CHAPTER 41 CAPACITORS AND INDUCTORS
EXERCISE 191, Page 432
1. Find the charge on a 10 F capacitor when the applied voltage is 250V.
Charge, Q = C × V = 10 106  250  2.5 103 C = 2.5 mC
2. Determine the voltage across a 1000 pF capacitor to charge it with 2 C.
Q = CV hence, voltage, V =
Q
2 106
= 2000 V or 2 kV

C 1000 1012
3. The charge on the plates of a capacitor is 6 mC when the potential between them is 2.4 kV.
Determine the capacitance of the capacitor.
Q = CV hence, capacitance, C =
Q 6 103

 2.5 106 = 2.5 μF
V 2.4 103
4. For how long must a charging current of 2 A be fed to a 5 F capacitor to raise the p.d. between
its plates by 500 V.
Charge Q = I  t and
from which,
Q=CV
time, t =
hence I  t = C  V
C  V 5  106  500

= 1.25 ms
I
2
5. A direct current of 10 A flows into a previously uncharged 5 F capacitor for 1 ms. Determine the
p.d. between the plates.
P.d. between plates, V =
Q I  t 10 1103
= 2000 V or 2 kV


C
C
5  106
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6. A capacitor uses a dielectric 0.04 mm thick and operates at 30 V. What is the electric field
strength across the dielectric at this voltage?
Electric field strength, E =
V
30

 750 kV/m
d 0.04  103
7. A charge of 1.5 C is carried on two parallel rectangular plates each measuring 60 mm by
80 mm. Calculate the electric flux density. If the plates are spaced 10 mm apart and the voltage
between them is 0.5 kV determine the electric field strength.
Electric flux density, D =
Q
1.5  106
= 312.5 C / m 2

6
A 60  80  10
Electric field strength, E =
V 0.5  103

= 50 kV/m
d 10  103
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EXERCISE 192, Page 434
1. A capacitor consists of two parallel plates each of area 0.01 m2, spaced 0.1 mm in air. Calculate the
capacitance in picofarads.
Capacitance, C 
0  r A 8.85  1012  1 0.01

since, for air, ε r = 1
d
0.1 103
= 885 1012 = 885 pF
2. A waxed paper capacitor has two parallel plates, each of effective area 0.2 m 2 . If the capaci-
-tance is 4000 pF determine the effective thickness of the paper if its relative permittivity is 2.
C
0 r A
  A 8.85 1012  2  0.2
 885 106 m
hence, thickness of the paper, d = 0 r 
C
4000  1012
d
= 0.885 mm
3. How many plates has a parallel plate capacitor having a capacitance of 5 nF, if each plate is
40 mm by 40 mm and each dielectric is 0.102 mm thick with a relative permittivity of 6.
C
0r A
Cd
5 109  0.102 103
=6

(n  1) from which, n – 1 =
0  r A 8.85 1012  6  40  40 106
d
Hence, the number of plates, n = 6 + 1 = 7
4. A parallel plate capacitor is made from 25 plates, each 70 mm by 120 mm, interleaved with mica
of relative permittivity 5. If the capacitance of the capacitor is 3000 pF determine the thickness of
the mica.
C
0r A
(n  1) from which, dielectric thickness,
d
d=
0  r A
8.85 1012  5  70 120 106
(n  1) 
(25  1) = 0.00297 m = 2.97 mm
C
3000 1012
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5. A capacitor is constructed with parallel plates and has a value of 50 pF. What would be the
capacitance of the capacitor if the plate area is doubled and the plate spacing is halved?
If the plate area is doubled, so is the capacitance (i.e. direct proportion).
If the plate spacing is halved, then the capacitance is doubled (i.e. inverse proportion).
Hence, capacitance of capacitor = 4 × 50 = 200 pF
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EXERCISE 193, Page 435
1. Capacitors of 2 F and 6 F are connected (a) in parallel and (b) in series. Determine the
equivalent capacitance in each case.
(a) In parallel, equivalent capacitance, CT = 2 + 6 = 8 μF
(b) In series, equivalent capacitance, CT =
2  6 12
= 1.5 μF

26 8
2. Find the capacitance to be connected in series with a 10 F capacitor for the equivalent
capacitance to be 6 F
For series connection,
1
1
1


C1 C2 CT
i.e.
1
1 1


10 C2 6
and
C2 =
from which,
1 1 1
   0.06666666
C2 6 10
1
= 15 F
0.06666666
3. What value of capacitance would be obtained if capacitors of 0.15 F and 0.10 F are connected
(a) in series and (b) in parallel.
(a) In series, equivalent capacitance, CT =
0.15  0.10 0.015
= 0.06 μF

0.15  0.10 0.25
(b) In parallel, equivalent capacitance, CT = 0.15 + 0.10 = 0.25 μF
4. Two 6 F capacitors are connected in series with one having a capacitance of 12 F. Find the
total equivalent circuit capacitance. What capacitance must be added in series to obtain a
capacitance of 1.2 F?
430
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Two 6 F capacitors in series has a total capacitance of
6 6
= 3 F. (Two equal value capacitors
66
in series will have a total capacitance of half the value of one of the capacitors).
3 F in series with 12 F has a total capacitance of
3 12
= 2.4 F = total circuit capacitance.
3  12
Let new capacitance be CX then if new total capacitance is to be 1.2 F then
1
1
1


1.2 2.4 CX
from which
1
1
1


 0.41666
CX 1.2 2.4
Hence, capacitance to be added, CX =
1
= 2.4 F
0.41666
5. For the arrangement shown below find (a) the equivalent circuit capacitance and (b) the voltage
across a 4.5 F capacitor.
(a) Three 4.5 F capacitors in series gives 1.5 F and two 1 F capacitors in series gives 0.5 F
1.5 F and 0.5 F capacitors in parallel gives 1.5 + 0.5 = 2 F
2 F in series with 3 F gives:
23 6
 = 1.2 F = equivalent circuit capacitance
23 5
 3 
(b) The equivalent circuit is shown below where V1  
  500  = 300 V = voltage across three
 23
4.5 F capacitors in series. Hence, voltage across each 4.5 F capacitor = 300/3 = 100 V.
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(Alternatively, to find V1 :
Since CT = 1.2 F then QT  CT  V  1.2 106  500  600 C . This is the charge on each
capacitor of the circuit shown below. Hence, V1 
QT 600 106

= 300 V)
C1
2 106
6. In the circuit below, capacitors P, Q and R are identical and the total equivalent capacitance of
the circuit is 3 F. Determine the values of P, Q and R.
3.5 F and 4.5 F in parallel gives an equivalent capacitance of 3.5 + 4.5 = 8 F
2 F in series with 8 F gives
2  8 16

= 1.6 F
2  8 10
Let the equivalent capacitance of P, Q and R in series be CX
Then 1.6 + CX = 3 from which, CX = 3 – 1.6 = 1.4 F
Thus,
1
1
1
1
3
(since CP  CQ  CR )




1.4 CP CQ CR CP
i.e.
CP  3  1.4 = 4.2 F = CQ  CR
432
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EXERCISE 194, Page 436
1. When a capacitor is connected across a 200 V supply the charge is 4 C. Find (a) the
capacitance and (b) the energy stored.
(a) Q = CV from which, capacitance, C =
(b) Energy stored, W =
Q 4 106

= 20 nF or 0.02 F
V
200
1
1
CV 2   0.02 106  2002 = 400 J or 0.4 mJ
2
2
2. Find the energy stored in a 10 F capacitor when charged to 2 kV
Energy stored, W =
1
1
CV 2  10 106  20002 = 20 J
2
2
3. A 3300 pF capacitor is required to store 0.5 mJ of energy. Find the p.d. to which the capacitor
must be charged.
Energy, W =
1
CV 2 from which, p.d., V =
2
 2  0.5  103 
2W
 
= 550 V
12 
C
 3300 10 
4. A bakelite capacitor is to be constructed to have a capacitance of 0.04 F and to have a steady
working potential of 1 kV maximum. Allowing a safe value of field stress of 25 MV/m find (a)
the thickness of bakelite required, (b) the area of plate required if the relative permittivity of
bakelite is 5, (c) the maximum energy stored by the capacitor and (d) the average power
developed if this energy is dissipated in a time of 20 s.
V
from which,
d
V
1000
thickness of dielectric, d =

 40 106 m  40 103 mm = 0.04 mm
E 25  106
(a) Field stress, E =
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© John Bird Published by Taylor and Francis
0 r A
from which,
d
C d 0.04 106  0.04 103

 0.03616 m 2 = 361.6 cm 2
cross-sectional area, A =
0  r
8.85 1012  5
(b) Capacitance, C =
(c) Maximum energy, Wmax =
1
1
CV 2   0.04 106 10002 = 0.02 J
2
2
(d) Energy = power  time, hence, power, P =
energy
0.02 J

= 1000 W or 1 kW
time
20 106 s
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EXERCISE 195, Page 40
1. Find the e.m.f. induced in a coil of 200 turns when there is a change of flux of 30 mWb linking
with it in 40 ms.
Induced e.m.f., E =  N
 30  103 
d
= - 150 V
 200 
3 
dt
 40  10 
2. An e.m.f. of 25 V is induced in a coil of 300 turns when the flux linking with it changes by
12 mWb. Find the time, in milliseconds, in which the flux makes the change.
E N
d
N d 300  12 103

= 0.144 s or 144 ms
from which, time for change, dt =
dt
E
25
3. An ignition coil having 10000 turns has an e.m.f. of 8 kV induced in it. What rate of change of
flux is required for this to happen?
E N
E 8  103
d
d
= 0.8 Wb/s
from which, rate of change of flux,
=

dt
N 10 000
dt
4. A flux of 35 mWb passing through a 125-turn coil is reversed in 25 ms. Find the magnitude of
the average e.m.f. induced.
Magnitude of induced e.m.f., E = N
 0.35  103   0.35  103 
d
 125 
 = 3.5 V
dt
25 103


(Note that since the flux is reversed, it changes from 35 mWb to - 35 mWb, which is a change of
35 - - 35, i.e. 70 mWb).
5. Calculate the e.m.f. induced in a coil of inductance 6 H by a current changing at a rate of 15 A/s.
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© John Bird Published by Taylor and Francis
E.m.f. induced, E =  L
dI
 15 
  6    = - 90 V
dt
1
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EXERCISE 196, Page 441
1. An inductor of 20 H has a current of 2.5 A flowing in it. Find the energy stored in the magnetic
field of the inductor.
Energy stored, W =
1 2 1
2
LI   20   2.5  = 62.5 J
2
2
2. Calculate the value of the energy stored when a current of 30 mA is flowing in a coil of
inductance 400 mH.
Energy stored, W =
2
1 2 1
LI   400 103   30 103  = 0.18 mJ
2
2
3. The energy stored in the magnetic field of an inductor is 80 J when the current flowing in the
inductor is 2 A. Calculate the inductance of the coil.
Energy, W =
1 2
LI
2
from which, inductance, L =
2W 2  80

= 40 H
2
I2
 2
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EXERCISE 197, Page 442
1. A flux of 30 mWb links with a 1200 turn coil when a current of 5 A is passing through the coil.
Calculate (a) the inductance of the coil, (b) the energy stored in the magnetic field, and (c) the
average e.m.f. induced if the current is reduced to zero in 0.20 s.
(a) Inductance of coil, L =
(b) Energy stored, W =
N  1200  30 103
= 7.2 H

I
5
1 2 1
2
LI   7.2   5  = 90 J
2
2
(c) Induced e.m.f., E = L
dI
 50
 7.2 
 = 180 V
dt
 0.20 
2. An e.m.f. of 2 kV is induced in a coil when a current of 5 A collapses uniformly to zero in
10 ms. Determine the inductance of the coil.
Induced e.m.f., E = L
dI
dt
from which, inductance, L =
E
2000
2000 10 103


=4H
dI
50
5
dt 10 103
3. An average e.m.f. of 60 V is induced in a coil of inductance 160 mH when a current of 7.5 A is
reversed. Calculate the time taken for the current to reverse.
Induced e.m.f., E = L
from which,
dI
dt
hence, 60 = 160 103 
time, t = 160 103 
 7.5  7.5
t
15
= 0.04 s or 40 ms
60
4. A coil of 2500 turns has a flux of 10 mWb linking with it when carrying a current of 2 A.
Calculate the coil inductance and the e.m.f. induced in the coil when the current collapses to zero
in 20 ms.
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© John Bird Published by Taylor and Francis
Inductance, L =
N 2500 10 103
= 12.5 H

I
2
Induced e.m.f., E = L
dI
 20 
 12.5 
= 1.25 kV
3 
dt
 20 10 
5. When a current of 2 A flows in a coil, the flux linking with the coil is 80 Wb. If the coil
inductance is 0.5 H, calculate the number of turns of the coil.
If L =
N
LI
0.5  2
then number of turns, N =

= 12,500
 80 106
I
EXERCISE 198, Page 442
Answers found from within the text of the chapter, pages 429 to 442.
EXERCISE 199, Page 443
1. (a)
13. (d)
2. (b)
3. (c)
4. (a)
5. (b) 6. (b) 7. (c)
8. (c)
9. (c) 10. (d) 11. (c) 12. (b)
14. (a)
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