Nelson Physics 11 Solutions
Nelson Physics 11 Solutions
Nelson Physics 11 Solutions
Knowledge
1. (b)
2. (a)
3. (d)
4. (c)
5. (b)
6. (c)
7. (b)
8. (c)
9. (b)
10. (a)
11. (c)
12. (b)
13. (d)
14. (b)
15. (c)
16. False. Voltage is a measure of the amount of
electrical potential energy associated with each
charge.
17. False. In a circuit, electrons flow from negative
to positive.
18. True
19. True
20. True
21. False. Then two like magnetic poles are
brought close one another, they repel.
22. True
23. False. Adding a soft-iron core will increase the
strength of a DC motor.
24. True
25. True
26. False. A step-up transformer increases the
voltage in the secondary circuit.
OR
A step-up transformer decreases the current in the
secondary circuit.
27. True
28. (a) (v)
(b) (viii)
(c) (i)
(d) (ii)
(e) (iii)
(f) (ix)
(g) (vii)
(h) (iv)
(i) (vi)
Understanding
29. The nuclear power plant has an efficiency of
32 %, so 32 % of the total power is transformed
into electrical energy. 32 % of 14 000 MW is:
0.32 14 000 MW = 4500 MW
So, the power plant produces 4500 MW of
electrical power.
30. Given: P = 35 W; t = 220 h
Required: E
Analysis: P =
!E
!t
E = Pt
Solution: Convert time to seconds to get the
answer in joules:
3600 s
1h
!t = 792 000 s
!t = 220 h "
1 kWh
= 7.7 kWh
3.6 ! 106 J
!E
!t
U5-3
!t =
!E
Q
!E
Q=
V
!E
Solution: Q =
V
540 J
=
120 V
Q = 4.5 C
Analysis: V =
Q
!t
60 s
1 min
!t = 90 s
I parallel = I 2 + I 3
Q
I=
!t
0.65 C
=
90 s
I = 7.2 " 10#3 A
8.5 mA = 2.1 mA + I 3
I 3 = 6.4 mA
Q
I
4.0 C
=
0.50 A
!t = 8.0 s
1000 mA
1A
I = 7.2 mA
I parallel = I 2 + I 3
I parallel = I 3 + I 3
I parallel = 2I 3
I3 =
Q
!t
Q
!t =
I
Analysis: I =
So I3 is 6.4 mA.
(b) The current in a series circuit is constant and
the same as the source current. From part (a),
Isource = 8.5 mA. The amount of current entering a
junction is equal to the amount of current exiting
the junction. Letting I2 = I3, this can be used to find
I3:
I parallel
2
8.5 mA
=
2
I 3 = 4.2 mA
So I3 is 4.2 mA.
1A
1000 mA
I = 0.50 A
U5-4
V
I
1A
1000 mA
I = 7.2 ! 10"4 A
V
I
V = IR
R=
1000 mV
1 V
V = 650 mV
1
1
+
11.4 ! 32.2 !
Rparallel = 8.42 !
U5-5
I=
1.8 ! 109 W
5.0 ! 104 V
= 3.6 ! 104 A
I = 36 kA
U5-6
V = 180 kV; I = 35 A
P = VI
= (180 kV)(35 A)
P = I2R
= 6.3 ! 106 W
P = 6.3 MW
= (3.0 ! 10 A) (0.40 !)
3
= 3.6 ! 106 W
P = 3.6 MW
P
R= 2
I
12.6 MW
=
(5.0 kA)2
=
1.26 ! 10 W
(5.0! 103 A)2
7
R = 0.50 !
1500 MW
0.30
Pin = 5000 MW
Pin =
U5-7
Rparallel
1
1
+
R2 R3
Rparallel = 12 !
!E
!t
!E
!t =
P
3.0 " 103 J
=
J
54.0
s
= 55.56 s
!t = 56 s
Rtotal
I source =
Vsource
Rsource
15 V
17 !
= 0.8824 A (two extra digits carried)
=
I source
Rtotal = R1 + R2
= 40 ! + 15 !
= 55 !
V
.
R
Vsource
Rsource
30.0 V
55 !
= 0.55 A
=
Q
.
!t
Q
I=
!t
Q = I !t
= (0.55 A)(10.0 s)
Q = 5.5 C
V
.
R
I source
1
1
1
=
+
Rparallel 20.0 ! 30.0 !
P=
Rtotal
Q
.
!t
Q
!t
Q = I !t
= (0.8824 A)(8.0 s)
I=
Q = 7.1 C
U5-8
8.0 V
0.50 A
R3 = 16 !
= 15 V ! 4.412 V
V2 = 10.59 V (two extra digits carried)
V
.
R
V
I2 = 2
R2
I=
8.0 V
40.0 !
I 2 = 0.20 A
=
10.59 V
=
20.0 !
I 2 = 0.5295 A
Q
!t
Q = I !t
= (0.5295 A)(10.0 s)
I=
Q = 5.3 C
V3
I3
V2 = Vsource ! V1
R3 =
V
.
I
Q
.
!t
Q
!t
Q
!t =
I
15 C
=
0.70 A
!t = 21 s
I=
U5-9
rise
run
!V
m=
!I
slope =
21 V " 10 V
8.4 # 10"4 A " 4.1 # 10"4 A
m = 2.6 # 104 !
So the resistance of the circuit is 2.6 104 .
72. (a) First apply KVL to the determine V2, V3,
and V4. Since the first two resistors are in parallel,
V1 = V2, so V2 = 3.0 V.
=
Vsource = V1 + V3
V3 = Vsource ! V1
= 12 V ! 2.0 V
V3 = 9.0 V
V
.
R
V3
R3
9.0 V
250 !
I 3 = 0.036 A
=
1A
1000 mA
I1 = 8.4 ! 10"4 A
I5 = 0.41 mA !
1A
1000 mA
I5 = 4.1 ! 10"4 A
V
.
R
V4
R4
9.0 V
300 !
I 4 = 0.03 A
=
U5-10
V
.
I
V1
I1
3.0 V
0.033 A
R1 = 91 !
=
! 12 cm = 240 windings
1 cm
15 windings
L = 16 cm
U5-11
I s Vp
=
I p Vs
Vp I p
Is
Solution:
Vs =
Vp I p
Is
(250 V)(5.0 A)
10.0 A
Vs = 125V (one extra digit carried)
=
=
=
Np
Ns
Vp
Ns
=
=
Np
Ns
Analysis:
Vp
Vs
250 V
Vp
Vs
Vs =
Np
Ns
Vp N s
Np
Solution:
Vp N s
Vs =
Np
(80.0 V)(160)
100
Vs = 128V
=
Is N s
Np
(10.0A)(160)
100
I p = 16.0A
=
Vp
Vs
Ns =
Np
Ns
Vs N p
Vp
Solution:
Ns =
Vs
Vp = 250 V; Vs = 125 V
Np
Vs N p
Vp
(6.0 ! 101 V )(150)
3.0 ! 102 V
N s = 30
125 V
=2
U5-12
V
and the
R
Vp
Ip =
Rp
Rp
3.0 ! 10 V
10.0!
3.0kV
750!
I p = 3.0 ! 101 A
3.0 ! 103 V
750!
I p = 4.0A
Ip N p
Ns
Is =
Vp
Vs
Np
Ns
Vp N s
Vs
Solution:
Np =
=
V
and the
R
Vp N s
Vs
(3.0kV)(60)
1.0 ! 102 V
=
=
I pVp
Vs
(4.0 A)(3.0kV)
1.0 ! 102 V
(4.0A)(3.0! 103 V )
1.0 ! 102 V
I s = 120 A
V
and the
I
Vs
Is
1.0 ! 102 V
120 A
Rs = 0.83 !
=
U5-13
" P%
0.0060 ! P = $ ' R
#V &
P2
R = 0.0060 ! P
V2
P2
P
R = 0.0060 !
2
P
V P
P
R = 0.0060
V2
0.0060V 2
P=
R
P = VI
P
I=
V
14 MW
=
160 kV
14! 106 W
=
1.6 ! 105 V
I = 87.5 A (one extra digit carried)
0.0060V 2
R
0.0060(240 kV)2
=
0.50 !
0.0060(2.4 ! 105 V)2
=
0.50 !
P = 690 MW
P=
R = 15 !
Vp
Np
Vp =
Vs
Ns
Vs N p
Ns
(250 kV)(1000)
6000
Vp = 4.2 ! 104 V
=
" P%
0.0060 ! P = $ ' R
#V &
U5-14
Evaluation
91. (a) To make an ammeter, place a galvanometer
in parallel with a resistor with lower resistance
than the galvanometer (to keep the current away
from the galvanometer). The galvanometer
measures the current passing through the resistor.
(b) To make a voltmeter, place a galvanometer in
series with a resistor with a very high resistance (to
keep the current away from the galvanometer).
Then, knowing the resistance of the path the
galvanometer is on, calculate the voltage by
multiply the current by the resistance.
92. (a) Household circuits are set up with an AC
power source and the devices plugged in are in
parallel.
(b) Since the circuit is in parallel and all power
strips and additional plugs create more parallel
circuits, as more devices are plugged in the overall
resistance decreases.
(c) The voltage of an outlet remains constant, so
using Ohms Law, as the resistance decreases there
is a corresponding increase in current, which could
melt or fray the wires. To prevent this, houses have
safety devices such as circuit breakers, fuses,
GFCIs, and AFCIs.
93. Yes, it is possible to create a DC motor without
permanent magnets. Instead of magnets, add two
solenoids to the circuit. Orient the solenoids so
they have north and south magnetic poles in the
former position of the magnets. Resistors are
needed to control the amount of current in the
circuit. If all three components are in parallel, then
they will have the same voltage. By controlling the
amount of current with the resistors you can
control the strength of the magnetic fields and
power of the motor.
94. (a) The current induced from a magnet in a coil
or loop cannot result in a magnetic field that
attracts the magnet because this violates the laws
of conservation of energy. If the current created an
attraction to the magnet then work could be done
without any work put in. For example, we could
lift a bar magnet through a horizontal loop and
then the induced current would pull the weight of
the magnet upward without any additional work
being put in.
(b) We can control the current in a coil to attract or
repel a magnet because in both cases the energy
required to do so and thus the work done is
supplied by an external power source. The power
source must do work to create the potential
difference necessary to maintain the current in the
coil.
Research
98. Answers may vary. Students may choose to
write about James Watt, James Prescott Joule,
Charles-Augustin de Coulomb, Alessandro Volta,
Andr-Marie Ampre, or Georg Simon Ohm.
Biographies should include details about when and
where the scientist lived and what contributions he
made to science.
99. (a) Answers may vary. The United States,
China, and Japan rank among the worlds top
energy consumers with most of the energy being
derived from fossil fuels. The United States uses
22 % of the total global energy, China uses 20 %,
while Japan uses 7 %. Students may mention
France, which gets 80 % of its power from nuclear
energy but only uses 3 % of the total global
energy.
(b) Answers may vary. Switzerland and Denmark
rank among the greenest industrialized nations by
experts at Yale and Columbia universities. These
rankings are based on carbon emission reductions,
reforestation efforts, and use of hydropower and
geothermal energy. Japan is ranked as one of the
most energy efficient. Other countries students
may mention include Hong Kong, Ireland, and the
U.K.
U5-15
U5-16