the amount of water present or the presence of water, in
trace amounts in the soil. Plant growth is restricted by low and high levels of soil
moisture. Good soil moisture improves nutrient uptake.
Water is a primary component of photosynthesis. Water also provides the pressure
to move a root through the soil. Among water’s most critical roles is that of a
solvent for minerals moving into the plant and for carbohydrates moving to their
site of use or storage. By its gradual evaporation, water from the surface of the
leaf, near the stomata, helps stabilize plant temperature.
Warm air can hold more water vapor than cold air. If the amount of water in the
air stays the same and the temperature increases the relative humidity decreases.
Water vapor will move from an area of high relative humidity to one of low relative
humidity. The greater the difference in humidity the faster water will move.
Self-check
Explain the importance of moisture to plant life
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14.1.4 Supply of mineral nutrients
Activity 14.5 To find out how the supply of mineral nutrients

affects the plant growth
Materials
• Science encyclopedia or Internet articles
Steps
1. Conduct a research from the internet and reference books about the supply
of mineral nutrients and how they can affect the growth of a plant.
2. In your research, name some mineral nutrients that are needed by a plant.
Plant nutrition refers to the needs and uses of the basic chemical elements in
the plant. Nutrient availability affects the soil's inherent fertility and its ability to
hold nutrients.
People confuse plant nutrition with plant fertilization. Fertilization is the term
used when these materials are supplied to the environment around the plant.
Plant nutrients are composed of single elements (for example, phosphorus (P)) or
compounds of elements (for example, ammonium nitrate (NH4NO3)). In either
case, the nutrients are all composed of atoms.
Plants need 18 elements for normal growth. Carbon, hydrogen, and oxygen are
found in air and water. Nitrogen, phosphorus, potassium, magnesium, calcium,
and sulfur are found in the soil. The latter six elements are used in relatively
large amounts by the plant and are called macronutrients. There are nine other
elements that are used in much smaller amounts; these are called micronutrients
or trace elements. The micronutrients, which are found in the soil are iron, zinc,
molybdenum, nickel, manganese, boron, copper, cobalt, and chlorine. All 18
elements, both macronutrients and micronutrients are essential for plant growth.
Most of the soil nutrients that a plant takes up must be in a soluble form (in other
words, mixed with water). Most of the nutrients that a plant needs are dissolved
in water and then absorbed by the roots. Ninety-eight percent of these plant
nutrients are absorbed from the soil solution and only about 2% are actually
extracted from the soil particles by the root. Most of the nutrient elements are
absorbed as charged ions or pieces of molecules. Ions may be positively charged
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cations or negatively charged anions. Positive and negative are equally paired so
that there is no overall charge. For example, nitrogen may be absorbed as nitrate
(NO3–) which is an anion with one negative charge. A potassium ion (K+) is a
cation with one positive charge. Potassium nitrate (KNO3–) has one potassium
ion and one nitrate ion. Calcium nitrate (Ca(NO3)2) has one calcium cation that
has two positive charges and two negative, single charge, nitrate ions to match
the two positive charges of the calcium.
Adequate water and oxygen must be available in the soil. Water is required for
nutrient movement into and throughout the roots. Oxygen is required in the soil
for respiration to occur to produce energy for growth and the movement of mineral
ions into the root cells across their membranes. This is an active absorption process
utilizing energy from respiration. Oxygen is not transported to roots from the shoot.
Without adequate oxygen from the soil environment there is no energy produced
for nutrient absorption. This also stops active absorption in which water flows into
the cell due to the higher concentration of nutrients that were actively absorbed.
Exercise 14.3
Multiple choice questions 1- 4, Select the most appropriate answer.
1. How many chemical elements are known to be important to a plants growth
survival?
A. 18 B. 13
C. 3 D. 1
2. The letters NPK represent in order;
A. Nitrate, phosporicium and potassium
B. Nitrogen, potassium and phosphate
C. Nitrogen, phosphorus and potassium
3. The ability of the soil to hold essential elements in the soil so plants can
access them during plant growth is called _____________?
A. Nutrient level
B. fertilization level
C. pH
D. cation exchange capacity or CEC
4. _____________ is required for nutrient movement in plants and _____________
is required for respiration to occur to produce energy throughout the roots
for the growth and movement of mineral ions into the root cells across their
membranes.
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A. Humus, oxygen B. Water, oxygen

C. Water, humus D. Minerals, oxygen
5. Name five macronutrients and five micronutrients of plants
Macronutrients Micronutrients
6. Give the nutrient name for the following elemental abbreviations;

(a) P___________ (b) Fe__________
(c) Mg_________ (d) K___________
14.1.5 Composition of the Earth's atmosphere
Activity 14.6 To describe the composition of the Earth's

atmosphere
Materials
• Science encyclopedia • Internet articles
Steps
1. Conduct research on the internet and reference books about the composition
of the atmosphere and how it affects plant growth and development.
2. Name the gases that are found in the atmosphere.
4. Share your findings with your friends and then with the whole class.
Scientists believe the Earth was formed about 4.5 billion years ago, and that its
early atmosphere was probably created from the gases escaping from the Earth’s
interior. The atmosphere we have today is very different from the Earth’s early
atmosphere. When the planet first cooled down 4.4 billion years ago, volcanoes
spewed out steam, carbon dioxide and ammonia, and it was 100 times as dense
as today’s atmosphere.
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The atmosphere consists of many gases.

The Earth’s atmosphere is composed of
the following molecules: nitrogen (78%),
oxygen (21%), argon (1%), and then trace
amounts of carbon dioxide, neon, helium,
methane, krypton, hydrogen, nitrous oxide,
xenon, ozone, iodine, carbon monoxide, and
ammonia. Lower altitudes also have quantities
of water vapor. The amount of carbon
dioxide in the atmosphere is maintained
through a balance between processes such as Fig. 14.1 : Composition of gases.
photosynthesis, respiration and combustion. But human activities are polluting
the atmosphere.
Photosynthesis is a key process in the evolution of the Earth’s atmosphere. It is
clear that the main gas is nitrogen. Oxygen - the gas that allows animals and plants
to respire, and fuels to burn - is the next most abundant gas. These two gases are
both elements and account for about 99% of the gases in the atmosphere. The
remaining gases, such as carbon dioxide, water vapor and noble gases such as
argon, are found in much smaller proportions.
The composition of the atmosphere, among other things, determines its ability to
transmit sunlight and trap infrared light, leading to potentially long-term changes
in climate.
14.1.6 Soil properties and their impacts on plant and growth
14.1.6.1 Soil structure and plant growth
Activity 14.7 To describe different types of soil
Materials
• Clay soil • Silt • Sand soil
Steps:
1. Collect different types of soil with the help of your teacher from different
areas including swamps.
2. Get different small heaps of each type and moisten them with water.
3. Rub them between the thumb and fingers.
4. Which one is sand soil, silt and clay?
5. Outline the characteristics of each type
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Soil structure refers to the arrangement of soil particles into aggregates (or peds)
and the distribution of pores in between them.
Aggregates are groups of soil particles held together by organic matter or chemical
forces. The arrangement of soil aggregates into different forms gives a soil its
structure. The natural processes that aid in forming aggregates are:
(i) wetting and drying,
(ii) freezing and thawing,
(iii) microbial activity that aids in the decay of organic matter,
(iv) activity of roots and soil animals, and
(v) absorbed cations.
The wetting/drying and freezing/thawing action as well as root or animal activity
push particles back and forth to form aggregates. Decaying plant residues and
microbial by-products coat soil particles and bind particles into aggregates.
Absorbed cations help form aggregates whenever a cation is bonded to two or
more particles.
Aggregates are described by their shape, size and stability. Aggregate types are
used most frequently when discussing structure ( Fig 14.1 and Table 14.1).
Fig. 14.2 : Soil structural types.
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Table 14.1 Structure type and description
Type Description
Granular Rounded surface
Crumb Rounded surfaces but larger than granular
Cube-like with flattened surfaces and
Subangular blocky
rounded corners
Cube-like with flattened surfaces and sharp
Blocky
corners
Rectangular with a long vertical dimension
Prismatic
and flattened top
Rectangular with a long vertical dimension
Columnar
and rounded top
Rectangular with a long vertical horizontal
Platy
dimension
Single grain No aggregation of coarse particles when dry
Structureless No aggregation of fine particles when dry
Water movement and drainage are poor in soils having blocky, prismatic, columnar
and platy structures. These structured soils especially the platy type are most
suitable for aquaculture.
Soil structure is defined in terms of grade, class and type of aggregates.
Grade: is the degree of aggregation and expresses the differential between cohesion
within aggregates and adhesion between aggregates.
Class: The class of structure describes the average size of individual aggregates.
Type: describes their form or shape. The various class divisions are: very fine or
very thin, fine or thin, medium, coarse or thick and very coarse or very thick.
The spaces in the soilare called pores .The pores between the aggregates are usually
large (macro-pores), and their large size allows good aeration, rapid infiltration of
water, easy plant root penetration, and good water drainage, as well as providing
good conditions for soil micro-organisms to thrive. The smaller pores within the
aggregates or between soils particles (micropores) hold water against gravity
(capillary action) but not necessarily so tightly that plants cannot extract the water.
A well-structured soil forms stable aggregates (aggregates that don't fall apart
easily) and has many pores (Figure 3.6a). A well-structured soil is friable, easily
worked and allows germinating seedlings to emerge and to quickly establish a
strong root system.
A poorly structured soil has either few or unstable (readily broken apart) aggregates
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and few pore spaces. A poorly structured soil can result in unproductive compacted
or waterlogged soils that have poor drainage and aeration. Poorly structured soil
is also more likely to slake and to become eroded.
Importance of soil structure plant growth
• Reduced erosion due to greater soil aggregate strength and decreased overland flow
• Improved root penetration and access to soil moisture and nutrients
• Retention and availability due to improved porosity.
• Increases infiltration of water, thus reducing runoff and erosion and increases
the amount of plant available water.
• Improves seedling emergence due to reduced crusting of the surface
• Large continuous pores increase permeability.
Maintaining soil structure
• Growth of legumes will also give the soil more microorganisms which give
certain beneficial fungi which will stabilize peds.
• plant cover crops in fall and winter
• plant more grasses
• turn under crop residue
• Add manure.
• Till soil only at the proper moisture contents. Never till when the soil is too
wet. This will cause the soil to become cloddy. Aggregates are easily destroyed.
• Add the proper amounts of lime and fertilizer. Proper plant growth will lead
to the development of good soil structure.
• Grow grasses and legumes. These plants may help form unstable aggregates
and their organic matter will help stabilize the aggregate.
Exercise 14.4
For questions 1 - 5, select the most appropriate answer from the choices
given.
1. An ideal soil should contain ____________?
A. Equivalent portions of sand, silt, clay and organic matter
B. Mostly clay
C. Mostly sand D. Don’t know
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2. Soil structure is _____?

A. Perforating the soil with small holes to allow air circulation, water and
nutrients to penetrate the grass roots
B. Arrangement of soil particles and pores in the soil and ability of par-
ticles to form aggregates
C. The strength of the acidity or alkalinity of soil
D. An activity of a living organism that affects another living organism
within its environment
3. The following are the natural processes that aid in the formation of ag-
gregates apart from
A. Wetting and drying B. Activity of roots and animals
C. Freezing and thawing D. Soil structure
4. __________ is not a way of maintaining a good soil structure
A. Add manure
B. Plant more grasses
C. Add the proper amounts of lime and fertilizer
D. Till when the soil is too wet
5. Pores are______?
A. Degree of aggregation
B. Average size of individual aggregates
C. Spaces in the soil
D. Class of soil
6. Discuss ways of maintaining a good soil structure for a long period of time
for better growth and development of plant life.
7. Explain the importance of soil structure for the growth of plants.
8. Discuss the difference between a well structure soil and a poorly structured
soil to plant growth and development.
9. Define soil structure and aggregation.
14.1.6.2 Soil aeration
Activity 14.8 To compare aeration in different soils
Materials
Sand soil, clay soil, water and two beakers
Steps:
1. Put sand and clay soils in two different beakers.
2. In each type of soil put some water and observe what happens. Which soil
gives out most air bubbles? Explain why.
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2. Discuss with your partner what soil aeration means

3. Describe why soil aeration is important to plant growth and development
Aeration involves perforating the soil with small holes to allow air circulation,
water and nutrients to penetrate the grass roots. This helps the roots grow deeply
and produce a stronger, more vigorous lawn. The main reason for aerating is to
alleviate soil compaction.
Compact soils of high bulk density and poor structure are poorly aerated.
Pore space is occupied by air and water so the amount of air and water are inversely
proportional to the amount of oxygen in the soil. On well drained soils, oxygen
content is not likely to be limiting to plant growth.
Plants vary widely in their sensitivity to soil oxygen. Oxygen is required by
microbe and plants for respiration. Oxygen taken up and carbon dioxide evolved
are stoichiometric. Under anaerobic conditions, gaseous carbon compounds
other than carbon dioxide are evolved. Root elongation is particularly sensitive
to aeration. Oxygen deficiency disturbs metabolic processes in plants, resulting
in the accumulation of toxic substances in plants and low uptake of nutrients.
Importance of soil aeration

1. Plant and root growth: Soil aeration is an important factor in the normal
growth of plants. The supply of oxygen to roots in adequate quantities and
the removal of CO2 from the soil atmosphere are very essential for healthy
plant growth.
When the supply of oxygen is inadequate, the plant growth either retards
or ceases completely as the accumulated CO2 hampers the growth of plant
roots. The abnormal effect of insufficient aeration on root development is
most noticeable on the root crops. Abnormally shaped roots of these plants
are common on the compact and poorly aerated soils. The penetration and
development of root are poor. Such undeveloped root system cannot absorb
sufficient moisture and nutrients from the soil
2. Microorganism population and activity: The microorganisms living in
the soil also require oxygen for respiration and metabolism. Some of the
important microbial activities such as the decomposition of organic matter,
nitrification, Sulphur oxidation etc depend upon oxygen present in the soil
air. The deficiency of air (oxygen) in soil slows down the rate of microbial
activity.
For example, the decomposition of organic matter is retarded and nitrification
arrested. The microorganism population is also drastically affected by poor
aeration.
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3. Formation of toxic material: Poor aeration results in the development of

toxins and other injurious substances such as ferrous oxide, H2S gas, CO2
gas etc. in the soil.
4. Water and nutrient absorption: A deficiency of oxygen has been found to
check the nutrient and water absorption by plants. The energy of respiration
is utilized in absorption of water and nutrients. Under poor aeration condition
(this condition may arise when soil is water logged), plants exhibit water and
nutrient deficiency
5. Development of plant diseases: Insufficient aeration of the soil also leads to the
development of diseases. For example, wilt of gram and dieback of citrus and peach.
Exercise 14.5
For questions 1 - 5, select the most appropriate response from the choices
given.
1. This constituent gives the highest aeration quality to a soil.
A. Silt B. Sand C. clay
2. Which gas allows the respiration of microorganisms living in the soil?
A. Sulphur dioxide B. Nitrogen
C. Oxygen D. Hydrogen sulphide
3. Oxygen deficiency disturbs metabolic processes and results in accumulation of
toxic substances in plants apart from?
A. Ferrous oxide B. Hydrogen sulphide
C. . Carbon dioxide D. neon
4. Which one of the following is a disadvantage of poor soil aeration
A. development of plant diseases B. competition for food
C. improved root penetration D. reduces erosion
5. soil aeration is _______?
A. Perforating the soil with small holes to allow air circulation, water and nutri-
ents to penetrate the grass roots
B. Arrangement of soil particles into aggregates
C. The strength of the acidity or alkalinity of soil
D. An activity of a living organism that affects another living organism within
its environment.
6. Discuss the importance of soil aeration to plant life
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14.1.6.3 Soil texture

Soil texture is the relative distribution of the different sized particles in the soil.
It is determined by the proportions of sand, silt, and clay in the soil. Soil texture
can also refer to the relative proportions of particles of various sizes such as sand,
silt and clay in the soil. It is a stable property of soils and, hence, is used in soil
classification and description.
Each soil separate represents a distinct physical size group.
The term soil separate refers to a specific size of a particle and not the composition
of that particle. However, certain minerals will tend to dominate or make up the
various separates.
Sands: generally made up of quartz
Silts: commonly composed of quartz and feldspars
Clays: secondary minerals, clay minerals, and Fe oxides
The organic matter is not a part of the soil's texture.
When they are wet, sandy soils feel gritty, silt soils feel smooth and silky, and
clayey soils feel sticky and plastic, or capable of being moulded. Soils with a high
proportion of sand are referred to as 'light', and those with a high proportion of
clay are referred to as 'heavy'.
The names of soil texture classes are intended to give you an idea of their textural
make-up and physical properties. The three basic groups of texture classes are
sands, clays and loams.
A soil in the sand group contains at least 70% by weight of sand. A soil in the clay
group must contain at least 35% clay and, in most cases, not less than 40%. A
loam soil is, ideally, a mixture of sand, silt and clay particles that exhibit light and
heavy properties in about equal proportions, so a soil in the loam group will start
from this point and then include greater or lesser amounts of sand, silt or clay.
Additional texture class names are based on these three basic groups. The basic
group name always comes last in the class name. Thus, loamy sand is in the sand
group, and sandy loam is in the loam group.
Soil texture influences many soil physical properties, such as water-holding
capacity and water infiltration rates. Coarse-textured sandy soils generally have
high infiltration rates but poor water-holding capacity, whereas a fine-textured
clay soil generally has a low infiltration rate but a good water-holding capacity.
Soil texture also influences the soil's inherent fertility. More nutrients can be
absorbed by a gram of clay particles than by a gram of sand or silt particles,
because the clay particles provide a much greater surface area for absorption.
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Exercise 14.6
For the questions 1 - 10, choose the most appropriate answer.

1. Soil with a high water holding capacity has a texture which is
A. Spongy and loose B. Gritty and granular
C. Wet and muddy D. Clay-like and dense
2. Nsengimana is studying different types of soil. He is comparing clay, loam, sand
and silt. Which soil will Nsengimana find to be made of the largest particles?
A. Clay B. Loam
C. Sand D. silt
3. What determines the texture of the different kinds of soil?
A. The size of the grains and the amount of space between grains.
B. Texture depends on what kinds of crops have been grown in the soil.
C. Soil texture is determined by the kind of rock that lies underneath the
soil.
D. Texture depends on the age of the soil, older soil is much finer than new
soil.
4. If Mutoni wanted to buy two thousand acres of land to use as farmland,
what kind of soil should she look for?
A. Clay B. humus
C. Loam D. sandy
5. Which is the largest particle size found in soil?
A. Gravel B. Clay
C. Silt D. sand
6. How could you test a soil texture?
A. Sift it B. Smell it
C. Add water to it D. Add plants to it
7. Kirezi dug her hands into the soil of her garden and found an earth worm,
a few plants roots, a beetle, and a spider mixed in with the soil. Why are all
these organisms important to kirezi’s garden?
A. The soil is too cold for the organisms
B. The soil is mostly made of sand
C. They destroy the vegetables in the garden
D. They recycle energy and nutrients in the soil
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8. What type of soil would Nshozamihigo use if he wants to grow plant that
needs a lot of water
A. Sandy soil
B. Dark soil that has clay in it
C. Rocky soil made of pebbles
D. Salty soil made of silt
9. Why is fertile soil the best soil for growing crops?
A. It is lighter in color and can absorb more sunlight
B. It has more fresh water
C. It has more salt water
D. It is rich with minerals and matter from dead organisms
10. Which type of soil would retain the most water?
A. Sand B. Silt
C. Loam D. Clay
11. In Fig. 14.3, match the following characteristics to the specific soil types.
silt 1. feel gritty, often dry, and fast draining

clay 2. feel smooth, silky and have a tendency
to form crust
sand 3. high water holding capacity, sticky
and capable of being moulded
Fig. 14.3
14.1.6.4 Soil reaction
Soil pH or reaction is a measure of the concentration hydrogen ion
concentration. Soil pH or soil reaction is an indication of the acidity or
alkalinity of soil and is measured in pH units.
The pH scale goes from 0 to 14 with pH 7 as the neutral point. As the amount
of hydrogen ions in the soil increases the soil pH decreases thus becoming
more acidic. From pH 7 to 0 the soil is increasingly more acidic and from pH 7
to 14 the soil is increasingly more alkaline or basic.
Effects of soil reaction on plant growth
(i) The effect of soil pH is great on the solubility of minerals or nutrients.

Fourteen of the seventeen essential plant nutrients are obtained from the
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soil. Before a nutrient can be used by plants it must be dissolved in the soil
solution. Most minerals and nutrients are more soluble or available in acid
soils than in neutral or slightly alkaline soils. A pH range of approximately 6
to 7 promotes the most ready availability of plant nutrients.
(ii) The soil pH can also influence plant growth by its effect on activity of
beneficial microorganisms Bacteria that decompose soil organic matter are
hindered in strong acid soils. This prevents organic matter from breaking
down, resulting in an accumulation of organic matter and the tie up of
nutrients, particularly nitrogen, that are held in the organic matter.
Exercise 14.7
For the following questions, choose the most correct answer

1. Which of the following pH values represents the most acid condition?
A. 1.0 B. 5.55 C. 7.0
D. 10 E. 100
2. Hydroxyl ion concentration is greatest in a soil solution with a pH a value of?
A. 10 B. 4.0
C. 5.0 D. 6.5
3. A measure of the acidity or alkalinity of the soil is called?
A. Leaching B. A Soil test
C. Soil pH D. Don’t know
4. An acid soil will have a pH ____________ while alkaline soil will have a pH
____________
A. Above 7.0 below 7.0 B. Below 7.0, above 7.0
C. Above 1.2, below 13.5 D. None of the above
5. Aluminium sulphate is used to make the soil pH
A. More alkaline B. More basic
C. More nutrient rich D. More acidic
14.2 Biotic factors
Activity 14.9 To find out what a biotic factors are
Materials
• Refence books or Internet
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Steps
1. Conduct research from the internet and reference books about biotic factors
and abiotic factors
2. In your research, name and explain some biotic and abiotic factors.
3. From the knowledge of biology, define an ecosystem.
Biotic factors refer to the living organisms, both plants and animals and the
activities they do that influence plant growth.
The effects of these living factors on plant may be advantageous or disadvantageous,

depending on how they interact with the plant.
Examples of Biotic Factors and their effects on plant growth:

Mutualism : is a species-to-species interaction in which both the biotic factor
and the plant benefit by the relationship.
Example of mutualism between plants and other organisms include:
The symbiotic relationship the Rhizobium bacteria and leguminous plants.
In which The bacteria live and obtain the supplies of energy in the roots of
the plant and in exchange, it fixes atmospheric nitrogen into the plant
Role of insects, birds and a number of animals as agents of their pollination
as they feed on nectar and leaves on plants.
Role of animals in the dispersal of fruits and seeds: Many animals, such as
birds, bats, monkeys, act as important agents for disseminating the seeds,
fruits and spores and thus they play important role in the migration of plant
Herbivory: plant-eating organisms called herbivore, such as ruminant animals,

rodents, insects, and molluscs feed on plant parts. This damage to the plant
ranging from death of the entire plant or organs, reduced root, stem, leaf
mass, total defoliation, bores and holes on plant parts
Parasitism: is an interaction between two organisms in which one organism,

called parasite, is benefited but causes harm to another, called host. Parasites
retards the growth of plants by consuming the food and minerals in the
plants. Some parasites such as fungi, bacteria and virus causing diseases to
the plants.
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Competition: Plants compete with one another for light, water and essential
minerals. For this reason, some plants in the forest grow thin and tall as
they seek light while others develop fibrous roots to tap the limited water.
Human activities: Man affect plant growth through such activities like improved
cultivation methods, deforestation, causing fires and pollution
Exercise 14.8
For questions 1 -5, select the most appropriate answer

1. What is a biotic factor?
A. Something in an ecosystem that is non living
B. Something in an ecosystem that is living
C. All the above
2. Abiotic factors are living things in an ecosystem
A. True B. False
3. Which is a biotic
A. A lake B. The air C. A tree
4. Temperature, light, air water, soil and climate are all __________ parts of the
environment
A. Biotic B. Abiotic
C. Living D. boreal
5. What is an ecosystem?
A. All the interacting organisms that live in an environment and the abiotic
parts of the environment that affects the organisms
B. A person who observes and studies the interactions between the biotic
and abiotic parts of the environment
C. The relationship among the biotic parts of the environment
D. The relationship between all the biotic elements of a pond
6. Categorise the following factors as biotic of abiotic:
salt concentration, herbivory, sunlight, diseases, soil pH water depth, rain, wind,
temperature, attitude, pollution, parasitism, competition for food.
Biotic factors Abiotic factors
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Environmental (abiotic) factors are all external conditions and influences affecting
the life and development of an organism. Examples of environmental factors
include light, temperature, moisture supply and composition of the atmosphere etc.
• Light; is factor that influences photosynthesis, phototropism, abscission,
mineral absorption, photo morphogenesis, translocation, and stomata
movement.
• Temperature influences plant growth processes such as respiration,
transpiration, seed germination, protein synthesis, breaking of seed dormancy,
and photosynthesis.
• Moisture supply(relative humidity); moisture supply affects the opening and
closing of the stomata.
• Composition of the atmosphere; oxygen and carbon dioxide in the air are of
particular importance to the physiology of plants i.e. Oxygen is essential in
respiration for the production of energy and carbon dioxide is a raw material
in photosynthesis.
• Soil structure refers to the arrangement of soil particles into aggregates (or
peds) and the distribution of forces in between.
• Soil aeration is the process perforating the soil with small holes to allow air
circulation, water and nutrients to penetrate the grass roots.
• Soil texture is the relative distribution of the different sized particles in the
soil.
• Biotic factors refer to the living organisms, both plants and animals and the
activities they do that influence plant growth. Such factors include mutualism,
parasitism, competition and human activities.
Unit Test 14
For questions 1 - 20, select the most appropriate response from the
options given
1. What does biotic mean?
A. The same thing as antibiotic B. Mammal-like
C. Plant-like D. Living
E. nonliving
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2. What will happen to a plant left in the dark?

A. It will grow green and healthy
B. It will turn yellow and spindly
C. It will go to sleep
3. What will happen to a plant that is left un-watered?
A. It will move to where there is water
B. It will grow well and healthy
C. It will wilt and eventually die
4. Which three things do plants need to make food?
A. Water, heat, oxygen B. Water, carbon dioxide, sunlight.
C. Water, sugar, air
5. The more water a plant is given, the better it will grow. Is this statement true
or false?
A. True B. False
C. It is impossible to say
6. The warmth a plant is given, the better it will grow. Is this statement true or
false?
A. True B. False
C. It is impossible to say
7. Grass doesn’t grow as quickly during the winter as it does during the summer.
What could be a reason for this?
A. It is colder in the winter
B. It doesn’t rain as much in the winter
C. The grass is sulking
8. What job does the stem doesn’t do for the plant?
A. Breathes for the plant
B. Holds the plant upright
C. Carries water and minerals up to the leaves
9. Why would keeping one plant in a fridge and one on a windowsill not be a
fair test of how temperature affects the growth of plants?
A. Because it will be cold and dark in the fridge
B. Because the plant in the fridge will have access to food
C. Because the temperature will be the same in both positions
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10. What part makes food for the plant?

A. Roots B. Petals
C. leaves
11. Sandy soils are more easily compacted than clay soils.
A. True
B. False
12. Soil is made up of?
A. Mineral matter B. Organic matter
C. Air D. Water
E. All the above
13. An ideal garden soil is?
A. Fertile B. Well drained
C. Fairly high in organic matter D. All the above
14. Poor drainage in clay soils is probably the most common soil problem which
causes difficulty in growing plants.
A. True B. false
15. The four principle components of soil are ______________?
A. Colloids, water, oxygen and compost
B. Air, water, minerals and organic matter
C. Sand, rocks, organic matter and air
16. The three textural classes of soil are?
A. Sand, silt, and clay
B. Water, minerals and air
C. Organic matter, loam and subsoil strata
D. Grass clippings, rhizomes and runners
17. ____________ is the area in the soil that water is stored so plants can absorb it.
A. Pockets B. Water table
C. Pore space D. Tube area
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18. Soil texture and soil structure are important in plant growth because they
influence the amount of air and water available to plants.
A. True B. False
19. _____________ soils hold relatively small amounts of water because their
large pore spaces allow water to drain and ____________ soils hold relatively
large amounts of water due to its majority of small pore spaces.
A. Sandy, clay B. Sandy, loam
C. Loam, clay D. Silt, humus
20. Retaining more water, reduced pore spaces, reduced oxygen flow, and
restricted root growth are the effects of
A. Fertilization B. Drainage
C. Soil compaction D. Drought
21. Use the words in the table below to fill in the blank spaces. A word can be
used twice.
Macronutrients, hydrogen, micronutrients, oxygen, ammonium nitrate,
soluble, phosphorous
Several inorganic minerals are essential for plant growth and these are usually
obtained by roots from the soil. Plant nutrients are composed of single
elements for example, _________ or compounds of elements for example,
__________. In either case, the nutrients are all composed of atoms. Plants
need 18 elements for normal growth. Carbon, _________, and _______
are found in air and water. Nitrogen, phosphorus, potassium, magnesium,
calcium, and sulfur are found in the soil. The latter six elements are used
in relatively large amounts by the plant and are called ____________. There
are nine other elements that are used in much smaller amounts; these are
called ___________. The micronutrients, which are found in the soil are iron,
zinc, molybdenum, nickel, manganese, boron, copper, cobalt, and chlorine.
All 18 elements, both _____________ and micronutrients are essential for
plant growth. Most of the soil nutrients that a plant takes up must be in a
_________ form.
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Environmental phenomena and related physics concepts
Environmental phenomena and

UNIT 15 related physical concept
Key Unit Competence

By the end of this unit the learner should be able to explain environmental
phenomena and related physics concepts
Learning objectives

• Recall modes of heat transfer.
• Describe the basic thermodynamics and relate to the environment.
• Describe the basic composition, structure and dynamics of the atmosphere.
• Explain principle of hydrologic cycle and the mechanisms of water transport in the
atmosphere and the ground.
• Outline environmental problems such as noise pollution, ozone depletion and
global warming.
Skills
• Apply the basic thermodynamics to environment.

• Explain basic composition, structure and dynamics of the atmosphere.
• Evaluate the principle of hydrologic cycle as mechanisms for water transport in the
atmosphere and the ground.
• Carryout investigation on environmental problems such as noise pollution, ozone
depletion and global warming in the context of a dynamics atmosphere.
Attitude and value

• Appreciate that all environmental processes are interdependent.
• Appreciate applications of principles and laws of physics environments issues.
• Appreciate the need to think scientifically about the environment.
• Recognize the value of adapting scientific method in analyzing the environment.
• Appreciate applications of Physics principles and laws in environmental laws.
• Be aware of human activities that affect environment.
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Introduction
Unit Focus Activity
In recent years, there has been an outcry in the world on climate change and
pollution.
1. In what ways has the global climate changed, and the causes and impact
of climate change in the world, and ways in which we can control climate
change
2. Causes and impact of noise and air pollution and how the pollution can be
minimised.
3. How laws in physics, mainly laws of thermodynamics and modes of heat
transfer govern processes in our environment
4. Let one of your groups representatives present your findings to the whole
class during the class discussion.
Environmental processes influence our lives on a daily basis. For example, climate
change is the main cause of prolonged drought and flooding in parts of the world.
It is therefore important for everyone in the world to learn how environmental
processes take place and the physics behind them, in order to understand and be
involved in conserving our environment for our sake. This is what you will gain
as we discuss various environmental phenomena and the physics laws governing
them in this unit.
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15.1 Environment and energy transfer
Activity 15.1 To explain the laws that govern heat transfer in the
environment
1. Use your knowledge of physics to explain the following environmental
phenomena:
(a) How heat energy flows from hot to a cold region and not the reverse.
(b) How heat travels from the sun through a vacuum in outer space to reach
the earth.
(c) How sea and land breezes occur.
2. Present your findings to the whole class during the class discussion.
Think!
How can you make good use of incident solar radiation to improve the lives
of people in your community?
The natural environment consist of all living and non-living things occurring
naturally on earth. These things interact with each other in the environment
All living organisms require a regular supply of energy for survival. The main
source of earth’s energy is the Sun. However, the amount of energy available from
sunlight varies widely over the earth's surface due to variation in the number of
sunlight hours experienced at different places.
Out of the total energy supplied by the sun, only about 1% is absorbed and used
by plants in photosynthesis and then stored as chemical energy. The remaining
99% percent of the energy is lost as heat energy.
The flow and exchange of heat energy between objects in the environment is
governed by the laws of thermodynamics and the modes of heat transfer. Let us
briefly look of how this happens
15.2 Application of laws of thermodynamics in energy

transfer in the environment
As we learnt in Unit 6, the first law of thermodynamics relates to conservation of
energy. It states that energy cannot be created or destroyed but can be transformed
from one form into another.
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As such all the energy transformation in the environment are governed by this law.
Some example of energy transformation processes in the environment include:
• Conversion of light energy into chemical energy in plants during
photosynthesis.
• Conversion of kinetic energy into sound, light and electrical energy during
thunderstorms.
• Conversion of heat energy into kinetic energy during heating of liquids and
gases in the environment.
• Conversion of chemical energy to heat energy during burning of objects in
the environment.
The second law of thermodynamics states that energy in all forms tends to
transform itself spontaneously into a more dispersed, random, or less organised
form. This law is sometimes stated as "entropy increases" -- entropy being the
random, unavailable energy. Whenever energy is converted from one form into
another, some of it is given off as heat, which is the most random form of energy.
The following example illustrates how the second law of thermodynamics governs
heat transfer in the environment. When a hot frying pan is removed from the
stove, at first, the heat energy is concentrated near the pan, which is, relative to
the rest of the room in a non-random state. However with time, the pan cools to
room temperature with the heat radiated throughout the room. In this state, heat
energy is now dispersed and unavailable for cooking; the heat energy flow between
the pan and the room has gone towards equilibrium, become more random, and
entropy has increased. This is what happens during many heat exchange processes
in the environment
15.3 Modes of heat transfer in the environment
Activity 15.2 To explain the laws that govern heat transfer in the
environment
Materials
• Metallic rod • Heating source • Water in a beaker
• Ink • Candle wax • Small nails
• Retort stand • Tripod stand • Steel wire
Steps
1. Stick the candle wax at different points towards one end of the metallic rod.
Fix the nails on each candle wax as shown in Fig. 15.1.
2. Now hold the rod from the other end and start heating it as shown in Fig
15.1.
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Fig. 15.1: Heating a metal rod with nails attached to it using candle wax
3. Heat the metallic rod for some time. What will happen to the nails on the
candle wax? Explain.
4. Remove the rod from the Bunsen burner and move your hands near but
beside the burning flame. What do you feel? Explain.
5. Now place the tripod stand over the flame and then the steel wire on the
tripod stand. What is the importance of the steel wire?
6. Put one drop of ink at one side of the water in the beaker. Place the beaker
on the tripod stand and start heating. What do you observe after 10 minutes
of heating? Explain.
7. While heating, move your hands near the Bunsen burner. What do you feel?
Explain.
8. When heat flows between two objects, does the temperature increase in one
object always equals to the temperature decrease of the other object? Discuss.
9. Describe how the thermal energy of an object changes when the temperature
of an object changes.
10. Explain the three modes of heat transferring the environment.
11. Note down your findings in your notebooks.
Conduction is the transfer of heat energy or movement of heat through a substance
without the movement of the particles of the substance. Conduction also takes
place between two bodies that are in contact with each other. Materials, which
conduct heat well, are called conductors of heat. Electrical conductors (such as
metals) are also good conductors of heat.
Materials, which do not conduct heat well, are called insulators. Electrical
insulators (for example, wood or glass) are usually good insulators of heat. Most
insulators are materials of low density like air or foamed plastic. Insulators are
used to prevent heat from moving from one object to another.
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Examples of heat transfer in the environment include heat flow through solid
metallic ores and rocks.
Convection is the transfer of heat energy in a
fluid by the movement of warmer and cooler
fluid particles from place to place. The hot fluid
particles reduces in density and moves up while
the cold particles being denser drops down to
occupy the space left by the rising hot particles
(Fig. 15.2). This movement sets up convectional
currents in the fluid and results in the heat
energy being transferred to all the fluid particles.
Fig. 15.2: convectional currents in water
In the environment, convection causes sea and land breezes (Fig 15.3) that affect
the weather around large water bodies e.g. causing convectional rainfall near
oceans.
Sea breeze Land breeze
Fig. 15.3: Sea breeze and land breeze
Convection of molten rock inside the earth crust is in some cases responsible
for some volcanic eruptions and movement of some plate tectonics. Through
convection, warm water around the equator moves towards the poles while
the cooler water at the poles moves towards the equator. This facilitates the
circulation of ocean water. Convectional currents of the air inside a rain cloud
cause thunderstorms. In our houses, convection through which hot air rises and
escapes through the chimney while cold air is drawn into house through doors
and widows provides air circulation in the house.
Radiation is the transfer of energy by electromagnetic waves. Radiation does not
necessarily require a material medium for the heat energy to flow through, as is the
case with convection and conduction. The heat energy in hot objects like the sun
and fires is emitted and propagated in the form of electromagnet radiations like
infrared and gamma rays. It is by radiation that heat energy moves through vacuum.
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Heat energy moves through radiation from the sun to reaches the environment.
Once in the environment, the radiations can be both useful and harmful. For
example, the warm it produces facilitates quick growth of plants. However, some
radiations like gamma rays are known to cause problems like cancer of the skin.
Exercise 15.1

1. The second law of thermodynamics states that heat cannot flow from a colder
to a hotter temperature unless work is done, and it cannot be converted
completely into work.
A. True B. False
2. According to the first law of thermodynamics, the increase in the thermal
energy of a system is equals to the work done on the system and the amount
of heat added to the system.
A. True B. False
3. (a) Explain how the thermal energy of a closed system changes with time.
(b) What law of nature prevents heat from spontaneously flowing from a
lower temperature to a higher temperature?
(c) How does heat flow from the ground to the atmosphere to satisfy the
first law of thermodynamics?
4. Explain which has the greater amount of thermal energy, one litre of water
at 50°C or two litres of water at 50°C.
5. Suppose a beaker of water is heated from the top, predict what is more likely
to occur in the water-heat transfer.
6. Explain whether or not the following statement is true: if the thermal energy
of an object increases, the temperature of the object must also increase.
7. Explain the three different ways heat is transferred from one point to another
in the different states of matter.
8. Explain why the air temperature near the ceiling of a room tends to be
warmer than one near the floor.
9. Explain why materials that are good conductors of heat are poor insulators.
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15.4 Noise Pollution
Activity 15.3 To identify causes and effects of noise pollution
Materials
• Radio • Reference books
• Internet enabled computers
Steps
1. Switch the radio on and tune it to full volume for 2 minutes. Now, tune it to
low volume.
2. Briefly discuss the meaning of noise pollution.
3. Outline the causes of noise pollution in your environment and name some
places in Rwanda where noise pollution is highly experienced.
4. Conduct a research from the Internet and reference books about the effects
of noise pollution to the general public.
Noise pollution is undesired sound that is disruptive or dangerous and can cause
harm to life, nature, and property.
The hazardous effects of noise depend on its intensity (loudness in decibels),
duration, and frequency (high or low). High and low pitch noise is more damaging
than middle frequencies. Noise may be ambient (constantly present in the
background) or peak (shorter, louder sounds).
Some of the negatives impacts of noise pollution in the environment are:
1. Physiological problems – among other physiological damages and problems,
noise causes Noise-Induced Hearing loss (NIHL) in humans, damage to the
inner ear (loud, abrupt sounds can damage the eardrum, while sustained
lower volume noise can damage the middle ear). Noise also causes headaches
and feelings of fatigues.
2. Emotional problems – Such problems include irritability and nervousness.It
is very uncomfortable and stressing to stay in a noisy place.
3. Noise disrupts sleep and communication,
4. Noise disrupts the natural order of activities in an ecosystem; for example, the
feeding and breeding of livestock the wildlife on land and marine ecosystems
is disrupted by noise.
5. Structural damage to property like buildings and trees due to vibrations
induced by sound waves.
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Some causes of noise pollution in the environment include industrial and

construction activities,moving aircrafts and vehicles, household appliances like
radios and machines like lawn mowers and tractors.
In Rwanda, noise pollution is experienced in a few areas including near airports,
industries and roads. However, good planning by the government especially in
the location of airports and industries has reduced the number of people affected
by noise pollution.
Take care!
Hey!! It’s always better to limit the volume of your electronic devices especially
headphones, because hearing loss is a result of cumulative noise over time.
It’s also a good investment to get your ears cleaned; large amounts of wax can
cause an annoying ringing in the ears called tinnitus. Home remedies, such as
using Q-tips or ear candles are not recommended by medical professionals.
You can visit an audiologist or doctor for the service
Think!
Suggest some of the ways through which noise pollution can be minimized
in your area.
Exercise 15.2
For questions 1 - 5 select the most appropriate answer from the choices
given.
1. Which of the following jobs carries a high risk of exposing a worker to hearing
damage?
A. An airport employee B. Landscaping and lawn care worker
C. Office worker D. A and B
2. Why should you be concerned with the noise level of the activity you are
such as mowing the grass or listening to music, even if you are wearing ear
protection?
A. It may prevent you from getting tinnitus
B. Noise can still harm your ears, its depends on the pitch
C. You need to think about how the noise will affect others
D. All the above
3. ____________ is the unit used to measure the loudness of sounds
A. Decibel B. Meter
C. Pitch D. Acoustic frequency
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4. Which of the following is a way in which you can avoid noise pollution?
A. Wear earplugs when you go to music concerts or other loud sound
events.
B. Avoid listening to music through headphones or headsets at unsafe
levels
C. Wear ear plugs when you are using power equipment such as lawn
mower and leaf blowers
D. All of the above
5. _____ is an immediate and permanent loss of hearing caused by a short,
intense sound
A. Sound frequency B. Amplitude
C. Acoustic trauma D. Temporary threshold shift
6. What is noise pollution?
7. What are the most causes of noise pollution in your area?
8. Discuss the negative effects of noise pollution?
15.5 Air pollution

15.5.1 Definition of air pollution
Activity 15.4 To define air pollution
Materials
• Dry heap of litters in the school compound
• Matchbox
Steps
1. Collect the litter in your compound and make a heap in the pit far away from
the buildings.
2. Use the matchbox to lit the litter. Observe the smoke coming from of it.
3. Now take a keen look at the picture in fig. 15.4. What do you see?
Fig. 15.4: Air pollution
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4. Briefly discuss the meaning of air pollution.

5. Conduct a research from the Internet and reference books about the effects
of air pollution to the general public and note down your findings in your
notebooks.
6. Discuss your findings with the whole class and with the help of the teacher
note them down in your notebooks.
Air pollution is the introduction of gases, dust particles, fumes (or smoke) or odour
that are harmful to humans, animals and plants into the atmosphere.
15.5.2 Causes of air pollution
Activity 15.5 Fieldwork: To identify causes of air pollution

Your teacher will organize a trip to one of the industrial sites in Rwanda for you
to learn application of concepts and principles in physics.
While in the industrial area, in addition to observing the application of Physics
principles, observe and make a report on the ways industries pollute the
environment, and measures the industries have taken to minimize pollution.
Pollution can result from both natural occurrences and human activities.
Natural events that pollute the air include forest fires, volcanic eruptions,
wind erosion, pollen dispersal, evaporation of organic compounds and natural
radioactivity. Pollution from natural occurrences though is not very frequent and
is relatively hard to control.
Human activities that result in air pollution include:
1. Emissions from industries and manufacturing activities

Waste incinerators, manufacturing industries and power plants emit high levels
of carbon monoxide, organic compounds, and chemicals into the air.
2. Burning Fossil Fuels

Motor vehicles, trains, shipping vessels and airplanes all burn lots of fossil fuels to
get the energy that drives them. The fumes emitted from the combustion contain
dangerous gases such as carbon monoxide, oxides of nitrogen, hydrocarbons and
dust particulates. Some of these gases further react with gases in the environment
to form other toxic gases.
This is a major cause of pollution and one that is very difficult to manage because
humans heavily rely on such modes of transports for people, goods and services.
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3. Agricultural and household chemicals

Agricultural chemicals like fertilizers, insecticides and pesticides, household fuel
chemicals like kerosene, fumigators homes, cleaning products and paints emit
harmful chemicals into the air that cause pollution.
Health matters!
When using household chemicals inside the house or offices, ensure the room
is well ventilated to avoid inhaling too much of the dangerous chemicals.
Use incinerators when burning waste that release dangerous gases and
chemicals to avoid them spreading all over the environment where people
are living
15.5.3 Effects of air pollution
Activity 15.6 To identify the effects of air pollution
Steps
1. Outline the problems that result from air pollution.
2. Discuss the methods that can be used to curb the problem.
The following are some major effects of air pollution:

1. Acidification: This is a chemical reaction involving air pollutants creating
acidic compounds in the atmosphere. When these compounds dissolve in
rain drops,they form in acidic rain. When acid rain falls over in an area, it
can corrode iron sheets; kill trees and harm livestock, wildlife and aquatic
animals like fish. In the soil, the acid changes the chemistry of the soil making
it unfit for microorganisms.
2. Eutrophication: This occurs when rain water dissolve chemicals like
nitrogenous compounds in the atmosphere then drains into water bodies
and soils. The compounds, being nutrients, result in excessive algae growth
that may cause death of living organisms due to lack of oxygen
3. Particulate matter: Air pollutants can be in the form of particulate matter,
which can be very harmful to our health. The effect usually depends on the
length of exposure time as well as the kind and concentration of chemicals
and particles one is exposed to.
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Short-term effects include eye irritation, nose, throat and upper respiratory
infections like bronchitis and pneumonia, headaches, nausea, and allergic reactions.
Long-term health effects include chronic respiratory disease, lung cancer, heart
disease, and even damage to the brain, nerves, liver, or kidneys and lung damage.
15.5.4 Prevention, monitoring and control of air pollution

Measures to prevent and control air pollution should be a collaborative effort
between governments (laws) and individual actions. In many big cities, monitoring
equipment has been installed at many points in the city.
The Government of Rwanda has over time put in place laws and polices to
prevent air pollution in the country. An example of this is the adoption of the
ministerial order N0003/16.01 of 15/07/2010 that prohibit activities that pollute
the atmosphere.
Exercise 15.3
Multiple choice questions 1 - 5 select the most appropriate answer
1. Which gas makes up the largest part of air?
A. Carbon dioxide B. Oxygen
C. Nitrogen D. Sulfur dioxide
2. Which of the following substances is not an atmospheric pollutant?
A. Carbon dioxide B. Helium
C. Oxides of nitrogen D. Sulfur dioxide
3. Why is carbon dioxide a deadly air pollutant?
A. Colourless, highly toxic with a pungent odour
B. Greenish in colour, highly toxic and odourless
C. Colourless, odourless and highly toxic
D. Ozone, is abundant in this layer.
4. A safe level of noise depends on?
A. level of noise and exposure to noise B. area
C. pitch D. frequency
5. Main sources of noise pollution are
A. Transportation equipment B. Musical instruments
C. Heavy machinery D. A and C both
6. Name some health effects of carbon dioxide as a pollutant
7. State two effects of noise pollution
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15.6 Structure and composition of the atmosphere
Activity 15.7 To describe the structure of the atmosphere
Materials
• Plain paper • Reference materials and internet
Steps
1. With the help of research from reference books or internet, name the
components that make up the atmosphere. Note them down on the plain
paper provided to you.
2. On the plain paper, draw the structure of the atmosphere.
3. Compare your findings with those of your classmates.
4. In which part do you think ozone gas would be found?
The atmosphere of Earth is the layer of gases, commonly known as air that
surrounds the planet Earth and is held in place by Earth's gravity. The air consists
of a mixture of nitrogen (78%), oxygen (21%), and other gases (1%) that surrounds
Earth. High above the planet, the atmosphere becomes thinner until it gradually
fades out in space (See Fig. 15.5)
Fig. 15.5: The structure of the atmosphere
The atmosphere is structured into 5 layers. It is thickest near the surface and thins
out with height until it eventually merges with space.
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• Troposphere - is the layer just above the Earth's surface and contains
about 75% of the atmospheric mass. Weather processes occurs in this layer.
Temperature and pressure decrease as you go higher up the troposphere.
• Stratosphere is the layer immediately above the troposphere. It is warmer at
the top than the bottom, the direct opposite to the situation in the troposphere.
The stratosphere contains the Ozone Layer, which is a thin layer of ozone
molecules with three oxygen atoms. This layer forms a protective layer that
shields life on Earth from the harmful ultraviolet radiation from the sun.
Many aircrafts fly through the stratosphere to avoid turbulence that is high
in the troposphere.
• Mesosphere is the layer immediately above the stratosphere. It s a cold layer
in which temperature generally decrease with increase in altitude. The layer
is highly rarefied (with very low oxygen) but is thick enough to slow down
meteors that move at very high speeds in the atmosphere where they burn
up, leaving fiery trails in the sky at night.
• Thermosphere is the layer immediately above the mesosphere. The density
of air in this layer is very low. In this layer, the temperature is very high in that
the few molecules present in it receive very high amounts of energy from the
Sun.
The gases in the thermosphere are not uniformly distributed but are stratified
(exist in layers) according to their molecular masses.
• Exosphere is the uppermost region of Earth's atmosphere. It gradually fades
into the space. Air in this layer is extremely thin, almost the same as the
airless vacuum of space.
Most satellites orbit the earth in the thermosphere and the exosphere.
Importance of atmosphere
• The atmosphere through the ozone layer shields life on Earth from the
harmful ultraviolet radiation from the sun. It traps heat giving the earth a
comfortable temperature.
• The gases in the atmosphere are necessary for life on earth. For example,
animals breathe in oxygen from the atmosphere for their respiration, plants
absorb carbon dioxide from the atmosphere and use it in photosynthesis.
Nitrogen is fixed in the soil by bacteria and is used in protein manufacture.
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15.7 Climate change science

15.7.1 Introduction to climate change
Activity 15.8 To differentiate between climate and weather
Steps
1. Look outside your classroom. Is there sunshine, rain, cloud or what do you
see in the atmosphere?
2. Briefly discuss the difference between weather and climate and note it down
your notebooks.
3. Discuss your facts with your friend and then with the whole class
Weather is the short-term state of the atmosphere, which may be hot or cold, wet
(rainy) or dry, calm or stormy, clear or cloudy. These weather conditions occur
in the troposphere, and may change one hour to the next or from one day to the
other. For example it may be sunny in one day and rainy the following day
Climate is the average of the prevailing weather conditions in a region over a
long period of time, usually a year. It is determined by averaging measurements
of temperature, air pressure, humidity, precipitation, sunshine, cloudiness, and
winds throughout the year and averaged over a series of years
Activity 15.9
To define climate change
1. Discuss in your group the meaning of climate change

2. Identify eight causes of climate change.
3. Discuss the effects of climate change and the measures that can be taken to
reverse and minimise climate change
4. Present your findings to the whole class in a class discussion.
Climate change refers to the large-scale changes in the long-term averages of

weather patterns.
The following are some natural causes of climate change
Continental drift:The continents as established today were formed millions of years
ago when the earth’s landmass began gradually segmenting and drifting apart. The
drifting of the continents continues even today, and continues to impact on the flow
of ocean currents and winds and position of water bodies which affect the climate.
Volcanoes: A volcanic erupts and throws large volumes of sulphur dioxide (SO2),
water vapour, dust, and ash into the atmosphere. The gases and dust particles
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partially block the incoming rays of the sun, leading to cooling. Sulphur dioxide
combines with water to form tiny droplets of sulphuric acid that may stay in the
atmosphere for a long period reflecting back the rays of sunlight blocking them
from reaching the earth leading to cooling.
The earth's tilt: The earth’s axis is tilted at an angle of 23.5° to the plane of the
orbit along which it revolves round the sun. In one half of the year, the northern
hemisphere tilts towards the sun hence it is summer in that region. The reverse
happens in the other half year causing winter. This is how the earth’s tilt causes
seasons. Changes in the tilt of the earth can affect the severity of the seasons -
more tilt means warmer summers and colder winters and vice versa
The earth's axis is actually not fixed, but moves, at a rate of about half a degree
each century. This gradual change in the direction of the earth's axis, called
precession is partially responsible for climate change.
Ocean currents: Winds push horizontally against the sea surface and drive ocean
currents. The currents move vast amounts of heat across the planet. Much of the
heat that escapes from the oceans is in the form of water vapour, which is the most
abundant greenhouse gas on Earth. The water vapour leads to the formation of
clouds that have a net cooling effect. However, ocean currents have been known
to change direction or slow down in speed, which in one way or another affects
climate.
15.7.2 Human causes of climate change
Human causes of climate change involves engaging in activities that lead to
• Ozone layer depletion
• Green house effect
Let us discuss each of these two causes of climate change in details’
15.7.2.1 The ozone layer depletion
(a) Description of ozone layer
Activity 15.10
To describe ozone layer
Materials
• Science encyclopedia • Internet articles on ozone layer
1. Using the Internet and reference books, briefly describe the ozone layer.
2. Where is ozone found in the atmosphere?
3. What role does the ozone layer play in the atmosphere?
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4. What human activities can cause the depletion of the ozone layer?
5. What should the government of Rwanda do about the protection of the
ozone layer?
6. Share your findings with the whole class
Ozone is a molecule made up of three oxygen atoms. It is formed when high-
energy ultraviolet (UV) light collides with the oxygen gas molecule (O2) causing
it to split into two oxygen atoms (O1). These atoms are unstable on their own and
bind themselves with the unsplit oxygen molecules forming ozone.
O1(atom) + O2(oxygen gas) ⇒ O3(ozone)
This process is reversed when ozone absorbs the UV rays emitted by the sun. The
energy of the UV rays splits ozone molecules (O3) into one free oxygen atom (O1)
and one molecule of oxygen gas (O2).
O3(ozone) ⇒ O1(atom) + O2(oxygen gas)
This is the process through which ozone protects the earth from the harmful
effects of UV radiations from the sun by absorbing them, allowing only a small
amount to reach the Earth's surface. This accounts for the increase in temperature
with altitude in the stratosphere where the Ozone layer is located (upper region
of stratosphere).
(b) Causes of ozone layer depletion

Ozone layer is being destroyed by a group of manufactured chemicals, containing
chlorine and/or bromine. The chemicals are called "ozone-depleting substances"
(ODS).
ODS are very stable, nontoxic and environmentally safe in the lower atmosphere.
However, their stability allows them to float up, intact, to the stratosphere where
they are split apart by the ultraviolet light, releasing chlorine and bromine. Chlorine
and bromine vigorously bombard ozone, by plucking off one atom from the ozone
molecule .A single molecule of chlorine split thousands of molecules of ozone.
The main Ozone-Depleting Substances (ODS) are:
Chlorofluorocarbons (CFCs): Are the most widely used ODS and account for
over 80% of ozone depletion. They are used as coolants in refrigerators, freezers
and air conditioners. They are found in industrial solvents, dry-cleaning agents
and hospital sterilants and in foam products such as soft-foam padding
Halons: Are used in some fire extinguishers
Methyl Chloroform: Is used mainly in industries for vapour degreasing, in aerosols
and adhesives.
Carbon Tetrachloride: Is used in solvents and some fire extinguishers.
Hydrofluorocarbons (HCFCs): They have become suitable substitutes for CFCs.
They cause less stratospheric ozone depletion than CFCs.
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O3
Fig. 15.6: Destruction of ozone layer by CFC’s
(c) Impact of ozone depletion

Continued Ozone layer depletion allows more UV rays to reach the earth where
they will cause serious impacts on humans, animals and plants including:
• Harm to human and animal health: - more skin cancers, sunburns and
premature aging of the skin, cataracts, blindness and other eye diseases and
weakening of the human immune system.
• Adverse impacts on agriculture, forestry and natural ecosystems: reduced
growth, photosynthesis and flowering of plants
• Damage to marine life:
• Materials: degrading of materials like wood, plastic, rubber, fabrics and many
construction materials.
(d) Measures taken globally and by Rwanda to protect the ozone layer
Many nations of the world Rwanda included have so far signed an international
agreement known as the Montreal Protocol on Substances that Deplete the Ozone
Layer. This nations have committed to discontinue production of CFCs, halons,
carbon tetrachloride, and methyl chloroform (except for a few special uses), and
develop more "ozone-friendly" substitutes.
Rwanda also adopted the ministerial order 0006/2008 of 15/08/2008 regulating
the importation and exportation of ozone layer depleting substances, products
and equipment containing such substances ozone depleting substances (ODS).
Key institutions like Rwanda environment management authority (REMA)
and Rwanda standard board (RSB) serve as watchdogs to ensure effective
implementation of the foregoing order and general national environmental policy.
Commendation: Rwanda’s enormous contribution to the preservation of the
ozone layer earned it a 2012 Ozone Protection Award by the United Nations
environment program (UNEP).
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Exercise 15.4
1. Ozone in the troposphere is a harmful pollutant. True or false?
A. True B. False
2. Much of the X-ray and UV radiation from the sun is absorbed in the ________
A. Stratosphere B. Thermosphere
C. Troposphere D. Mesosphere
3. Where in the atmosphere is ozone found?
A. Close to the earth
B. High up in the atmosphere
C. There is no place
4. Which of the following represents the ozone gas molecule?
A. O4 B. O3
C. O9 D. O6
5. The ozone layer is mainly found at ________ above the surface of the earth
A. 20 to 30 km B. 30 to 40 km
C. 10 to 20 km D. 40 to 60 km
6. Ozone layer absorbs the sun’s rays. Which of the following radiation does it
prevents from falling on the earth’s surface;
A. Gamma radiation B. X-rays radiation
C. Infrared radiation D. UV radiation
7. What is the number of atoms in the ozone molecule?
A. 2 B. 3
C. 4 D. 1
8. Which of the following is true?
A. Cars are solely responsible for ozone pollution
B. Only chemical industries are responsible for smoke
C. Emissions from cars and industries contribute to ozone pollution
9. Which one of the following is the largest ozone depletion substation that is
emitted through human activities?
A. Nitrous oxide B. Carbon monoxide
C. Atomic bromine D. Atomic chlorine
10. (a) Discuss how the ozone layer is formed in the atmosphere?
(b) Why do we care about atmospheric ozone?
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(c) Is total ozone uniform over the globe? Discuss.

11. Write a brief report on ozone: how it is formed, its importance and effects to
animals and plants.
12. What are the principal steps in stratospheric ozone depletion caused by
human activities?
13. What emissions from human activities lead to ozone depletion?
14. What are the reactive halogen gases that destroy stratospheric ozone?
15. Does depletion of the ozone layer increase ground level ultraviolet radiation?
16. Is depletion of the ozone layer the principal cause of climate change? Explain.
15.7.2.2 Greenhouse effect and global warming
(a) Introduction to the greenhouse effect
Activity 15.11 To explain the greenhouse effect
Materials
• A greenhouse structure • Reference books • Internet
Steps
1. Your teacher will organize a trip to a greenhouse structure where horticultural
plants are grown
2. While there, ask the greenhouse attendant how the green house accelerates
the growth of plants. Note your findings in a note book
3. With the help of the Internet and reference books, describe greenhouse
effect.
4. Explain the causes of the green house effect.
5. Share your findings with your class mates
When sunlight reaches the Earth’s surface, it is absorbed by the Earth or is reflected
back into the atmosphere. Once absorbed, the earth releases some of the energy
back into the atmosphere as heat in form infrared radiation. Greenhouse gases
like water vapor (H2O), carbon dioxide (CO2), and methane (CH4) and Nitrous
oxide (N2O)in the atmosphere absorb the heat energy from the earth and radiate
it back to the earth (Fig. 15.5). This slows or prevents the loss of heat to space and
makes the earth warmer than it would otherwise be. This process is commonly
known as the “greenhouse effect.”
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Fig. 15.7: Greenhouse effect
(b) Causes of greenhouse effect

Some human activities greatly increase the release of greenhouse gases to the
atmosphere, increasing the green house effect.
Some of these human activities include:
• Burning coal, oil and gas – this produces carbon dioxide and nitrous oxide.
• Cutting down forests (deforestation). Trees help to regulate the climate
by absorbing CO2 from the atmosphere. So when they are cut down, that
beneficial effect is lost and the carbon stored in the trees is released into the
atmosphere, adding to the greenhouse effect.
• Increasing livestock farming. Cows and sheep produce large amounts of
methane when they digest their food.
• Use of fertilisers containing nitrogen produce nitrous oxide emissions.
• Production of fluorinated gases - these gases produce a very strong warming
effect, many times greater than that by CO2.
(c) Impacts of the greenhouse effect
The greenhouse effect is the main cause of global warming which is the rise in
the average global temperatures. This is gradually affecting life on earth.
Some of the effects of global warming in the world today include:
• Changes in rainfall averages and patterns: Some areas especially in the
northern hemisphere are now experiencing excessive rainfall leading to
flooding while others experiencing low rainfall through out the year leading
to drought. Shifts in seasons have been also been observe in many parts of the
world.
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• Melting of glaciers, sea ice and ice sheets: Glaciers in the high altitude
regions in the world including the Alps, Atlas and Himalayas, are melting and
shrinking at alarming rates. Sea ice in the Arctic regions has been melting and
decreasing at an increasing rate. The Greenland and Antarctic ice sheets that
store the most of the world's fresh water, are both shrinking at an alarming
rate.
• Raising sea levels: Due to the melting of ice sea levels are rising globally. The
rate of rising has increased in recent decades.
(d) Measures that reduce the greenhouse effect and global warming
The following measures can help bring down the emissions of greenhouse gases
in the atmosphere hence minimize the greenhouse effect and global warming.
1. Reducing the use of fossil fuels: Burning of fossil fuels like wood or coal
produces more carbon emissions than other sources of energy. Reducing
the use of fossil fuels directly reduce the release of greenhouse gases to the
atmosphere.
2. Use of Green energy. This is energy that comes from renewable sources
such as sunlight, wind, rain, tides, plants, algae and geothermal heat. These
energy resources are renewable, meaning they're naturally replenished. These
sources do not release greenhouse gases to the environment
3. Afforestation: Trees absorb carbon dioxide gas from the atmosphere and
release oxygen. Absorbing carbon dioxide gas reducing the quantity of
greenhouse gases in the atmosphere.
4. Using carbon efficient technologies: For example developing vehicles that
emit negligible amounts of carbon dioxide into the atmosphere.
(e) Useful application of the greenhouse effect
The greenhouse effect is applied in structures such as greenhouses, which are
house-like structures that are fully covered with light trapping polythene paper.
Horticultural plants are then planted inside the structure. The structures trap the
sun’s energy and keep the plants warm, even in cold times. The warmth makes
the plants grow much more faster.
Exercise 15.5
1. Greenhouse effect means-
A. Release of heat from the plants during daytime
B. Slowing down the release of heat into space at night
C. Quickening the release of heat into space
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2. The extent to which greenhouse gas warms the earth depends on the length
of time it remains in the atmosphere and its ability to absorb energy.
A. True B. False
3. Which greenhouse gas has the highest global warming potential?
A. Carbon dioxide B. Methane
C. Nitrous oxide D. Sulfur hexafluoride
4. What is the largest source of greenhouse gas emissions in Rwanda?
A. Agriculture B. Transportation
C. Electricity production D. Home heating
5. Which of the following gases are greenhouse gases?
A. Carbon dioxide and methane
B. Oxygen and nitrogen
C. Carbon dioxide and oxygen
D. Methane and oxygen
15.8 The hydrosphere and hydrologic Cycle

15.8.1 Description of the hydrosphere
Activity 15.12 To describe the hydrosphere
Materials
• Reference books
Steps
1. With the help of reference books, describe the hydrosphere.
2. Present your findings to the rest of the class in a class discussion.
The hydrosphere is the combined mass of water found on, under, and over the
surface of a planet or is the liquid water component of the Earth. It includes the
oceans, seas, lakes, ponds, rivers and streams water vapour in the atmosphere,
groundwater held in soil and rocks and ice on lakes and higher on the mountains.
The hydrosphere covers about 70% of the surface of the Earth and is the home
for many plants and animals. The study of water is known as hydrology, and the
scientists who study water are called hydrologists. The chemical formula for water
is H2O
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15.8.2 The hydrologic cycle
Activity 15.13 To describe the hydrological cycle
Materials
• Manila paper • Reference books
• Internet
Steps
1. Using Internet and reference books, identify the elements in the hydrological
cycle.
2. On the manila paper illustrate the hydrological cycle with a diagram.
3. Describe fifteen uses of water
4. Present your findings to the whole class in a class discussion.
The water cycle, also known as hydrological cycle, is the natural sequence through
which water passes into the atmosphere as water vapour, precipitates to earth in
liquid or solid form, and ultimately returns to the atmosphere through evaporation.
Fig. 15. 8 shows an illustration of the water cycle.
Figure 15.8: The hydrological cycle
The hydrologic cycle does not have a particular starting point. But let us begin
at the surface water bodies. Heat from the sun heats the water at the surfaces of
water bodies like oceans, lakes, dams and rivers and make it to evaporate into the
atmosphere. Some little amount of water vapour is added into the atmosphere
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through transpiration by plants and evaporating directly from the soil in the
processes collectively referred to as evapotranspiration. A relatively small amount of
moisture sublimates from snow and ice into vapour that rises into the atmosphere
As the water vapour rises, it cools and eventually condenses to form clouds. Air
currents move the clouds around the globe. The clouds continue growing in size
and colliding as they move and eventually fall down from the sky as precipitation.
Some precipitation falls in the form of ice and snow and accumulates as glaciers on
mountains and ice sheets, Most precipitation falls as rain into the oceans or onto
land, where much of it flows by gravity to the ocean and surface runoff through
streams and rivers. A portion of the runoff groundwater seeps (percolates) into the
ground and is either stored as ground water in aquifers (saturated rock), absorbed
by plants or later emerges onto the surface as fresh water springs and streams.
From the oceans, glaciers, ground and inside the plants, the cycle begins once
again and continues for ever.
15.8.3 The importance of water

Certainly you must have heard the saying “water is life”. This saying is so true
because water is so important that without it humans, plants and animals cannot
survive on earth. The following are just a few uses of water; the list is endless.
1. Water in our bodies – Over 60% of our body is water. It is a key component
of all body fluids, and regulates body temperature.
2. Water in plants – it is a key component in all plant tissues and is a requirement
in photosynthesis and a medium for transport in plants.
3. Water for domestic use – In our homes, we use water for drinking cooking
and washing.
4. Water for industrial use: - In industries, water is used for cleaning, cooling
and as ingredient in many industrial products
5. Water for power generation – Water is used to generate hydroelectric power.
6. Water for transport – Water is used in transport of people and goods using
canoes, boats and ships.
7. Water as an habitat for marine animals- Water is the home to all marine
animal including fish, amphibians and hippos.
8. Water for recreation – Water supports many recreational activities like
swimming, diving, skating etc.
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Exercise 15.6
For questions 1- 6, select the most appropriate response from the options
given
1. What is the other name for water cycle?
A. Hydrologic cycle B. Happy fun time cycle
C. Tectonic cycle D. Hydraulic cycle
2. What is the most important step in the hydrologic cycle
A. Evaporation B. Condensation
C. Precipitation D. All of them
3. The water cycle starts at
A. Condensation B. Evaporation
C. It has no beginning or end D. Trees
4. Clouds form through this process
A. Transpiration B. Condensation
C. Infiltration D. Evaporation
5. Which of the following is a process that occurs in the water cycle?
A. Precipitation B. Condensation
C. Evaporation D. All the above
6. ____________ is the only substance on earth that commonly exists in all of
the three states of matter(solid, liquid, gas)
A. milk B. oil C. water D. paraffin
7. Fill in the gaps
As water vapor cools below 100 degrees Celsius, it condenses to __________
8. Define the term hydrosphere.
9. State five uses of water.
10. Briefly describe the water cycle
11. Discuss four ways in which we can conserve water.
15.9 Clouds
Activity 15.14 To explain formation of clouds and identify types of

clouds
Materials
• Open ground • Manila paper
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Steps
1. On a cloudy day or session, move out of the classroom in groups into an
open ground.
2. Look high up into the sky and observe the clouds. Identify different types
of clouds based on their colours and heights above the ground. Record your
observations in a note book,
3. Move to the classroom. Use the Internet and reference books to identify
the names of the different types of clouds you observed based on their
characteristics. Prepare a table on your manila paper and fill this information.
4. Discuss in your group how clouds form and record your findings.
5. Share your findings with your friends and then with the whole class
6. Present your findings and the clouds table to the whole class during the class
discussion time.
A cloud is visible mass of condensed water vapour floating in the atmosphere, high
above the ground. As learnt earlier in the section of hydrological cycle, clouds are
formed when the rising water vapour condenses due to the low temperatures in
the atmosphere.
There are very many ways to classify and name clouds based on their size, shape,
location etc. The most widely used method of classification classifies clouds into
four types: cirrus, cumulus, stratus and nimbus.
Type of cloud and features Appearance
Cirrus
These are whitish, wispy and hair-like
clouds. They a contain ice crystals.
Located in very high altitude
Cumulus
These are detached clouds that look like
white fluffy cotton balls. They are dense
in appearance and have sharp outlines.
The bases of cumulus are generally flat,
They occur at the high altitude.
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Stratus
These are broad and fairly widespread
and appear like a blanket. Their edges
are are diffuse (spread out over a large
area.
Nimbus
These are the dark rain clouds. They
have the greatest vertical height.
Exercise 15.7
1. What are clouds?
2. Name the different types of clouds.
3. What types of clouds are shown in Fig. 15.9. Give a reason for your answer.
Fig. 15.9:Clouds
4. Explain how clouds are formed.

5. Explain the term dew point temperature.
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15.10 Cyclone and anticyclones
Activity 15.15 To explain the formation of cyclones and

anticyclones
Materials
• Water • Basin
• internet • An atlas
Steps
1. Half fill the basin with water.
2. Dip your forearm into the water in the basin and whirl in the clockwise
direction while inside the water. Observe the movement of the water in the
basin.
3. Reverse the direction of whirling the water to anticlockwise Observe the
movement of the water in the basin.
4. Using your observation in steps 2 and 3, describe what cyclones and
anticyclones are.
5. Use the Internet or an atlas to confirm your facts in step 4 and also identify
areas on the earth surface where cyclones and anticyclones occur.
A cyclone (Fig. 15.10) is a large-scale, atmospheric wind-and-pressure system

characterized by low pressure at its center and by circular wind motion,
counterclockwise in the Northern Hemisphere, clockwise in the Southern
Hemisphere.
Fig. 15.10: A cyclone in the northern hemisphere

An anticyclone is the reverse of a cyclone i.e, it is a large-scale circulation of winds
around a central region of high atmospheric pressure, clockwise in the Northern
Hemisphere, counterclockwise in the Southern Hemisphere".
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Distinctive weather patterns tend to be associated with both cyclones and

anticyclones. Cyclones (commonly known as lows, L) generally are indicators of
rain, clouds, and other forms of bad weather. Anticyclones (commonly known as
highs, H) are predictors of fair weather.
Fig. 15.11 illustrates cyclones and anticyclones in the northern and southern
hemispheres.
Fig. 15.11: Cyclones and anticyclone
Cyclones, also known as hurricanes or typhoons, are mainly caused by high ocean
temperatures, broad-scale wind systems and clustered thunderstorms which
liberate the heat energy from the ocean surface and transfer it to the cyclone
For a cyclone to form, the following conditions are required:
• The ocean water must be at least 26°C.
• The atmosphere must have a low air pressure
• A thunderstorm needs to be produced
Cyclones occur in the tropical regions on either side of the equator (between the
Tropics of Cancer and Capricorn) where the above conditions are met
Due to the high temperatures in the tropics, warm air above the oceans rises
carrying water vapour. As it rises, it cools. Since the cool air can't hold as much
moisture as warm air, some water gets squeezed out of the condensing air and a
cloud begins to form. If the warm air rises very quickly, this creates an updraft.
The water in the cloud builds up quickly and starts falling back to the ground
as rain, drawing cool air down with it as a downdraft. As the simultaneous warm
updraft and cool downdraft continues, the cloud grows and eventually become
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a large thunderstorm cloud. Since there is low air pressure in the atmosphere in
the tropics, the thunderstorm clouds begin to rotate due to the Earth spinning
on its axis in a circular motion and a cyclone is created.
15.11 Global convectional currents and wind patterns.
Activity 15.16 To identify and explain the formation of global wind

patterns
Materials
• An atlas
• Internet
Steps
1. From the knowledge you acquired in Unit 5, describe heat transfer by
convection.
2. Use an atlas and Internet to describe the global convection currents and
note down the facts in your notebook?
3. Present your facts to the rest of the class in a class discussion.
Convection is the transfer of heat energy by the movement of a liquid or gas

particles. The hot fluid particles rise while the cold ones sinks. The movement
of these fluid particles, known as convectional currents, spreads the heat to the
entire fluid.
Globally, giant convectional currents in the atmosphere are caused by temperature
differences between the equator and the poles.
The Earth’s curvature causes some parts to receive the Sun’s rays more directly
than other parts. For example, the Sun shines more directly on the surface at the
equator than at the poles. As the warmer air over the equator rises, colder air from
the poles rushes toward the equator to take its place. This steady exchange of warm
and cold air between the equator and the poles produces global wind belts. The
Earth’s rotation causes the direction of the winds to bend slightly: toward the right
in the Northern Hemisphere and toward the left in the Southern Hemisphere.
Global winds push air masses around Earth and bring changes in the weather.
There are six major global wind belts namely Polar and Tropical Easterlies, the
Prevailing Westerlies and the Intertropical Convergence Zone, Trade Winds,
Doldrums, and Horse Latitudes (See Fig. 15.12).
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Fig. 15.12: Global wind belts
Polar Easterlies - Polar Easterlies are found at the north and south poles, between
60° and 90° latitudes in in the southern and northern hemispheres. They are cold
and dry because of the high latitudes of those regions. They form when the cool
air at the poles moves towards the equator.
Tropical Easterlies - Tropical Easterlies flow east to west due to the rotation of the
Earth. They form as the warm air from the equator rises, and on cooling down it
comes back down to the equator. They are located between 0° and 30° latitudes
latitude in both hemispheres.
Prevailing Westerlies - These wind belts are located between the 30° and 60°
latitude in the northern and southern hemispheres. They blow from west to east.
Intertropical Convergence Zone (ITCZ) - This is also known as Equatorial
Convergence Zone or the Intertropical Front. It is formed when southeast and
northeast trade winds converge in a low-pressure zone near the equator. It usually
appears as a band of clouds and comes with thunderstorms, which are short but
produce extreme amounts of rain.
Horse Latitudes - They occur about 30°-35° degrees north and south of the
equator. It is a region with weak winds because of high pressure and decreasing
dry air.
Trade Winds - These winds blow from the horse latitudes to the low pressure of
the ITCZ. The get their name from their ability of blowing trade ships across the
ocean very quickly. In the northern hemisphere, they blow from the northeast,
and are called Northeast Trade winds. In the Southern hemisphere, they blow
from the southeast, and called the Southeast Trade Winds.
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Doldrums - They are also called Intertropical Convergence Zone, They occur 50°
north and south of the equator the equator between the two belts of trade winds,
due to the convergence of trade winds.
The importance of wind include:
1. Production of energy using wind turbines
2. It increases the rate of evaporation hence facilitates rainfall.
3. It is an agent of pollination and seed dispersal in plants.
4. Winds regulate the temperature in a region
15.12 Thermoregulation and the physics laws that govern it

15.12.1. Definition of thermoregulation
Activity 15. 17 To define thermoregulation
Materials
• Reference books
• Internet
Steps
1. Using the knowledge of science and biology, define the term thermoregulation.
2. How does thermoregulation work?
3. Name the types of thermoregulation.
Thermoregulation is the process maintaining a constant body’s internal core

temperature despite temperature changes in the external environment.
A healthy person should have a temperature of between 37°C and 37.8°C. A
temperature above or below this range may be an indication of ill-health or
infection.Some factors that may raise the body temperature include ill-health or
infection, exercise, digestion and being outdoors in a hot weather can also increase
your temperature. Some factors that may lower the body temperature include being
exposed to cold weather, drug and alcohol, and metabolic conditions like diabetes.
Let us now discuss some physics laws that govern thermoregulation
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15.12.2 Laws of thermodynamics and thermoregulation

The first law of thermodynamics and thermoregulation
The first law of thermodynamics states that, “The total amount of energy in an
isolated system is conserved.” And is mathematically represented as
∆Q =∆U + ∆W
Where ∆Q is the heat supplied to or extracted from an closed system, ∆U is the

change in the internal energy of the system, and ∆W is the work done by the system
The energy transfers in the body metabolic processes are governed by the first
law of thermodynamics in order to maintain a constant body temperature
(thermoregulation), as explained below.
The total energy produced in the body is called the metabolic rate (∆M). It is
related to the total metabolic energy production of the body (∆H), and the external
work done by the body (∆W), by the expression:
∆M= ∆H+ ∆W
For example, if no mechanical work is done (∆W= 0), then the total chemical
energy input in form of food is converted to thermal energy, i.e. ∆H= ∆U. If the
human body does some work (i.e ∆W≠ 0), then the total chemical energy input
in form of food is converted to thermal energy and the work done by the body, i.e
∆M= ∆H+ ∆W
The second Law of thermodynamics and thermoregulation

The law states that the spontaneous change for an irreversible process in an
isolated system always proceeds in the direction of increasing entropy. In other
words, the entropy of any isolated system always increases.
Simply put the law states that ‘heat flows spontaneously from a hotter object to a
colder one, but not in the opposite direction; the reverse cannot happen without
the addition of energy.
This law governs both the direction and attainment of equilibrium in body
metabolic processes. The entropy change governs the direction in which a
metabolic process will take and determines whether a particular process will
occur or not. For example, in the oxidation of glucose, some amount of energy
is lost as heat. This ‘wasted’ energy determines the direction in which a metabolic
process should go, in this case, it raises the body temperature to maintains the
core body temperature.
493
15.12.3 Modes of heat transfer and thermoregulation

Convection is the transfer of thermal energy by the motion of fluid particles. In
a human body, the movement of body fluids e.g blood circulation facilitates heat
transfer and distribution in the body through convection.
A warm body warms the surrounding air. Convection occurs in the surrounding
air to diffuse the heat. This heat flow by convection facilitates faster cooling of
the body; as cold air flows towards the body. Since convection is increased by
wind, body movements increase the body’s heat loss by convection, for example
the ‘pendulum’ movement of the hands and limbs.
Conduction is the process by which heat energy is transferred between two points
in a material at different temperatures.
The human body is a good conductor of heat.This is why when the body is touched
with a hot object; the mind rapidly initiates a withdrawal mechanism to break the
flow of heat by conduction. Air, being a poor conductor of heat, is used in most
insulating systems, including clothing for human beings and fur and feathers for
animals that act by trapping air. The air insulates the animal and humans from
loosing heat to the environment.
Radiation is the movement of heat through electromagnetic radiations. It does not
necessarily require a medium. It plays an important role in the energy balance of
human beings. The warm human body emits heat in form of infrared radiation.
Similarly, the body absorbs heat from hot objects like hot walls and fire.
Evaporation is the process through which a liquid is transformed into vapour.
For evaporation to take place, the liquid molecules absorb heat energy from the
surrounding to change to gaseous state. As such evaporation causes cooling.
Evaporation plays key thermoregulation roles in animals and plants. As sweat
vaporizes from the skin, it produces a net cooling effect
on the body. Respiration (act of breathing) also produces
some cooling effect on animals. This is why humans
and animals sweat and pant heavily when involved in
high metabolic activities that result in high heat energy
emission e.g running. The rate of evaporation depends
on the surface area on which evaporation is taking place,
the temperature difference, the humidity, the rate of
sweating and the velocity of wind. Some animals in hotter
climates e.g the jackrabbit, (Fig. 15.13) have large ears
Fig. 15.13: Jackrabbit with
(with large surface area) over which evaporation takes large ears for cooling
place resulting in cooling.
494
15.12.4 Newton’s law of cooling

Newton’s law of cooling states that the rate at which energy is lost from a body is directly
proportional to the difference between the body’s temperature and the environmental
temperature. This means that the greater the difference, the faster the rate of heat
loss and vice versa. This is the reason why, animals loose heat and at a faster rate
in cold weather since the difference between their body temperature and that of
the environment is higher than is the case during in hot weather.
Exercise 15.8
For questions 1 - 2, select the most appropriate response from the options
given.
1. What is thermoregulation?
A. Control of body temperature B. Control of water in the body
C. Control of glucose in the body D. All the above
2. How does sweat keep us cool?
A. Sweat cools us down
B. As sweat evaporates it takes heat with it
C. We don’t sweat when hot
D. Heat is lost and gained.
3. Explain how evaporation causes cooling.
4. How does the first law of thermodynamics govern the maintenance of the
core body temperature
5. What factors influence the amount of energy that will be lost by the human
body and the environment?
6. With reference to the underlying Physics principles, mention two ways
through which a human being can prevent himself/from from loosing too
much body heat in a very cold region.
7. How does hibernation help animals to conserve heat energy?
8. Describe two ways through which animals dissipate excess heat.

The natural environment consist of all living and non-living things occurring
naturally on earth.
Laws of thermal dynamics and the environment
• The first law of thermodynamics states that energy cannot be created or
destroyed but can be transformed from one form into another. This law
governs energy transfer in the environment by ensuring that the total energy
in all closed systems in the environment remains a constant.
495
• The second law of thermodynamics states that energy in all forms tends
to transform itself spontaneously into a more dispersed, random, or less
organised form. This law governs energy transfer in the environment by
dictating the direction in which environmental processes will take i.e, to gain
or loose energy and in what form.
Modes of heat transfer
• The three modes of heat transfer are conduction, radiation and convection,
but they rely on different physical interactions to transfer heat.
• Conduction is the transfer of heat energy or movement of heat through a
substance without the movement of the particles of the substance. Example
of heat flow by conduction in the environment is the flow of heat from a hot
metal to the human body when one touches such a metal.
• Convection is the transfer of heat energy in a fluid by the movement of warmer
and cooler fluid particles from place to place. It causes sea and land breezes
and global convectional currents.
• Radiation is the transfer of heat through electromagnetic waves. It enables
energy to flow through the vacuum in space to reach the earth.
Noise pollution
• Noise pollution is undesired sound that is disruptive or dangerous and can
cause harm to life, nature and property.
• Effects of noise pollution include ear damage, headaches, stress, irritability,
disruption of sleep, communication and natural order in an ecosystem, and
structural damage to property like buildings.
Air pollution
• Air pollution is the introduction of gases, dust particles, fumes (or smoke) or
odour that are harmful to humans, animals and plants into the atmosphere.
• Causes of air pollution include emissions by industries, burning of fossil fuels,
agricultural and household chemicals
• Effects of air pollution include chronic respiratory diseases and formation of
acidic rain.
Climate change
Climate change refers to the large-scale changes in the long-term averages of
weather patterns.
• Some natural causes of climate change include, continental drift, volcanoes,
the earth's tilt and ocean currents.
• Some human causes of climate change include engaging in activities that lead
to ozone layer depletion and the greenhouse effect.
496
Ozone layer depeletion

• Ozone is a molecule made up of three oxygen atoms. The ozone layer is a layer
in the earth’s stratosphere that absorbs most of the ultraviolet radiation from
the sun and prevent it from reaching the earth.
• Ozone layer is being destroyed by a group of manufactured chemicals,
containing chlorine and/or bromine.The chemicals are called "ozone-depleting
substances" (ODS). They include chlorofluorocarbons (CFCs)halons, methyl
chloroform carbon tetrachloride and hydrofluorocarbons (HCFCs):
• Ozone depletion is the wearing out of the amount of ozone in the stratosphere
• Ozone depletion is due to chlorofluorocarbons (CFC’s), substances produced
by industries that manufacture soaps, insulating foams, air conditioners,
which are heavier than air. UV radiations break them after they are taken into
the stratosphere by wind and the breaking up releases chlorine, which reacts
with the ozone, starting a chemical cycle that destroys ozone.
Greenhouse effect.
• The greenhouse effect occurs as follows: The sunlight reaches Earth’s surface,
and is absorbed by the Earth or is reflected back into the atmosphere. Once
absorbed, the earth releases some of the energy back into the atmosphere
as heat in the form of infrared radiation. Greenhouse gases like water vapor
(H2O), carbon dioxide (CO2), and methane (CH4) and Nitrous oxide (N2O)
in the atmosphere are responsible for absorbing the heat energy from the
earth and radiating it back to the earth.
• Human activities that produce greenhouse gases include burning coal, oil
and gas, deforestation, use of fertilisers containing oxygen and production of
fluorinated gases.
• The main effect of greenhouse effect is global warming.
• Measures that reduce the greenhouse effect and global warming include
reducing the use of fossil fuels, Use of green energy, and afforestation:
Global warming
• Global warming is the increase of earth’s average surface temperature.
• Some results of global warming include changes in rainfall averages and
patterns and sea level rises
Hydrosphere and hydrological cycle
• The hydrosphere is the combined mass of water found on, under, and over
the surface of a planet or is the liquid water component of the Earth.
• The hydrological cycle is a conceptual model that describes the storage and
movement of water between the biosphere, atmosphere, lithosphere and the
hydrosphere.
497
• Some of the uses of water include being a component of all body fluids,
transport in plants, photosynthesis, industrial use, recreation and as are
habitat for marine animals
Clouds
• A cloud is a visible mass of condensed watery vapor floating in the atmosphere,
typically high above the general level of the ground.
• Clouds are generally classified into 4 broad categories cirrus, stratus,
cumulus,and nimbus.
Cyclones and anti-cyclones
• A cyclone is a large-scale, atmospheric wind-and-pressure system characterized
by low pressure at its center and by circular wind motion, counterclockwise in
the Northern Hemisphere, clockwise in the Southern Hemisphere.
• An anticyclone is a large-scale circulation of winds around a central region
of high atmospheric pressure, clockwise in the Northern Hemisphere,
counterclockwise in the Southern Hemisphere".
Convection currents and global wind patterns
• A convection current is the circular movement of air caused by the cycle of
warm air rising and cool air sinking. These currents cause the global wind
patterns.
Thermoregulation
• Thermoregulation is the process of maintaining a body’s constant internal
core temperature despite temperature changes in the external environment.
• Thermoregulation is governed by the following Physics laws and principles:
• The laws of thermodynamics
• Modes of heat transfer (conduction, convection, radiation)
• Evaporation
• Newton’s law of cooling and
• Gibbs free energy law
498
Unit Test 15
For questions 1-12 select the most appropriate answer
1. The energy of an isolated system
A. Is always decreasing B. Is always constant
C. Is always increasing D. None of the above
2. The term which can differentiate thermodynamics from other sciences is
____
A. Pressure B. Temperature
C. Mass D. None of the above
3. The thermodynamic work done by a system on the surrounding is considered
as __
A. Positive B. Negative
C. Neutral D. None of the above
4. When the heat transfer into the system is more than the work transfer out of
the system, then
A. The internal energy of the system remains constant
B. The internal energy of the system decreases
C. The internal energy of the system increases
D. None of the above
5. How does the heat transfer take place in metals?
A. Volumetric density
B. Transporting energy with free electrons
C. Unstable elastic collision
D. Random molecular collision
6. Mass transfer does not take place in
A. Conduction heat transfer B. Convection heat transfer
C. Radiation heat transfer D. None of the above
7. What is the condition for conduction mode of heat transfer between two
bodies?
A. The two bodies must be in physical contact
B. There must be a temperature gradient between the bodies
C. Both A and B
D. None of the above
8. Ozone layer is present in which of the following layers of earth
A. Stratosphere B. Ionosphere
C. Troposphere D. Mesosphere
499
9. What is the earth’s hydrosphere?

A. The gases in the air
B. The solid, rocky part of the earth
C. All of the water on the planet
D. The study of the earth’s atmosphere
10. During evaporation, water goes from a ___________ to a ____________.
A. Liquid, solid C. Solid, gas
B. Liquid, gas D. Liquid, plasma
11. Clouds form through the process of
A. Transpiration C. Infiltration
B. Evaporation D. Condensation
12. ______ is the only substance on the earth that exists in all of the three states of
matter
(a) Milk (b) Oil (c) Water (d) Paraffin
13. Describe the causes of global warming and suggest strategies that could be
adopted to overcome it.
14. What are climate change and global warming and how are they related?
15. (a) What is ozone and where is it found in the atmosphere?
(b) How is ozone formed in the atmosphere?
(c) What emissions from human activities lead to ozone depletion?
16. Explain the statement “the energy of an isolated system is always constant.”
17. Name the chemicals, which are used in refrigerators and air conditioners
and damage ozone layer when released in air.
18. What do CFCs stand for?
19. Name some devices where CFCs are used.
20. Why are CFCs considered as pollutants?
21. It’s said “carbon dioxide contribute to global warming” explain.
22. Write down the principle steps in stratospheric ozone depletion caused by
human activities in the right order.
500
Appendix AP1
Quantities and Units of Magnetism
Magnetic flux (Φ): is a measure of the quantity of magnetism, being the total number of
magnetic lines of force passing through a specified area in a magnetic field. It represents the
strength of the magnetic field over a given area. The SI unit is weber (symbol Wb).
Magnetic flux density (B) or magnetic induction is the amount of magnetic flux through
a unit area taken perpendicular to the direction of the magnetic flux, a vector quantity
measuring the strength and direction of the magnetic field around a magnet or an electric
current. It is equal to magnetic field strength times the magnetic permeability in the region
in which the field exists. The SI unit is tesla (symbol T), or N/(A·m) expressed in SI base
unit.
Magnetic field strength (H) is the intensity of a magnetic field at a given location, a vector
quantity indicating the ability of a magnetic field to exert a force on moving electric charges.
It is equal to the magnetic flux density divided by the magnetic permeability of the space
where the field exists. SI derived unit is amperes per meter (symbol A/m).
Magnetomotive force, or magnetic potential difference, is any physical cause that

produces magnetic flux. It is analogous to electromotive force. The SI unit is ampere
(symbol A). The obsolete unit ampere turn.
Reluctance— (The opposition to magnetic field flux through a given volume of space or
material. Analogous to electrical resistance.
Permeability (µ) The specific measure of a material’s acceptance of magnetic flux. Its SI
units are Newtons per square metre
Inductance (L) or self inductance is the property of a electrical circuit to oppose a change
in current. The moving magnetic field produced by a change in current causes an induced
voltage to oppose the original change. The ratio of the magnetic flux produced to the current
is called the inductance. SI unit is henry (symbol H), or Weber per ampere.
501
Appendix AP2
Relationship between linear and angular displacements
and velocities
AP2.1 Angular displacement and angular velocity
Angular displacement
Consider a particle moving along a circular path. B

As the particle moves along the arc of the circle r Arc
from A to B (Fig. AP2.1) the line OA (radius θ
O A
r) joining the particle to the centre of the circle r
sweeps through an angle θ. The angle swept is
called angular displacement. It is measured in
radian.
Fig. AP2.1: Angular displacement
In Fig. AP2.2(a), the length of the arc AB is equal to the radius r of the
circle. The angle subtended by this arc at the centre of the circle is equal
to one radian. One radian is the angle subtended at the centre of a circle
by an arc of length equal to the radius of the circle. If the length of the
arc is 2 times the radius, then the angular displacement is 2 radians. For
the whole circle, the length of the arc is its circumference, i.e (2πr). The
angular displacement is therefore 2π radians (Fig. AP2.2(b)).
B
r 2 π radian
r
O
1 radian
r 360°
A
(a) (b)
Fig. AP2.2: The radian measure
Note: θ radians is also denoted as θ rad or θ c

The angle at the centre of a circle is 2π radians. It is also equal to 360º.
Therefore,
2π radians = 360°
From this, we see that,
502
360º
1 radian = 2π = 57.3°
If the angle at the centre of a circle is 1 radian, then the length of the arc is
r units. If the angle at the centre is θ radians (Fig. AP2.3), the length l of the
arc AB of the circle is given by
r
l = × θ rad = rθ
1 rad
B
r
O θ Arc l
r
Fig. AP2.3: Arc length l A
Arc length l = rθ
Example AP2.1
The radius of a particle moving along a circular path sweeps through an angle
of 60° at the centre of the circle. Calculate the angular displacement of the
particle in radians.
Solution
2π
360° = 2π rad ⇒ 1° = rad
360°
Hence, 60° = 2π × 60° = π radians

360° 3
π
Angular displacement of the particle = radians or 1.05 rad.
3
Angular velocity
A body moving from point A to point B in a straight line (Fig. AP2.4) has
linear velocity. Linear velocity (v) is defined as the rate of change of linear
displacement.
linear displacement x
Linear velocity v = =
t
time
A body moving from point A to B in a circular motion (Fig. AP2.5) has
A B
Fig. AP2.4: Linear velocity

angular velocity (ω). Angular velocity is B
defined as the rate of change of angular Arc AB
displacement. Consider a particle moving θ
A
O
along a circular path covering an arc of length
AB in a time, t (Fig. AP2.5). The angular
displacement of the radius OA is θ in the same
time t. Hence Fig. AP2.5: Angular velocity
503
angular displacement θ
angular velocity, ω = = t
time
Angular velocity is expressed in radians per second (rad/s)
Relationship between angular velocity and frequency

Relationship between angular velocity and frequency
For one complete circular motion, θ = 360° = 2π radians and the time taken
t = T, (referred to as the periodic time).
θ 2π
Hence, ω = =
t T
1
Since the frequency of revolution f = , we have ω = 2πf.
T
Therefore: angular velocity = 2π × frequency or ω = 2πf
Relationship between the angular velocity and the linear speed

We have seen that the arc length l = rθ.
Dividing both sides by t, we have,
l rθ
= But
t t l/t is the linear speed v of the rotating particle and θ/t is its
angular velocity, ω. Therefore
Linear speed, v = radius(r) × angular velocity(ω)

v = rω
504
Glossary
GLOSSARY
• Linear motion – is a motionalong a straight line, and can therefore be
described mathematically using only one spatial dimension.
• Uniform motion – is the kind of motion in which a body covers equal
distances in equal intervals of time. It does not matter how small the time
intervals are, as long as the distances covered are equal. If a body is involved
in rectilinear motion and the motion is uniform, then the acceleration of the
body must be zero.
• Uniform motion – is the kind of motion in which a body cover unequal
distances in equal distances of time, no matter how small the time intervals.
• Momentum – is the quantity of motion that an object has. A sports team that
is on the move has the momentum. If an object is in motion (on the move)
then it has momentum. Momentum can be defined as "mass in motion.
• Inertia – is property of matter by which it continues in its existing state of
rest or uniform motion in a straight line, unless that state is changed by an
external force.
• Elastic collision – is defined as one in which there is no loss of kinetic energy
in the collision. An inelastic collision is one in which part of the kinetic
energy is changed to some other form of energy in the collision.
• Inelastic collision – is a collision in which kinetic energy is not conserved
due to the action of internal friction. In collisions of macroscopic bodies,
all kinetic energy is turned into vibrational energy of the atoms, causing a
heating effect, and the bodies are deformed.
• Coefficient of friction – is the ratio between the force necessary to move one
surface horizontally over another and the pressure between the two surfaces.
• Kinetic energy – is the energy possessed by a body by virtual of its motion.
It depends on the mass and the speed of the object.
• Kinetic theory – is the explanation of the way molecules behave- states that
all matter is made up of small particles called atoms or molecules which are
constantly moving and collide without losing energy
• Law of conservation of energy – states that energy can neither be destroyed
nor created but changes from one form to another.
• Non-renewable resources – are sources of energy that will run out or will not
be replenished in our life time. They include fossil fuels
• Atmospheric pressure – is the pressure exerted by the weight of the air
column, which at sea level has a mean value of 101,325 Pascal’s (roughly
14.6959 pounds per square inch).
505
Glossary
• Heat – is thermal energy. It can be transferred from one place to another by

conduction, convection and radiation. Conduction and convection involve
particles, but radiation involves electromagnetic waves.
• Convection – transfer of heat in a fluid by the movement of warmer and
cooler fluid from one place to another.
• Conduction – is the transfer of heat energy through a substance without the
movement of the substance from a place of higher temperature to a place of
lower temperature.
• Radiant energy – energy carried by an electromagnetic wave.
• Thermal expansion – is the tendency of matter to change in shape, area, and
volume in response to a change in temperature. Temperature is a monotonic
function of the average molecular kinetic energy of a substance. When a
substance is heated, the kinetic energy of its molecules increases.
• Heat capacity – is the number of heat units needed to raise the temperature
of a body by one degree.
• Second law of thermodynamics – states that heat flows from a hotter object
to a colder one, but not in the opposite direction, the reverse cannot happen
without the addition of energy.
• Solar collector – device used in an active solar heating system that absorbs
radiant energy from the sun.
• Sublimation – is the process of a solid changing directly to vapor without
forming a liquid.
• Temperature – measure of the average kinetic energy of all the particles in an
object.
• Thermal energy – sum of the kinetic and potential energy of the particles in
an object.
• Thermodynamics – is the study of the relationship between thermal energy,
heat and work.
• Electromagnetic induction – is a process where a conductor placed in
a changing magnetic field (or a conductor moving through a stationary
magnetic field) causes the production of a voltage across the conductor.
• A transformer – is an electrical device that transfers electrical energy between
two or more circuits through electromagnetic induction
• Eddy currents (also called Foucault currents) – are loops of electrical current
induced within conductors by a changing magnetic field in the conductor,
due to Faraday's law of induction. Eddy currents flow in closed loops within
conductors, in planes perpendicular to the magnetic field.
• Electric field – is a region where a charged body experiences a force
506
Glossary
• Electric field intensity – is the measure of the strength of an electric field at

a specified point.
• Electric field line – is a path along which a unit positive charge would tend
to move in the electric field.
• First law of thermodynamics – states that the increase in internal energy of
the system equals the work done on the system plus the heat added to the
system.
• A light-emitting diode (LED) – is a two-lead semiconductor light source.
It is a p–n junction diode that emits light when activated. When a suitable
voltage is applied to the leads, electrons are able to recombine with electron
holes within the device, releasing energy in the form of photons.
• In electrical engineering, ground or earth – is the reference point in an
electrical circuit from which voltages are measured, a common return path
for electric current, or a direct physical connection to the Earth.
• A lightning arrester – is a device used on electrical power systems and
telecommunications systems to protect the insulation and conductors of the
system from the damaging effects of lightning. The typical lightning arrester
has a high-voltage terminal and a ground terminal.
• Electricalimpedance – is the measure of the opposition that a circuit presents
to a current when a voltage is applied. In quantitative terms, it is the complex
ratio of the voltage to the current in an alternating current (AC) circuit..
• Climate – is the average of the prevailing weather conditions in a region over
a long period of time, usually a year.
• Climate change – is the large-scale changes in the long-term averages of
weather patterns
• Cloud – is a visible mass of condensed water vapor floating in the atmosphere
high above the ground
• Decibel – unit for sound intensity; abbreviated as dB
• Fossil fuels – are non-renewable sources of energy made mainly of carbon.
They include coal, oil and natural gas
• Geothermal energy – is heat from the earth, can be converted into electrical
energy by power plants.
• Generator – is a device that converts mechanical energy into electrical energy
• Global warming – is the rise of the average global temperatures.
• Ozone – is a molecule made up of three oxygen atoms.
• Loudness – the human perception of sound intensity.
• Melting point – the temperature at which a substance (solid) changes into a
liquid at that temperature.
507
Glossary
References
• REB, R. (2015). Ordinary Level Physics Syllabus, Kigali: Ministry of

Education.
• Abbot A. F., (1977), Ordinary Level Physics, Heinemann Educational
Publishers, 3rd Edition.
• Atikinson A., Sinuff H (1987), New Complete Junior Physics, Longhorn
Publishers, Nairobi.
• Tom, D. (2000). Advanced Physics. London: Hodder Education.
• Duncan T (1990 ), Physics for Today and Tomorrow, Trans-Atlantic
Publications, Inc.; 2nd edition.
• Malawi Institute of Education (2013), Malawi Junior Secondary Syllabus
for Form 3 and 4 Physics, Domasi
• Nelkon M (1990), Principles of Physics, Longman, 8th Edition
• Kariuki C (2004), Longhorn Secondary Physics Form 3, Longhorn
Publishers , Nairobi.
• Apple Dictionary.
508

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PHYSICS

For Rwanda Schools

Longhorn Publishers (Rwanda) Ltd

Longhorn Publishers (Kenya) Ltd

Longhorn Publishers (Uganda) Ltd

Longhorn Publishers (T) Ltd

© N. Kaboyo, C. Kariuki, P. Kimani 2017

All rights reserved. No part of this publication may be reproduced, stored

First published 2017

ISBN 978 9997 77 112 4

Printed by Ramco Printing Works Ltd,

Key unit competence........................................................................................ 1

2. Friction Force and Newton’s Laws of Motion ............................. 24

Key unit competence ..................................................................................... 24

3. Applications of Atmospheric Pressure....................................... 57

Key unit competence...................................................................................... 57

5. Heat Transfer and Quantity of heat..........................................107

6. Laws of thermodynamics .......................................................167

8. Electrical Power Transmission ............................................... 228

9. Electric Field Intensity ......................................................... 253

10. House Electric Installation .................................................... 277

11. Basic alternating current circuits ........................................... 299

12 Refraction of light .................................................................331

13. Telecommunication Channels................................................ 406

14. Properties of physical processes affecting plant growth.............. 432

15. Environmental phenomena and related physical concept........... 459

Key Unit Competence

Knowledge and understanding

• Describe graphs of uniform rectilinear motion.

Attitude and value

Unit Focus Activity

Fig. 1.1: A bouncing ball

1.1 Uniform and non-uniform linear motion

Activity 1.1 To identify uniform and non-uniform linear motion

(a) Distinguish between uniform and non-uniform motion.

Time (s) Time (s)

Time (s) Time (s)

(a) Distance-time graph of an object moving with uniform speed

Activity 1.2 To plot a distance-time graph for an object moving

Note: When drawing graphs determine a convenient scale that will

Determination of Speed/Velocity from distance-time graph

(b) Distance-time graph for an object moving with non-uniform speed

Activity 1.3 To plot a distance-time graph of an object with

The motion of a body moving with non-uniform speed can be presented on a

Distance (m) 0 5 12.5 25 45

Fig. 1.4: Distance-time graph for non-uniform motion

Determination of speed/velocity from the graph

1.2.2 Velocity-time graphs

Activity 1.4 To show the variation of velocity with time using a

The velocity-time graph for a body moving at constant velocity is a straight

∴ Velocity × Time = Displacement

Fig. 1.6: Velocity-time graph for a body with constant velocity

(b) Velocity – time graph for an accelerating/decelerating object.

Activity 1.5 To determine the distance moved by an accelerating

Fig. 1.7: To determine uniform velocity using a ticker-tape timer

Fig. 1.8: Tape chart for constant acceleration

Fig. 1.10: Graph of velocity against time

Change in velocity between t1 and t2 = v2 – v1