Y JL 5 Sa X8 WCML4 DJsktaz
Y JL 5 Sa X8 WCML4 DJsktaz
Y JL 5 Sa X8 WCML4 DJsktaz
C H A P TE R
12 ELECTRICITY
Syllabus
Electric current, potential difference and electric current, Ohm’s law; Resistance, Resistivity,
Factors on which the resistance of a conductor depends. Series combination of resistors,
parallel combination of resistors and its application in daily life, Heating effect of electric
current and its applications in daily life, Electric power, Interrelation between P, V, I and R.
Trend Analysis
2018 2019 2020
List of Concepts
OD/D OD D OD D
Electric Current, Potential
difference, Galvanometer, ammeter, 1 Q (1 M) 2 Q (1 M) 2 Q (3 M)
voltmeter
Ohm’s Law, Resistance, Resistivity,
Factors on which the resistance of a 1 Q (3 M)
conductor depends, alloys
Series combination of resistors, 1 Q (5 M)
1 Q (3 M) 1 Q (5 M) 1 Q (5 M) 1 Q (1 M)
parallel combination of resistors Or 1 Q (5 M)
Heating Effect of Electric Current,
Or 1 Q (3 M) 1 Q (1 M) Or 1 Q (5 M) 1 Q (5 M)
Electric Power.
TOPIC - 1 TOPIC - 1
Static and Current Electricity: Static electricity deals with the electric charges at
rest while the current electricity deals with the electric charges in motion. Resistance in Series and
Electric Current: The electric current is defined as the rate of flow of electric Parallel Combination,
charge through any cross section of a conductor. Electric Power and
Charge Q Heating Effect
Electric current = or I =
Time t
Electric current is a scalar quantity.
Ampere: It is the SI unit of current. If one coulomb of charge flows through any cross-section of a conductor in
one second, then current through it is said to be one ampere.
Electric circuit: The closed path along which an electric current flows is called an ‘electric circuit’.
Conventional direction of current: Conventionally, the direction of motion of positive charges through the
conductor is taken as the direction of current. The direction of conventional current is opposite to that of the
negatively charged electrons.
Electrochemical or voltaic cell: It is a device which converts chemical energy into electrical energy.
Galvanometer: It is a device to detect current in an electric circuit.
Ammeter: It is a device to measure current in a circuit. It is a low resistance galvanometer and is always connected
in series in a circuit.
Voltmeter: It is a device to measure the potential difference. It is a high resistance galvanometer and is always
connected in parallel to the component across which the potential difference is to be measured. Symbol is,
Ohm’s Law: The current through a conductor between two points is directly proportional to the voltage across
the two points provided external conditions remain constant.
(i) Mathematical expression for Ohm’s law:
I ∝ V
V = IR (Where R = Resistance)
(ii) V-I graph for Ohm’s law:
Resistance (R): It is the property of a conductor to resist the flow of charges through it.
(i) S.I. unit of resistance is Ohm (Ω).
1 volt
(ii) 1 ohm =
1 ampere
When potential difference is 1 V and current through the circuit is 1 A, then resistance is 1 ohm.
Rheostat: Rheostat is a variable resistor used to regulate current without changing the source of voltage.
Factors on which the Resistance of a Conductor depends: Resistance of a uniform metallic conductor is,
(i) Directly proportional to the length of conductor
(ii) Inversely proportional to the area of cross-section
(iii) Directly proportional to the temperature
(iv) Depends on nature of the material.
Resistivity (r): It is defined as the resistance offered by a cube of a material of side 1 m when current flows
perpendicular to its opposite faces.
(i) Its S.I. unit is ohm-metre (Ωm).
(ii) Resistivity does not change with change in length or area of cross-section but it changes with change in
temperature.
(iii) Range of resistivity of metals and alloys is 10–8 to 10–6 Ωm.
(iv) Range of resistivity of insulators is 1012 to 1017 Ωm.
(v) Resistivity of alloy is generally higher than that of its constituent metals.
(vi) Alloys do not oxidize (burn) readily at high temperature, so they are commonly used in electrical heating
devices.
(vii) Copper and aluminium are used for electrical transmission lines as they have low resistivity.
Mnemonics
Interpretation: Interpretation:
To find V=Multiply I and R I=Q/t
To find I=Divide V and R Q=I x t
To find R=Divide V and I t=Q /I
Q. 3. What is the function of a galvanometer in a circuit ? pure metal ? A [CBSE Board Outside Delhi, 2019]
R [CBSE Board Delhi- Set- I, 2019]
Ans. Due to high resistivity of alloys rather than its
Ans. Detect the presence or direction of current. 1 constituting metals. 1
[CBSE Marking Scheme, 2019] [CBSE Marking Scheme, 2019]
Q. 4. Name and define the SI unit of current. Q. 7. Should the resistance of a voltmeter be low or
R [CBSE Board Delhi, Set- II, 2019] high? Give reason.
A [Board Outside Delhi, Set- II, 2019]
Ans. Ampere. ½
Flow of 1 coulomb of charge per second. Ans. High. In parallel connection, less current passes
1 coulomb through high resistance. ½+½
1 ampere =
1 second ½ [CBSE Marking Scheme, 2019]
[CBSE Marking Scheme, 2019] Q. 8. Name the device that helps to maintain a potential
difference across a conductor.
Q. 5. Write the function of voltmeter in an electric R [Board Term I, 2016]
circuit. R [CBSE Board Delhi, Set-II, 2019]
Ans. A battery is used to maintain potential difference
Ans. To measure potential difference across two points. across a conductor. 1
[CBSE Marking Scheme, 2019] 1 Q. 9. Write SI unit of resistivity.
R [Board Term I, Set-2, 2015] [DDE 2017]
Q. 6. Why are the heating elements of electric toasters Ans. Ohm metre (ohm m).
and electric irons made of an alloy rather than a
Q. 1. In the experiment to study the dependence of current (I) on the potential difference (V) across a resistor, a
student obtained a graph as shown,
What does the graph depict about the dependence of current on the potential difference?
Find the current that flows through the resistor when the potential difference across it is 2.5V.
A [CBSE Board Delhi, 2019]
Topper Answer, 2019
Q. 1. A V-I graph for a nichrome wire is given below. diffidence (V) established across it. This is ohm’s
What do you infer from this graph? Draw a law.
labelled circuit diagram to obtain such a graph. Resistance of the wire can be calculated as :
[CBSE 2020] V
R =
I
0.8
= 4 ohm
0.2
This means nichrome wire has a constant value of
the resistance 4 ohm. 2
Circuit diagram:
ANSWERING TIP
AE [SQP - 2018]
Q. 5. (a) List the factors on which the resistance of a conductor in the shape of a wire depends.
(b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity ? Give reason.
(c) Why are alloys commonly used in electrical heating devices ? Give reason. U [Delhi /Outside Delhi 2018]
Ans. (a) Factors on which resistance of a conductor depends :
(i) Length of conductor [or R ∝ l]
(ii) Area of cross-section of the conductor
1
[or R ∝ ] 1
A
(b) Metals are good conductor of electricity – as they have low resistivity/have free electrons. Glass is a bad conductor
of electricity – as it has high resistivity/have no free electrons. 1
(c) Reason: Alloys have high resistivity/high melting point/alloys do not oxidize (or burn) readily at high tempera-
tures. (Any one) 1
[CBSE Marking Scheme, 2018]
Detailed Answer:
Q. 6. State Ohm’s Law. Draw a circuit diagram to Ans. Statement of Ohm’s Law,
verify this law indicating the positive and nega- Circuit diagram with polarity of battery, ammeter
tive terminals of the battery and the meters. Also and voltmeter
show the direction of current in the circuit. Direction of current by arrow.
R [Board Term I, 2016] [CBSE Marking Scheme, 2016] 3
etailed Answer:
D
Ohm’s Law: It states that “Physical conditions ANSWERING TIP
remaining same, the current flowing through a
conductor is directly proportional to the potential Candidates should write Ohm’s law
difference across its two ends”. correctly and should draw the correct
i.e., I ∝ V diagram to verify Ohm’s law.
V = IR
where the constant of proportionality R is called the Q. 7. Calculate the resistance of a 1 km long copper wire
electrical resistance. of area of cross section 2 × 10–2 cm2. The resistivity
Diagram to verify Ohm’s Law: of copper is 1.623 × 10–8 ohm-meter.
A [Board Term I, 2016]
l
Ans. R = ρ
A
1.623 × 10 −8 × 1000
=
2×10 −2×10 −4 m 2
= 0.81 × 10 Ω = 8.1 Ω. 3
Graph :
[CBSE Marking Scheme, 2016]
1 Joule 1J
1 volt = or 1 V = 2
1 Coulomb 1C
Q. 2. (i) Name an instrument that measures electric (iii)
current in a circuit. Define unit of electric current.
(ii) What are the following symbols mean in an electric
circuit. [CBSE Marking Scheme, 2016] 2 + 3
Q. 3. (i) Draw a labelled circuit diagram to study (ii) V = 1.0 V
a relationship between potential difference
I = 0.25 A
(V) across the two ends of a conductor and the
current (I) flowing through it. State the formula to V = IR
show how I in a conductor varies when V across it V
R =
is increased step wise. Show this relationship also I
on a schematic graph.
1
(ii) Calculate the resistance of a conductor if the = 4 1+2+2
0.25
current flowing through it is 0.25 A when the
applied potential difference is 1.0 V. [CBSE Marking Scheme, 2016]
U [Board Term I, 2016]
Q. 4. Draw a labelled circuit diagram to study the
relationship between the current (I) flowing
Ans. (i) through a conductor and the potential difference
(V) applied across its two ends. State the formula
co-relating the I in a conductor and the V across it.
Also show their relationship by drawing a diagram.
What would be the resistance of a resistor if the
current flowing through it is 0.15 A when the
potential difference across it is 1.05 V ?
A [Board Term I, 2015]
The formula states that the current passing
through a conductor is directly proportional to the Ans.
potential difference across its ends, provided the
physical conditions like temperature, density, etc.
remain unchanged. This is Ohm‘s law.
V
I α V or I =
R
V = IR.
Try Yourself. See Question No. 3(i) of Long Answer
Type Question.
V
I µ V or I =
R
V = IR.
V = 1.05 V
I = 0.15 A
Putting in equation
V = IR
or, 1.05 = 0.15 × R
1.05
\ R = = 7 W 1+2+2
0.15
TOPIC - 2
Resistance in Series and Parallel Combination, Electric
Power and Heating Effect
Revision Notes
Resistances in series: When two or more resistances are connected end to end so that same current flows through
each one of them in turn, they are said to be connected in series. Here, the total resistance is equal to the sum of
the individual resistances.
Rs = Rl + R2 + R3 + ........
Resistances in parallel: When two or more resistances are connected across two points so that each one of
them provides a separate path for current, they are said to be connected in parallel. Here, the reciprocal of their
combined resistance is equal to the sum of the reciprocals of the individual resistances.
1 1 1 1
= + + + ....
RP R1 R 2 R 3
Heating effect of current : When an electric current is passed through a conductor, heat is produced in it. This is
known as heating effect of current.
Joule’s law of heating: It states that the heat produced in a conductor is directly proportional to (i) the square of
the current I through it, (ii) its resistance R and (iii) the time t, for which current is passed. Mathematically, it can
be expressed as :
I 2 Rt
H = I 2 Rt joule = cal
4.18
VIt
Or H = VIt joule = cal
4.18
Practical application of the heating effect of electric current: It is utilised in the electrical
heating appliances such as electric iron, room heaters, water heaters etc. The electric heating
is also used to produce light as in an electric-bulb.
Electric energy: It is the total work done in maintaining an electric current in an electric circuit for a given time.
Electric energy, W = VIt = I2Rt joule
Electric Fuse: It is a safety device that protects our electrical appliances in case of short circuit or overloading.
(i) Fuse is made up of pure tin or alloy of copper and tin.
(ii) Fuse is always connected in series with live wire.
(iii) Fuse has low melting point.
(iv) Current capacity of fuse is slightly higher than that of the appliance.
Electric Power: The rate at which electric energy is consumed or dissipated in an electric circuit :
P = VI
V2
P = I2R =
R
S.I. unit of power = Watt (W)
1 Watt = 1 Volt × 1 ampere
Commercial unit of electric energy = Kilo Watt hour (KWh)
1 kWh = 3.6 × 106 J
1 kWh = 1 unit of electric energy
Electrical power: Electrical power is the rate at which electric energy is consumed by an appliance.
Watt: It is the SI unit of power. The power of an appliance is 1 watt if one ampere of current flows through it on
applying a potential difference of 1 volt across its ends.
1 joule
1 watt = = 1 volt × 1 ampere
1 second
or 1 W = 1Js–1 = 1VA
1 kilowatt = 1000 W.
Kilowatt hour: It is the commercial unit of electrical energy. One kilowatt hour is the electrical
energy consumed by an appliance of 1000 watts when used for one hour.
1 kilowatt hour (kWh) = 3·6 × 106 J
Power rating: The power rating of an appliance is the electric energy consumed per second
by the appliance when connected across the marked voltage of the mains.
Efficiency of an electrical device: It is the ratio of the output power to the input power.
Output power
Efficiency, h =
Input power
Mnemonics
1
A Multiple Choice Questions
⇒R=
25
Ω
Q. 1. At the time of short circuit, the electric current in Q. 5. Which of the following represents voltage?
the circuit: Work done
(a) vary continuously (b) does not change (a)
Current ×Time
(c) reduces substantially (d) increases heavily
[CBSE 2020, Delhi] (b)
Work done × Charge
Ans. Correct option : (d) Work done × Time
(c)
Q. 2. Two bulbs of 100 W and 40 W are connected in Current
series. The current through the 100 W bulb is 1 A.
The current through the 40 W bulb will be: Work done × Charge
(d) [NCERT Exemp.]
(a) 0.4 A (b) 0.6 A Time
(c) 0.8 A (d) 1 A [CBSE 2020] Ans. Correct option : (a)
Ans. Correct option : (d) Explanation : As we know that,
Q. 3. What is the maximum resistance which can be made
Work done = Charge × Potential difference
1
using five resistors each of Ω? ⇒Work done = (Current × Time) × Potential
1 5 Difference
(a) Ω (b) 10 Ω
5 [Q Charge = Current × time]
(c) 5 Ω (d) 1 Ω Workdone
Potential difference
[NCERT Exemp.] Current×Time
Ans. Correct option : (d) Q. 6. If the current I through a resistor is increased
Explanation : The highest resistance is always given by 100% (assume that temperature remains
by connecting the resistors in series. Here, the highest unchanged), the increase in power dissipated will be
1 (a) 100% (b) 200%
resistance would be 5 × =1 ohm.
5 (c) 300% (d) 400%
Therefore, the maximum resistance is 1 ohm. [NCERT Exemp.]
Q. 4. What is the minimum resistance which can be made Ans. Correct option : (c)
1 Explanation : If I is current and R is resistance then,
using five resistors each of Ω? Power, P =I2R
1 5 1
(a) Ω. (b) Ω. Power in first case, P1 = I2R
5 25 100% increase in current means that current becomes 2I
Power in second case, P2 = (2I)2R = 4I2R
(c) 1 Ω. (d) 25 Ω.
10 [NCERT Exemp.] Now, increase in dissipated power = P2 – P1
Ans. Correct option : (b) = 4I2R – I2R
Explanation : Minimum resistance is obtained when = 3I2R
resistors are connected in parallel combination. 3P
Percentage increase in dissipated power = 1 × 100
Thus, equivalent resistance obtained by connecting P1
1 = 300%
five resistors of resistance Ω each, parallel to Q. 7. In an electrical circuit three incandescent bulbs
each other : 5
A, B, and C of rating 40 W, 60 W, and 100 W, re-
1 1 1 1 1 1 1 5 spectively are connected in parallel to an electric
= + + + + ⇒ =
R 1 1 1 1 1 R 1 source. Which of the following is likely to happen
5 5 5 5 5 5
regarding their brightness?
1 25
⇒ = (a) Brightness of all the bulbs will be the same.
R 1 (b) Brightness of bulb A will be the maximum.
(c) Brightness of bulb B will be more than that of A. Explanation : It is quite clear that in a battery
(d) Brightness of bulb C will be less than that of B. circuit, the point of lowest potential is the negative
[NCERT Exemp.] terminal of the battery and the current flows from
Ans. Correct option : (c) higher potential to lower potential.
Explanation : We know that power is defined as Q. 3. Assertion (A): Electric appliances with metallic
rate of doing work. A bulb consumes electric energy body have three connections, whereas an electric
and produces heat and light. Now, bulb with more bulb has a two pin connection.
power rating will produce more heat and light or Reason (R): Three pin connections reduce heating
we can say that power rating of bulb is directly of connecting wires.
proportional to the brightness produced by bulb. Ans. Correct option : (c)
Therefore, brightness of bulb B with power rating Explanation : The metallic body of an electrical
60 W will be more than the brightness of bulb A appliances is connected to the third pin which is
having power rating as 40 W. connected to the earth. This is a safety precaution
Q. 8. An electric kettle consumes 1 kW of electric power and avoids eventual electric shock. By doing this the
when operated at 220 V. A fuse wire of what rating extra charge flowing through the metallic body is
must be used for it? passed to earth and avoid shocks. There is nothing
(a) 1 A (b) 2 A such as reducing of the heating of connecting wires
(c) 4 A (d) 5 A by three pin connections.
[NCERT Exemp.] Q. 4. Assertion (A): The electric bulbs glow immediate-
Ans. Correct option : (d) ly when switch is ON.
Explanation : Given that, Reason (R): The drift velocity of electrons in a
power = P = 1 kW = 1000 W metallic wire is very high.
Voltage = V = 220 Ans. Correct option : (a)
P 1000 Explanation : In a conductor there are large
Now, I = = = 4.5 A numbers of free electrons. When we close the
V 220 circuit, the electric field is established instantly with
Now rating of fuse wire must be slightly greater the speed of electromagnetic wave which causes
than 4.5 A, that is, 5 A. electron drift at every portion of the circuit. Due
to which the current is set up in the entire circuit
B Assertions and Reasons Type Questions
instantly. The current which is set up does not
Directions: In the following questions, a statement wait for the electrons flow from one end of the
of assertion (A) is followed by a statement of conductor to another end. It is due to this, the bulb
glows immediately when switch is ON.
reason (R). Mark the correct choice as:
Q. 5. Assertion (A): Copper is used to make electric
(a) Both assertion (A) and reason (R) are true and wires.
reason (R) is the correct explanation of assertion Reason (R): Copper has very low electrical
(A). resistance.
(b) Both assertion (A) and reason (R) are true but Ans. Correct option : (a)
reason (R) is not the correct explanation of Explanation : A low electrical resistance of copper
assertion (A). makes it a good electric conductor. So, it is used to
make electric wires.
(c) Assertion (A) is true but reason (R) is false. Q. 6. Assertion (A): Silver is not used to make electric
(d) Assertion (A) is false but reason (R) is true. wires.
Q. 1. Assertion (A): Alloys are commonly used in Reason (R): Silver is a bad conductor.
electrical heating devices like electric iron and Ans. Correct option : (c)
heater. Explanation : Silver is a good conductor of electricity
but it is not used to make electric wires because it is
Reason (R): Resistivity of an alloy is generally expensive.
higher than that of its constituent metals but
the alloys have low melting points then their
constituent metals.
C Very Short Answer Type Questions
Ans. Correct option : (c) Q. 1. Two unequal resistances are connected in parallel.
If you are not provided with any other parameters
Explanation : Alloy are hold for electrical heating
(eg. numerical values of I and R), what can be said
devices due to their light restivity and high melting about the voltage drop across the two resistors?
point compared to constituent metals. Ans. Voltage-drop is same across both.
Q. 2. Assertion (A): In a simple battery circuit the point of Q. 2. What is meant by the statement. “The resistance of
lowest potential is positive terminal of the battery. a conductor is one ohm” ?
Reason (R): The current flows towards the point U [CBSE 2020,OD,Set-3]
of the lower potential as it flows in such a circuit Ans. The resistance of a conductor is said to be 1 ohm
from the positive to the negative terminal. if a current of 1 ampere flows through it when the
Ans. Correct option : (d) potential difference across it is 1 volt.
Q. 3. Write the mathematical expression for Joule’s law Ans. Cord is made up of copper wire whereas heating
of heating. R [CBSE 2020] element is made up of alloy.
Ans. Mathematical expression of Joule’s law of heating is [CBSE Marking Scheme, 2019] 1
: H = I2Rt
Where, H = Produced Heat Detailed Answer:
I = Current flowing through the device The cord of an electric oven is usually made of
copper or aluminium whose resistance is very low
t = Time of current flow
so it does not glow. Whereas, its heating element is
R = Resistance of the appliance made up of alloy which has very high resistance.
Q. 4. What does the cord of an electric oven not glow So, when current is passed through the heating
while its heating element does ? element it becomes very hot and glows red.
A [Board Outside Delhi, Set- III, 2019]
Q. 1. Consider the following circuit: What would be the readings of the ammeter and
the voltmeter when key is closed ? Give reason to
justify your answer.
R [CBSE Delhi Comptt. 31/1, 31/2, 31/3, 2018]
Ans. R = R1 + R2 + R3
R = 5 W + 8 W + 12 W = 25 W 1
V = 6 V
V = IR
V 6V
\ I = = = 0.24 A 1
R 25 W
6 (b) Current in 1st bulb,
Hence, Current through 12 W resistance is A P1 100 5
25 I1 = = = A or 0.45 A
V 220 11
= 0.24 A
Current in 2nd bulb,
6A×12W
V = IR = = 2.88 V 1 P2 60 3
25 I2 = = = A or 0.27 A 1+2
V 220 11
Q. 2. Calculate the total cost of running the following
Q. 4. Show how would you join three resistors, each
electrical devices in the month of September, if the
of resistance 9 W so that the equivalent resistance
rate of 1 unit of electricity is ` 6.00.
of the combination is (i) 13.5 W, (ii) 6 W ?
(i) Electric heater of 1000 W for 5 hours daily. U [Delhi /Outside Delhi, 2018]
(ii) Electric refrigerator of 400 W for 10 hours daily. Ans. (i)
R [CBSE Comptt. 31/1, 31/2, 31/3, 2018]
1000
Ans. P1 = 1000 W = kW, t1 = 5h
1000
400
P2 = 400 W = kW, t2 = 10h
1000 Two 9 ohm resistors in parallel connected to one 9
No. of days, n = 30 ohm resistor in series.
E1 =P1 × t1 × n ½ 1 1 1 2
= + =
= 1 kW × 5h × 30 = 150 kWh ½ Rp 9 9 9
E2 = P2 × t2 × n
9
400 \ Rp = W
= kW × 10 h × 30 2
1000
= 120 kWh ½ 9
R = 9 W + W = 13.5 W
\ Total energy = (150 + 120) kWh = 270 kWh ½ 2
\ Total cost = 270 × 6 = ` 1620 1 (ii)
Q. 5. Three resistors of 10 Ω, 15 Ω and 20 Ω are connected = 0.2 A
in series in a circuit. If the potential drop across the ∴ Current in the circuit = 0.2 A
15 Ω resistor is 3 V, find the current in the circuit ∴ Potential drop across 10 Ω resistor is.
and potential drop across the 10 Ω resistor. V = IR
= 0.2 A × 10 Ω
A [Board Term I, 2016]
= 2 V 1½ + 1½
Ans. In series circuit same current flows through all the [CBSE Marking Scheme, 2016]
resistors. Current through 15 Ω resistor, Q. 6. A circuit has a line of 5 A. How many lamps of
V 3V 1 rating 40 W, 200 V can simultaneously run on this
I = = =
R 15 W 5 line safely ? A [Board Term I, 2016]
Ans. Given, V = 200 V, P = 40 W, I = 5A, n = ? Q. 8. Give reason for the following:
(i) Why are copper and aluminium wires used as
nP = VI connecting wires ?
VI 200×5 (ii) Why is tungsten used for filament of electric lamps?
n = =
P 40 (iii) Why is lead-tin alloy used for fuse wires ?
100 AE [Board Term-I- 2015, 2016]
= = 25 lamps
4 Ans. (i) These are good conductors of electricity/low
resistance, low resistivity.
[CBSE Marking Scheme, 2016] 3
(ii) Very high melting point and high resistivity.
Q. 7. A bulb is rated at 200V – 40W. What is its (iii) Low melting point. 1+1+1
resistance ? 5 such bulbs are lighted for 5 hours. [CBSE Marking Scheme, 2016]
Calculate the electrical energy consumed ? Find
Q. 9. Show four different ways in which three resistors
the cost if the rate is 5.10 per KWh.
of ‘r ’ ohm each may be connected in a circuit.
A [Board Term I, 2016]
In which case is the equivalent resistance of the
Ans. V = 200 V, P = 40 W combination: C [DDE-2015]
P = VI (i) Maximum
P 40 1 (ii) Minimum [Board Term I, 2014]
I = = = A
V 200 5 Ans. (a) ½
V 200
R = =
I 1
½
5 (b)
= 200 × 5 = 1000 Ω
Total Power = 40 W × 5 = 200 W
Time = 5 hrs (c) ½
Electrical energy = 200 W × 5 hrs.
= 1000 Wh
= 1 KWh.
Cost of 1 KWh = ` 5.10 3 (d) ½
\ Total cost = ` 1 × 5.10 = ` 5.10
[CBSE Marking Scheme, 2016] (i) Circuit (a) has maximum resistance ½
(ii) Circuit (b) has minimum resistance ½
Q. 1. (a) Define Power and state its S1 unit. (b) Calculate the energy consumed by 3 such bulbs
(b) A torch bulb is rated 5 V and 500 mA. Calculate: if they glow continuously for 10 hours for complete
(i) Power (ii) Resistances (iii) Energy consumed month of November. AE
when it is lighted for 2½ hours.R [CBSE 2020 OD] (c) Calculate the total cost if the rate is Rs 6.50 per
Ans. (a) Power: It is the amount of electric energy unit. U [CBSE 2020 OD]
consumed in a circuit per unit time. Ans. (a) Given, V = 200 volts and P = 100 watt
P = W/t V2 V2 (200)2 40000
As P = or R = = = = 400 W
Its S.I unit is Watt (W). R R 100 W 100Ω
(b) V = 5V 2
I = 500 mA = 0.5 A 2 (b) Electrical energy consumed, E = number of units ×
(i) P = V × I = 5 × 0.5 = 2.5 W Power of each unit × time × total days
V 5 Here, n = 3, P = 100 W, t = 10 hours, Days = 30
(ii) Resistance R = = = 10 ohms
I 0.5 So, E = 3 × 100 W × 10 h × 30 = 90,000 Wh
= 90 kWh 2
(iii) Energy consumed = P × t
(c) Total cost of electricity = Total unit of energy
= 2.5 × 2.5
consumed × Cost per unit
= 6.25 Wh 3
= 90 kWh × 6.50 = ` 585 1
Q. 2. (a) An electric bulb is rated at 200 V-100 W. What is
Q. 3. In the given circuit, A, B, C and D are four lamps
its resistance ? U connected with a battery of 60 V. [SQP 2020]
(iv) Find out the total resistance of the circuit R
Ans. (i) The lamps are in parallel. 1
(ii) Advantages: If one lamp is faulty, it will not affect
the working of the other lamps. They will also be
using the full potential of the battery as they are
connected in parallel. 1
(iii) The lamp with the highest power will glow the
brightest.
Analyse the circuit to answer the following P=VI In this case, all the bulbs have the same
questions. voltage. But lamp C has the highest current. Hence,
(i) What kind of combination are the lamps arranged for Lamp C
in (series or parallel)? U
P = 5 × 60 Watt = 300 W. (the maximum). 1
(ii) Explain with reference to your above answer, what
(iv) The total current in the circuit = 3+4+5+3 A = 15A
are the advantages (any two) of this combination
of lamps? C Voltage = 60V
(iii) Explain with proper calculations which lamp V= IR and hence R = V/I
glows the brightest? A = 60/15 A = 4A 1
Q. 4. A bulb is rated 40W, 220V. Find the current drawn by it, when it is connected to a 220V supply. Also find its
resistance. If the given bulb is replaced by a bulb of rating 25W, 220V,will be there be any change in the value
of current and resistance ? Justify your answer and determine the change. U
Ans. [CBSE Delhi, 2019]
Ans. P = 40 W, V = 220V (b) In an electric circuit two resistors of 12Ω each are
P = VI joined in parallel to a 6 V battery. Find the current
P 40 2 drawn from the battery.
\ I = = = A A [CBSE Board Delhi, Set- I, 2019]
V 220 11
From Ohm’s law,
V = IR Ans.
V 220
R = = = 1210W
I 2
when replaced by 25w, 220 v lamp
P 25 5
I = = = A
V 220 44
V 220
R = = = 1936W
I 5
Low power bulb draw less current since its 1
resistance is higher.
Q. 5. (a) With the help of a suitable circuit diagram prove From figure:
I = I1 + I2 + I3
that the reciprocal of the equivalent resistance of a
group of resistances joined in parallel is equal to V V V
I1 = , I2 = , I3 =
the sum of the reciprocals of the individual resist- R1 R2 R3
ances.
1 1 2
1 1 1 V
= + = = 1 /6 ohm
∴ I = V + + = 1 12 12 12
R1 R 2 R 3 R p
Rtotal = 6 ohm
1 V 6V
1 1 1
Hence, current I = = =1A
∴
= + + 1 RTotal 6W
Rp R1 R 2 R 3
Q. 6. An electric lamp of resistance 20 Ω and a conductor
(b) R1 = R2 = 12 Ω and V = 6 V
of resistance 4 Ω are connected to a 6 V battery as
1 1 1 1 1 shown in the circuit. Calculate:
∴ = + = + ½
Rp R1 R2 12 12 (a) the total resistance of the circuit.
∴ ½ (b) the current through the circuit.
Rp = 6 Ω
(c) the potential difference across the (i) electric lamp
V 6V and (ii) conductor, and
I=
1 = = 1A
Rp 6W (d) power of the lamp.
[CBSE Marking Scheme, 2013]
Detailed Answer:
(a)
A
ns. (a) R = R1 + R2 1
= 20 Ω + 4 Ω = 24 Ω
V
(b) I =
R
6V
= = 0.25 A 1
1+2+2 24 W
Let there be n resistance, each of value (c) (i) For electric lamp:
R1, R2 .... Rn respectively connected in parallel to V = IR
a battery of voltage V. 6
Let current I is sent to the circuit. = × 20 = 5 V 1
If the equivalent resistance is Req, then current 24
V (ii) For Conductor:
drawn I =
Req V = IR
According to the above circuit, 6
= ×4=1V 1
I = I1 + I2 + I3 + .... + In 24
(d) P = VI
V V V V V
= + + + .... + 6
Req R1 R2 R3 Rn = 5V × A = 1.25 W. 1
24
1 1 1 1 1 [CBSE Marking Scheme, 2019]
So, = + + + ...... +
Req R1 R2 R3 Rn Detailed Answer:
Given, Voltage of battery, V = 6 V
Therefore, the reciprocal of the equivalent
Resistance of electric lamp, R1 = 20 Ω
resistance of a group of resistances joined in
Resistance of series conductor, R2 = 4 Ω
parallel is equal to sum of the reciprocals of (a) Total resistance of circuit, RTotal = R1 + R2
individual resistances. = 20 Ω + 4 Ω
(b) In parallel combination, Rtotal is given as = 24 Ω
1 1 1 V
= + (b) Using Ohm’s law I =
RTotal R1 R2 RTotal
6V (b) Total resistance of the circuit =1
= = 0.25 A
24 W R = R1 + R2 + R3 = 5 + 10 + 15 = 30 ohm
(c) Potential difference across Potential difference across the circuit / By
(i) Electric lamp, V1 = IR1 ohm’s law
= 0.25 A × 20 W V 30 V
V = IR or I = 1A 1
= 5V R 30
(ii) Conductor, V2 = IR2 Potential difference across 15 ohm Resistor =
= 0.25 A × 4 W 1A × 15 Ω = 15 V
= 1V [CBSE Marking Scheme, 2019] ½
(d) Power of the lamp = I2R
= (0.25)2 × 20 W Detailed Answer:
= 1.25 W 1+1+2+1 (a) S
uppose the experimental set up comprises of three
resistors R1, R2 and R3 of three different values
COMMONLY MADE ERROR which are connected in series with an ammeter, key
and a battery of known voltage is given as below :
Many students commit errors in
substituting the correct values in formulas.
ANSWERING TIP
30
= = 1A
30
Potential difference across 15W resistor = IR
= 1 × 15 = 15 V 3+2
Q. 8. (a) Three resistors R1, R2 and R3 are connected
in parallel and the combination is connected
A [Board Outside Delhi, Set-I, 2019] to a battery, ammeter, voltmeter and key. Draw
suitable circuit diagram and obtain an expression
Ans. (a) (i) Join the three resistors of different values
for the equivalent resistance of the combination of
in series.
the resistors.
(ii) Connect them with battery, an ammeter
and plug key. (b) Calculate the equivalent resistance of the following
(iii) Plug the key and note the ammeter read- network:
ing.
(iv) Change the position of ammeter to
anywhere in between the resistors and
note the ammeter reading each time.
(v) The ammeter reading will remain same
every time. Therefore when resistors are R [Board Outside Delhi, 2019]
connected in series same current flows
ns. (a)
A Total current I = I1 + I2 + I3
through all resistors, when it is connected
to a battery. Let Rp be the equivalent resistance of R1,
Note: If explained with the help of diagram V
R2, R3. Then the total current I = 1
give full credit. ½×5 Rp
V2
(iii) P = .
R
If Reqv is less, power consumed will be more.
In the given case, Reqv is lesser in the parallel and
thus power consumed will be more.
[CBSE Marking Scheme, 2019]
½ Q. 10. What is meant by electric current ? Name and
define S.I. unit. In a conductor electrons are flowing
from B to A. What is the direction of conventional
current ? Give justification for your answer.
A steady current of 1 Ampere flows through a
V V V
I1 = , I2 = , I3 = ½ conductor. Calculate the number of electrons that
R1 R2 R3 flow through any section of conductor in 1 second.
1 1 1 V (Charge on electron = 1.6 × 10–19 C)
∴ I = V + + = ½
R1 R 2 R 3 R p R [DDE-2014] [Board Term-I, 2015]
Q. 1. Read the passage and answer any four questions (c) Two students perform experiments on two given
from (a) to (e). resistors R1 and R2 and plot the following V-I
Three resistors of 5 Ω, 10 Ω and 15 Ω are connected graphs. If R1> R2, which of the diagrams correctly
in series and the combination is connected to the represent the situation on the plotted curves? U
battery of 30 V. Ammeter and voltmeter are con-
(i)
nected in the circuit.
(a) Which of the following is the correct circuit
diagram to connect all the devices in proper correct
order. A
(i)
(ii)
(ii)
(iii)
(iii) (iv)
(d) The device used to measure the current:
(i) Ammeter (ii) Galvanometer
(iii) Voltmeter (iv) None of these
(iv) (e) Which of the following is connected in series in
circuit:
(i) Ammeter (ii) Voltmeter
(iii) Both of these (iv) None of these
Ans. (a) (iv)
(b) How much is the total resistance? AE
(i) 30Ω (ii) 20Ω
11 30
(iii) Ω (iv) Ω
30 11
(b) (i) 30Ω Tungsten 5.20 × 10–8
(c) (i) Nickel 6.84 × 10–8
Iron 10.0 × 10–8
Chromium 12.9 × 10–8
Mercury 94.0 × 10–8
Manganese 1.84 × 10–6
Alloys Constantan 49 × 10–6
(d) (i) ammeter
(e) (i) ammeter [alloy of Cu and Ni]
Q. 2. Observe the graph and answer any four questions Manganin 44 × 10–6
from (a) to (e). V-I graph for a conductor is as [alloy of Cu, Mn and
shown in figure. Ni]
Nichrome 100 × 10–6
[alloy of Ni, Cr, Mn
and Fe]
(a) Which is a better conductor: U
(i) Chromium (ii) Nickel
(iii) Mercury (iv) Iron
(b) Element used to make heating element of electric
(a) What do you infer from this graph ? AE geyser: AE
1 (i) Iron (ii) Silver
(i) V ∝ (ii) V ∝ I2 (iii) Nichrome (iv) tungsten
I
(c) Element used to make filament of incandescent
1 bulb:
(iii) V ∝ I
(iv) V ∝ AE
I2 (i) Copper (ii) Silver
(b) Name the physical quantity represented by the (iii) Nichrome (iv) tungsten
slope of this graph R (d) What happens to resistance of a conductor when
(i) Current its area of cross section is increased? A
(ii) Resistance (i) Resistance increases
(iii) Potential difference (ii) Resistance decreases
(iv) None of the above (iii) No change
(c) Ohm is the SI unit of: R (iii) Resistance doubles
(i) Potential difference (e) A given length of a wire is doubled and this
(ii) Resistance process is repeated once again. The resistance of
(iii) Current wire becomes: U
(iv) Resistivity (i) 1/4th of original resistance
(d) Which of the following law justify the above (ii) 16 times of original resistance
graph: A (iii) Double the original resistance
(i) Faradays Law (ii) Ohm’s Law (iv) Half of original resistance
(iii) Faradays Law (iv) Joule’s Law Ans. (a) (ii) Nickel
(e) Resistance of a conductor depends on: U (b) (i) Nichrome
(i) length of conductor (c) (iv) tungsten
(ii) area of cross-section (d) (ii) Resistance decreases as R ∝ 1 /A
(iii) temperature (e) (ii) 16 times of original resistance
(iv) all of the above Q. 4. Observe the following table and answer any four
Ans. (a) (iii) V ∝ I questions from (a) to (e). Electrical resistivities of
(b) (ii) Resistance some substances, at 20°C are given as follows:
(c) (ii) Resistance
Silver 1.60 × 10–8 W.m
(d) (ii) Ohm’s Law
(e) (iv) all of the above 1+1+1 Copper 1.62 × 10–8 W.m
Q. 3. Study the given table and answer any four
Tungsten 5.2 × 10–8 W.m
questions from (a) to (e):
Material Resistivity (Ω m) Mercury 94 × 10–8 W.m
Conductors Silver 1.60 × 10–8 Iron 10 × 10–8 W.m
Copper 1.62 × 10–8 Nichrome 10 × 10–6 W.m
Aluminium 2.63 × 10–8
(a) Which is a better conductor of electric current: U (ii) Ammeter reading will decrease
(i) Silver (ii) Copper (iii) Will decrease half
(iii) Tungsten (iv) mercury (iv) No change in ammeter reading
(b) Which element will be used for electrical (c) In a circuit two resistors of 5 Ω and 10 Ω are
U connected in series. Compare the current passing
transmission lines:
through the two resistors. U
(i) Iron (ii) Copper
(i) 1 : 2 (ii) 1: 3
(iii) Tungsten (iv) mercury
(c) Nichrome is used in the heating elements of (iii) 2: 1 (iv) 1:1
(d) The instrument used to measure current is
electric heating device because: AE
____________. R
(i) It has high resistivity
(i) Ammeter (ii) Voltmeter
(ii) It does not oxidise readily at high temperature
(iii) Galvanometer (iv) manometer
(iii) Both of the above
(e) When nichrome and copper wire of same length
(iv) None of the above and same radius are connected in series and
(d) Series arrangement is not used for domestic current I is passed through them. Which wire gets
circuits because: R
heated up more? A
(i) Current drawn is less
(i) Nichrome wire
(ii) Current drawn is more
(ii) Copper wire
(iii) Neither of the above
(iii) Both will heat up at the same temperature
(iv) Both of the above
(iv) None of the wire will get heated up.
(e) If the resistance is to be increased, then the
Ans. (a) (i) ammeter reading will decrease (becomes half)
resistors are to be increased in: A
Explanation: The ammeter reading will decrease
(i) Series (becomes half). This is because with the increase
(ii) Parallel in length, resistance of the circuit increases, hence
(iii) Mixed arrangement current decreases.
(iv) None of the above (b) (ii) The ammeter reading will increase (becomes
Ans. (a) (i) silver two times).
(b) (ii) copper Explanation: The ammeter reading will increase
(c) (iii) Both of the above (becomes two times). This is because as area
(d) (i) Current drawn is less increases, resistance decreases and hence current
(e) (i) Series increases.
Q. 5. In the given circuit, connect a nichrome wire of (c) (iv) 1:1
length ‘L’ between points X and Y and note the Explanation: In a series connection of resistors,
ammeter reading Answer any four questions from same current passes through all the resistors.
(a) to (e). Hence, current will be same. Ratio of the currents
will be 1 : 1.
(d) (i) Ammeter.
(e) (i) nichrome wire
Q. 6. Study the circuit shown in which three identical
bulbs B1, B2 and B3 are connected in parallel with
a battery of 4.5 V. Answer any four questions (a) to
(e).
Frictional Electricity: It is the electricity produced by rubbing two suitable bodies and flow of electrons from one
body to other.
Electricity: A fundamental form of energy observable in positive and negative forms that occurs naturally (as in
lightning) or is produced (as in a generator) and that is expressed in terms of the movement and interaction of
electrons.
Ohm: It is the SI unit of resistance. A conductor has a resistance of one ohm if a current of one ampere flows
through it on applying a potential difference of 1 volt across its ends.
1volt 1V
1 ohm = or 1 W =
1 ampere 1 A
Resistor: A conductor which has some appreciable resistance is called a ‘resistor’.
Semiconductors: Materials having resistivity between that of an insulator and a conductor are called semicon-
ductors. They are used in making integrated circuits.
Superconductors: These are certain materials that lose their resistivity at low temperature. Such materials
are called as superconductors. The phenomenon of complete loss of resistivity by substances below a certain
temperature is called superconductivity.
Fuse Wire: The wire which melts, breaks the circuit and prevents the damage of various appliances in the
household connections. It is connected in series with live line and its thickness determines the maximum current
that can be drawn. It is made of an alloy of aluminium, copper, iron and lead.
qqq