CHAPTER 33 AN INTRODUCTION TO ELECTRIC CIRCUITS
EXERCISE 156, Page 344
1. In what time would a current of 10 A transfer a charge of 50 C?
Q = I  t hence, time, t =
Q 50
=5s

I 10
2. A current of 6 A flows for 10 minutes. What charge is transferred?
Charge, Q = I  t = 6  (10  60) = 3600 C
3. How long must a current of 100 mA flow so as to transfer a charge of 80 C?
Q = I  t hence, time, t =
Q
80
800
min = 13 min 20 s

= 800 s =
3
I 100  10
60
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EXERCISE 157, Page 346
1. The current flowing through a heating element is 5 A when a p.d. of 35 V is applied across it. Find
the resistance of the element.
Resistance, R =
V 35

=7Ω
I
5
2. An electric light bulb of resistance 960  is connected to a 240 V supply. Determine the current
flowing in the bulb.
Current, I =
V 240 1

 = 0.25 A
R 960 4
3. Graphs of current against voltage for two resistors P and Q are shown below. Determine the value
of each resistor.
For resistor P, R =
V 16 V 16  106 V


= 2  103  = 2 m
3
I 8 mA
8  10 A
For resistor Q, R =
V 20 V 20 106 V


= 5  103  = 5 m
3
I
4 mA
4 10 A
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4. Determine the p.d. which must be applied to a 5 k resistor such that a current of 6 mA may
flow.
P.d., V = I  R = 6  103  5 103 = 30 V
375
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EXERCISE 158, Page 349
1. The hot resistance of a 250 V filament lamp is 625 . Determine the current taken by the lamp
and its power rating.
Current, I =
V 250

= 0.4 A
I 625
Power rating, P = V  I = 250  0.4 = 100 W
(or P =
V 2 2502

= 100 W
R
625
or
P = I 2 R   0.4   625 = 100 W)
2
2. Determine the resistance of an electric fire which takes a current of 12 A from a 240 V supply.
Find also the power rating of the fire and the energy used in 20 h.
Resistance, R =
V 240

= 20 
I
12
Power rating, P = V  I = 240  12 = 2880 W or 2.88 kW
Energy = power  time = 2.88 kW  20 h = 57.6 kWh
3. Determine the power dissipated when a current of 10 mA flows through an appliance having a
resistance of 8 k.
Power, P = I 2 R  10 103   8 103  = 0.8 W
2
4. 85.5 J of energy are converted into heat in nine seconds. What power is dissipated?
Power, P =
energy 85.5 J

= 9.5 W
time
9s
376
© John Bird Published by Taylor and Francis
5. A current of 4 A flows through a conductor and 10 W is dissipated. What p.d. exists across the
ends of the conductor?
Power, P = V  I from which, p.d., V =
P 10

= 2.5 V
I 4
6. Find the power dissipated when: (a) a current of 5 mA flows through a resistance of 20 k
(b) a voltage of 400 V is applied across a 120 k resistor
(c) a voltage applied to a resistor is 10 kV and the current flow is 4 mA
(a) Power, P = I 2 R   5 103   20 103  = 0.5 W
2
(b) Power, P =
V2
4002

= 1.33 W
R 120 103
(c) Power, P = V  I = 10  103  4 103  = 40 W
7. A d.c. electric motor consumes 72 MJ when connected to 400 V supply for 2 h 30 min. Find the
power rating of the motor and the current taken from the supply.
energy
72  106 J
Power =

= 8000 W = 8 kW = power rating of motor
time
2.5  60  60
P 8  103
Power, P = V  I, hence, current, I =

= 20 A
V
400
8. A p.d. of 500 V is applied across the winding of an electric motor and the resistance of the winding
is 50 . Determine the power dissipated by the coil.
Power, P =
V 2 5002

= 5000 W = 5 kW
R
50
377
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9. In a household during a particular week three 2 kW fires are used on average 25 h each and
eight 100 W light bulbs are used on average 35 h each. Determine the cost of electricity for the
week if 1 unit of electricity costs 15p
Energy in week = 3(2 kW  25 h) + 8( 100  103 kW 35 h) = 150 + 28 = 178 kWh
Cost = 178  15 = 2670p = £26.70
10. Calculate the power dissipated by the element of an electric fire of resistance 30  when a
current of 10 A flows in it. If the fire is on for 30 hours in a week determine the energy used.
Determine also the weekly cost of energy if electricity costs 13.50p per unit.
Power, P = I 2 R  10    30  = 3000 W or 3 kW
2
Energy = power  time = 3 kW  30 h = 90 kWh
Cost = 90  13.50p = 1215p = £12.15
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EXERCISE 159, Page 350
1. A television set having a power rating of 120 W and electric lawnmower of power rating 1 kW
are both connected to a 240 V supply. If 3 A, 5 A and 10 A fuses are available state which is the
most appropriate for each appliance.
Power, P = V  I hence, current, I =
For the television, I =
P
V
P 120
= 0.5 A, hence the 3 A fuse is the most appropriate

V 240
For the lawnmower, I =
P 1000

= 4.17 A, hence the 5 A fuse is the most appropriate
V 240
EXERCISE 160, Page 350
Answers found from within the text of the chapter, pages 341 to 350.
EXERCISE 161, Page 351
1. (d)
13. (b)
2. (a) 3. (c) 4. (b)
14. (a)
15. (c)
5. (d) 6. (d)
7. (b)
16. (b) 17. (d)
18. (d)
8. (c) 9. (a)
10. (a)
11. (c) 12. (c)
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