CHAPTER 2 AN INTRODUCTION TO ELECTRIC CIRCUITS
EXERCISE 5, Page 12
1. In what time would a current of 10 A transfer a charge of 50 C?
Charge, Q = I  t from which, 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

= 800 s =
min = 13.33 min = 13 min 20 s
3
I 100 10
60
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EXERCISE 6, Page 15
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 of element, R =
V 35

=7
I
5
2. A 60 W electric light bulb is connected to a 240 V supply. Determine (a) the current flowing in
the bulb and (b) the resistance of the bulb.
(a) Power, P = V  I, hence, current, I =
(b) Resistance, R =
P 60

= 0.25 A
V 240
V 240

= 960 
I 0.25
3. Graphs of current against voltage for two resistors P and Q are shown in Figure 2.9. Determine the
value of each resistor.
For resistor P, resistance, R =
V 16 10 6

 2 103  = 2 m
3
I
8 10
V 20 10 6

 5 103  = 5 m
For resistor Q, resistance, R =
3
I
4 10
4. Determine the p.d. which must be applied to a 5 k resistor such that a current of 6 mA may
flow.
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P.d., V = I  R = 6 103  5 103 = 30 V
5. A 20 V source of e.m.f. is connected across a circuit having a resistance of 400 . Calculate the
current flowing.
Current, I =
V 20

= 0.05 A or 50 mA
R 400
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EXERCISE 7, Page 17
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
R 625
Power rating, P = V  I = 250  0.4 = 100 W
V 2 2502

(or P =
= 100 W
R
625
or
P = I 2 R   0.4   625 = 100 W)
2
2. Determine the resistance of a coil connected to a 150 V supply when acurrent of (a) 75 mA
(b) 300 μA flows through it.
(a) Resistance, R =
V
150

 2000  = 2 k
I 75  103
(b) Resistance, R =
V
150

 500000   500 k or 0.5 MΩ
I 300 10  6
3. 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
4. Determine the power dissipated when a current of 10 mA flows through an appliance having a
resistance of 8 k.
Power, P = I2 R  10 103  8 103  = 0.8 W
2
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5. 85.5 J of energy are converted into heat in 9 s. What power is dissipated?
Power, P =
energy 85.5 J

= 9.5 W
time
9s
6. 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 hence p.d., V =
P 10

= 2.5 V
I 4
7. 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 = I2 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
8. A battery of e.m.f. 15 V supplies a current of 2 A for 5 min. How much energy is supplied in this
time?
Energy = power  time = (V  I)  t = (15  2)  (5  60) = 9000 J or 9 kJ
9. 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 = power  time from which, power =
enery 72 10 6 J

 8000W = 8 kW
time 150  60s
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Power = V  I from which, current =
P 8000

= 20 A
V 400
10. 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 dissipated by coil, P =
V 2 5002

= 5000 W = 5 kW
R
50
11. 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
12. 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 8, Page 19
1. A television set having a power rating of 120 W and electric lawnmower of power rating 1 kW
are both connected to a 250 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

= 0.48 A, hence the 3 A fuse is the most appropriate
V 250
For the lawnmower, I =
P 1000

= 4 A, hence the 5 A fuse is the most appropriate
V 250
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