CHAPTER 1 UNITS ASSOCIATED WITH BASIC ELECTRICAL
QUANTITIES
EXERCISE 1, Page 5
1. What quantity of electricity is carried by 6.24 1021 electrons?
1 coulomb = 6.24 1018 electrons
Hence, 6.24 1021 electrons =
6.24 1021
 1103 C = 1000 C
6.24 1018
2. In what time would a current of 1 A transfer a charge of 30 C?
Charge, Q = I  t from which, time, t =
Q 30 C

= 30 s
I
1A
3. A current of 3 A flows for 5 minutes. What charge is transferred?
Charge, Q = I  t = 3  (5  60) = 900 C
4. How long must a current of 0.1 A flow so as to transfer a charge of 30 C?
Q = I  t, hence, time, t =
Q 30 C

= 300 s or 5 minutes
I 0.1A
5. What force is required to give a mass of 20 kg an acceleration of 30 m/s2 ?
Force = mass  acceleration = 20 kg  30 m / s 2 = 600 N
6. Find the accelerating force when a car having a mass of 1.7 Mg increases its speed with a
constant acceleration of 3 m / s 2 .
Force = mass  acceleration = 1.7 Mg  3 m / s 2 = 1700 kg  3 m / s 2 = 5100 N or 5.1 kN
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7. A force of 40 N accelerates a mass at 5 m/s2. Determine the mass.
Force = mass  acceleration, hence mass =
force
40 N

= 8 kg
acceleration 5 m / s 2
8. Determine the force acting downwards on a mass of 1500 g suspended on a string.
Force = mass  acceleration = 1500 g  9.81 m / s 2 = 1.5 kg  9.81 m / s 2 = 14.72 N
9. A force of 4 N moves an object 200 cm in the direction of the force. What amount of work is done ?
Work done, W = force  distance = 4 N 
200
m=8J
100
10. A force of 2.5 kN is required to lift a load. How much work is done if the load is lifted through
500 cm?
Work done, W = force  distance = 2.5 kN  5 m = 12.5 kJ
11. An electromagnet exerts a force of 12 N and moves a soft iron armature through a distance of
1.5 cm in 40 ms. Find the power consumed.
Power consumed =
work force  dis tan ce 12 1.5 10 2


= 4.5 W
time
timne
40 10 3
12. A mass of 500 kg is raised to a height of 6 m in 30 s. Find (a) the work done, and (b) the power
developed.
(a) Work done, W = force  distance = (mass  acceleration)  distance
= (500 kg  9.81 m / s 2 )  6 m = 29.43 kN m
work done 29.43 10 3

(b) Power =
= 981 W
time taken
30
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13. Rewrite the following as indicated:
(a) 1 000 pF = ……… nF
(b) 0.02 F = ………. pF
(c) 5 000 kHz = ……… MHz
(d) 47 k = …….. M
(e) 0.32 mA = ……. A
(a) 1 000 pF = 1000 1012 F  1109 F = 1 nF
(b) 0.02 F = 0.02 106 F  20 109 F  20,000 1012 F = 20,000 pF
(c) 5 000 kHz = 5000 103 Hz  5 106 Hz = 5 MHz
(d) 47 k = 47 103   0.047 106  = 0.047 M
(e) 0.32 mA = 0.32 103 A  320 106 A = 320 A
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EXERCISE 2, Page 6
1. Find the conductance of a resistor of resistance (a) 10  (b) 2 k (c) 2 m
(a) If resistance, R =
1
1 1

then conductance, G =
= 0.1 S
G
R 10
(b) Conductance, G =
1
1

 0.5 102 S = 0.5 mS
3
R 2 10
(c) Conductance, G =
1
1

= 500 S
R 2  103
2. A conductor has a conductance of 50 S. What is its resistance?
1
1
1

then resistance, R =
= 20000  = 20 k
G 50 106
R
If conductance, G =
3. An e.m.f. of 250 V is connected across a resistance and the current flowing through the
resistance is 4 A. What is the power developed?
Power, P = V  I = 250  4 = 1000 W or 1 kW
4. 450 J of energy are converted into heat in 1 minute. What power is dissipated?
Energy = power  time, hence, power =
energy 450 J

= 7.5 W
time
60s
5. A current of 10 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
 =1V
I 10
6. A battery of e.m.f. 12 V supplies a current of 5 A for 2 minutes. How much energy is supplied
in this time?
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Energy = power  time = (V  I)  t = (12  5)  (2  60) = 7200 J or 7.2 kJ
7. A d.c. electric motor consumes 36 MJ when connected to a 250 V supply for 1 hour. Find the
power rating of the motor and the current taken from the supply.
Power =
energy 36 106 J

= 10000 W = 10 kW = power rating of motor
time
60  60
P 10 103

Power, P = V  I, hence current, I =
= 40 A
V
250
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