CHAPTER 15 SINGLE-PHASE SERIES A.C. CIRCUITS
Exercise 83, Page 236
1. Calculate the reactance of a coil of inductance 0.2 H when it is connected to (a) a 50 Hz,
(b) a 600 Hz and (c) a 40 kHz supply.
(a) Inductive reactance, XL  2 f L  2  50 0.2 = 62.83 
(b) Inductive reactance, XL  2 f L  2  600 0.2 = 754 
(c) Inductive reactance, X L  2 f L  2  40  103   0.2 = 50.27 k
2. A coil has a reactance of 120  in a circuit with a supply frequency of 4 kHz. Calculate the
inductance of the coil.
X L  2 f L hence, inductance, L =
XL
120

= 4.77 mH
2 f 2  4 103 
3. A supply of 240 V, 50 Hz is connected across a pure inductance and the resulting current is 1.2 A.
Calculate the inductance of the coil.
Inductive reactance, X L 
V 240

= 200 Ω
I 1.2
X L  2 f L hence, inductance, L =
XL
200
= 0.637 H

2 f 2  50 
4. An e.m.f. of 200 V at a frequency of 2 kHz is applied to a coil of pure inductance 50 mH.
Determine (a) the reactance of the coil, and (b) the current flowing in the coil.
(a) Inductive reactance, X L  2 f L  2  2 103  50 103  = 628 
(b) Current, I =
V 200

= 0.318 A
X L 628
© John Bird Published by Taylor and Francis
177
5. A 120 mH inductor has a 50 mA, 1 kHz alternating current flowing through it. Find the p.d.
across the inductor
Inductive reactance, X L  2 f L  2 1103 120 103  = 753.982 
P.d. across inductor, VL  I  X L  50 103  753.982 = 37.7 V
6. Calculate the capacitive reactance of a capacitor of 20 μF when connected to an a.c. circuit of
frequency (a) 20 Hz, (b) 500 Hz, (c) 4 kHz
(a) Capacitive reactance, X C 
1
1

= 397.9 
2 f C 2 20  20 10 6
(b) Capacitive reactance, X C 
1
1

= 15.92 
2 f C 2 500  20 10 6
(c) Capacitive reactance, X C 
1
1

= 1.989 
2 f C 2 4000  20 10 6
7. A capacitor has a reactance of 80  when connected to a 50 Hz supply. Calculate the value of
the capacitor.
Capacitive reactance, X C 
1
1
1
from which, capacitance, C =

2 f C
2 f XC 2  50 80 
= 39.79 F
8. Calculate the current taken by a 10 F capacitor when connected to a 200 V, 100 Hz supply.
Capacitive reactance, X C 
Current, I =
1
1

= 159.155 
2 f C 2100 10 106
V
200

= 1.257 A
X C 159.155
© John Bird Published by Taylor and Francis
178
9. A capacitor has a capacitive reactance of 400  when connected to a 100 V, 25 Hz supply.
Determine its capacitance and the current taken from the supply.
Capacitive reactance, X C 
1
1
1
from which, capacitance, C =

2 f C
2 f XC 2  25 400 
= 15.92 F
Current, I =
V 100

= 0.25 A
X L 400
10. Two similar capacitors are connected in parallel to a 200 V, 1 kHz supply. Find the value of
each capacitor if the current is 0.628 A.
XC 
i.e.
V
200

= 318.47 
I 0.628
1
1
 318.47 , hence, total capacitance, CT 
= 0.50 F
3
2 f C
2 10   318.47 
Since for parallel connection of capacitors, CT  C1  C2  2C1 , then C1 
0.50
= 0.25 F
2
i.e. each capacitor has a capacitance of 0.25 F
© John Bird Published by Taylor and Francis
179
Exercise 84, page 239
1. Determine the impedance of a coil which has a resistance of 12  and a reactance of 16 
Impedance, Z =
R 2  XL2  122  162 = 20 
2. A coil of inductance 80 mH and resistance 60  is connected to a 200 V, 100 Hz supply.
Calculate the circuit impedance and the current taken from the supply. Find also the phase angle
between the current and the supply voltage.
Inductive reactance, X L  2 f L  2 100   80 103  = 50.265 
Impedance, Z =
Current, I =
R 2  XL2  602  50.2652 = 78.27  (see impedance triangle in the diagram
below)
V
200

= 2.555 A
Z 78.27
From the impedance triangle, tan  
XL
R
 50.265 
hence the circuit phase angle,  = tan 1 
 = 39.95 lagging
 60 
3. An alternating voltage given by v = 100 sin 240t volts is applied across a coil of resistance 32 
and inductance 100 mH. Determine (a) the circuit impedance, (b) the current flowing, (c) the p.d.
across the resistance, and (d) the p.d. across the inductance.
(a) Inductive reactance, X L  2 f L  L   240  100 103  = 24 
Circuit impedance, Z =
R 2  XL 2  322  242 = 40 
© John Bird Published by Taylor and Francis
180
(b) Current flowing, I =
V 0.707 100

= 1.77 A
Z
40
(Note r.m.s. current = 0.707  maximum value)
(c) P.d. across the resistance, VR  I R  1.77  32 = 56.64 V
(d) P.d. across the inductance, VL  I XL  1.77  24 = 42.48 V
4. A coil takes a current of 5 A from a 20 V d.c. supply. When connected to a 200 V, 50 Hz a.c.
supply the current is 25 A. Calculate the (a) resistance, (b) impedance and (c) inductance of the
coil.
(a) From a d.c. circuit, resistance, R =
V 20

=4
I
5
(b) From an a.c. circuit, impedance, Z =
V 200

=8
I
25
(c) From the impedance triangle, Z2  R 2  X L 2
from which, X L  Z2  R 2  82  42 = 6.9282 
Also, X L  2 f L from which, inductance, L =
XL
6.9282
= 22.05 mH

2 f 2  50 
5. A resistor and an inductor of negligible resistance are connected in series to an a.c. supply. The
p.d. across the resistor is 18 V and the p.d. across the inductor is 24 V. Calculate the supply voltage
and the phase angle between voltage and current.
Supply voltage, V =
Tan ϕ =
VL 24

VR 18
VR 2  VL2  182  242 = 30 V
from which,
 24 
phase angle between voltage and current, ϕ = tan 1   = 53.13º lagging (i.e. current lags
 18 
© John Bird Published by Taylor and Francis
181
voltage in an inductive circuit)
6. A coil of inductance 636.6 mH and negligible resistance is connected in series with a 100 
resistor to a 250 V, 50 Hz supply. Calculate (a) the inductive reactance of the coil, (b) the
impedance of the circuit, (c) the current in the circuit, (d) the p.d. across each component, and
(e) the circuit phase angle.
The circuit is shown in the diagram below.
(a) Inductive reactance of coil, X L  2 f L  2  50   636.6 103  = 200 
(b) Impedance, Z =
(c) Current, I =
R 2  XL2  1002  2002 = 223.6 
(from impedance triangle)
V
250

= 1.118 A
Z 223.6
(d) Voltage across resistance, VR  I R  1.118 100 = 111.8 V
Voltage across inductance, VL  I XL  1.118  200 = 223.6 V
(e) From impedance triangle, tan  
XL
R
from which, circuit phase angle,  = tan 1
XL
 200 
 tan 1 
 = 63.43 lagging
R
 100 
© John Bird Published by Taylor and Francis
182
Exercise 85, page 241
1. A voltage of 35 V is applied across a C-R series circuit. If the voltage across the resistor is 21 V, find the
voltage across the capacitor.
Supply voltage, V =
VR 2  VC 2
i.e.
V 2  VR 2  VC 2
352  212  VC 2
i.e.
from which, voltage across the capacitor, VC  352  212 = 28 V
2. A resistance of 50  is connected in series with a capacitance of 20 F. If a supply of 200 V,
100 Hz is connected across the arrangement find (a) the circuit impedance, (b) the current
flowing, and (c) the phase angle between voltage and current.
The circuit diagram is shown below.
(a) Capacitive reactance, X C 
Impedance, Z =
(b) Current, I =
(c) tan  
XC
R
1
1

= 79.577 
2 f C 2 100   20 106 
R 2  XC2  502  79.577 2 = 93.98 
V
200

= 2.128 A
Z 93.98
from which, phase angle,  = tan 1
XC
 79.577 
 tan 1 
 = 57.86 leading
R
 50 
© John Bird Published by Taylor and Francis
183
3. A 24.87 μF capacitor and a 30  resistor are connected in series across a 150 V supply. If the current
flowing is 3 A find (a) the frequency of the supply, (b) the p.d. across the resistor and (c) the p.d. across the
capacitor.
(a) Impedance, Z =
V 150

= 50 Ω
I
3
Also, impedance, Z =
from which,
R 2  XC 2
502  302  X C 2
i.e. 50 =
and
302  XC 2
X C  502  302 = 40 Ω
1
from which,
2 f C
1
1

frequency, f =
= 160 Hz
2 X C C 2  40   24.87 106 
Capacitive reactance, X C 
(b) P.d across the resistor, VR  I R  3  30 = 90 V
(c) P.d across the capacitor, VC  I XC  3  40 = 120 V
4. An alternating voltage v = 250 sin 800t volts is applied across a series circuit containing a 30 
and 50 F capacitor. Calculate (a) the circuit impedance, (b) the current flowing, (c) the p.d.
across the resistor, (d) the p.d. across the capacitor, and (e) the phase angle between voltage and
current.
The circuit is shown below.
(a) Capacitive reactance, X C 
Impedance, Z =
1
1
1


= 25 
2 f C C  800   50 106 
R 2  XC2  302  252 = 39.05 
© John Bird Published by Taylor and Francis
184
(b) Current, I =
V 0.707  250

= 4.526 A
Z
39.05
(c) P.d across the resistor, VR  I R  4.526  30 = 135.8 V
(d) P.d across the capacitor, VC  I XC  4.526  25 = 113.2 V
(e) tan  
XC
R
from which, phase angle,  = tan 1
XC
 25 
 tan 1   = 39.81 leading
R
 30 
5. A 400  resistor is connected in series with a 2358 pF capacitor across a 12 V a.c. supply.
Determine the supply frequency if the current flowing in the circuit is 24 mA.
The circuit is shown below.
Impedance, Z =
V
12

= 500 
I 24 103
From the impedance triangle (as in the previous problem), Z2  R 2  X L 2
from which, capacitive reactance, X C  Z2  R 2  5002  4002  300 
1
1

Hence,
300 =
2 f C 2 f  2358 1012 
from which, supply frequency, f =
1
= 225 kHz
2  300   2358 1012 
© John Bird Published by Taylor and Francis
185
Exercise 86, Page 244
1. A 40 μF capacitor in series with a coil of resistance 8  and inductance 80 mH is connected to a 200 V,
100 Hz supply. Calculate (a) the circuit impedance, (b) the current flowing, (c) the phase angle between
voltage and current, (d) the voltage across the coil, and (e) the voltage across the capacitor.
The circuit diagram is shown below.
(a) Inductive reactance, X L  2 f L  2 100  80 103   50.265 
Capacitive reactance, X C 
1
1

 39.789 
2 f C 2 100   40 106 
XL  XC  50.265  39.789  10.476 
Impedance, Z =
R 2   XL  XC   82  10.4762 = 13.18 
2
(b) Current flowing, I =
(c) tan  
XL  XC
R
V 200

= 15.17 A
Z 13.18
from which,
phase angle,  = tan 1
XL  XC
 10.476 
 tan 1 
 = 52.63 lagging
R
 8 
(d) Zcoil  R12  X L1 2  82  50.2652  50.898 
Voltage across coil, Vcoil  I Zcoil  15.17 50.898 = 772.1 V
(e) Voltage across capacitor, VC  I XC  15.17 39.789  = 603.6 V
© John Bird Published by Taylor and Francis
186
2. Find the values of resistance R and inductance L in the circuit shown.
Circuit impedance, Z =
V
2400

 16035  (131  j91.772) 
I 1.5  35
Hence, resistance, R = 131 
and X L  XC  91.772
Hence,
Now, X C 
X L  79.577  91.772
1
1

= 79.577
2fC 2  50   40 106 
from which,
XL  91.772  79.577  171.349
2 f L  171.349 and inductance, L =
i.e.
171.349
= 0.545 H
2  50 
3. Three impedances are connected in series across a 100 V, 2 kHz supply. The impedances
comprise:
(i) an inductance of 0.45 mH and resistance 2
(ii) an inductance of 570 H and 5  resistance, and
(iii) a capacitor of capacitance 10 F and resistance 3 
Assuming no mutual inductive effects between the two inductances calculate (a) the circuit
impedance, (b) the circuit current, (c) the circuit phase angle and (d) the voltage across each
impedance. Draw the phasor diagram.
The circuit is shown below.
© John Bird Published by Taylor and Francis
187
Total resistance, R T = 2 + 5 + 3 = 10 
Total inductance, LT  0.45 mH  570 H  1.02 mH
The simplified circuit is shown below.
Inductive reactance, X L  2 f L  2  2000  1.02 103   12.818 
Capacitive reactance, X C 
1
1

 7.958 
2 f C 2  2000  10 106 
XL  XC  12.818  7.958  4.86 
(a) Impedance, Z =
(b) Current, I =
(c) tan  
R 2   XL  XC   102  4.862 = 11.12 
2
V 100

= 8.99 A
Z 11.12
XL  XC
R
from which,
phase angle,  = tan 1
XL  XC
 4.86 
 tan 1 
 = 25.92 lagging
R
 10 
(d) X L1  2  2000   0.45 103   5.655 
Z1  R12  X L1 2  22  5.6552  5.998 
Voltage across first impedance, V1  I Z1  8.99 5.998 = 53.92 V
X L2  2  2000   570 106   7.163 
Z2  R 2 2  X L2 2  52  7.1632  8.735 
Voltage across second impedance, V2  I Z2  8.998.735 = 78.53 V
© John Bird Published by Taylor and Francis
188
X C3  7.958  from earlier
Z2  R 2 2  X L2 2  32  7.9582  8.505 
Voltage across third impedance, V3  I Z3  8.998.505 = 76.46 V
4. For the circuit shown below determine the voltages V1 and V2 if the supply frequency is 1 kHz.
Draw the phasor diagram and hence determine the supply voltage V and the circuit phase angle.
X L1  2 f L  2 1000  1.91103   12 
Z1  R12  X L1 2  52  122  13 
and   tan 1
XL
 12 
 tan 1    67.38 lagging
R
 5
Voltage, V1 = I Z1   2 13 = 26.0 V at 67.38 lagging
X C2 
1
1

 32 
2 f C 2 1000   4.974 10 6 
Z2  R 2 2  XC2 2  102  322  33.526 
and   tan 1
XC
 32 
 tan 1    72.65 leading
R
 10 
Voltage, V2 = I Z2   2 33.526  = 67.05 V at 72.65 leading
The voltages are shown in the phasor diagram (i) below.
(i)
(ii)
© John Bird Published by Taylor and Francis
189
The supply voltage V is the phasor sum of voltages V1 and V2 . V is shown by the length ac in
diagram (ii).
In triangle abc, b = 180 - 72.65 – 67.38 = 39.97
Using the cosine rule, ac2  67.052  26.02  2  67.05 26.0  cos39.97
from which,
Using the sine rule,
ac = 50 V
26.0sin 39.97
26.0
50
 0.3340

from which, sin  
50
sin  sin 39.97
from which,   sin 1 0.3340  19.51 and from diagram (ii),   72.65 19.51  53.14 leading.
Hence,
supply voltage, V = 50 V at 53.14 leading
© John Bird Published by Taylor and Francis
190
Exercise 87, Page 247
1. Find the resonant frequency of a series a.c. circuit consisting of a coil of resistance 10  and inductance
50 mH and capacitance 0.05 μF. Find also the current flowing at resonance if the supply voltage is 100 V.
Resonant frequency, f r 
1

2 LC 2
1
At resonance, current, I =
V 100

= 10 A
R 10
50 10  0.05 10 
3
6
= 3.183 kHz
2. The current at resonance in a series L-C-R circuit is 0.2 mA. If the applied voltage is 250 mV at
a frequency of 100 kHz and the circuit capacitance is 0.04 F, find the circuit resistance and
inductance.
At resonance, current, I =
V
R
i.e. resistance, R =
At resonance, resonant frequency, f r 
Hence, inductance, L =
1
C  2f 
2

1
2 LC
V 250 10 3

= 1.25 k
I 0.2 10 3
i.e.
2f 
1
0.04  106  2 100  103 
2
1
LC
and
 2f 
2

1
LC
= 63.3 H
3. A coil of resistance 25  and inductance 100 mH is connected in series with a capacitance of
0.12 F across a 200 V, variable frequency supply. Calculate (a) the resonant frequency, (b) the
current at resonance and (c) the factor by which the voltage across the reactance is greater than
the supply voltage.
(a) Resonant frequency, f r 
1
1
= 1.453 kHz


2 LC 2 100 10 3  0.12 106 
(b) At resonance, current, I =
V 200

=8A
R 25
© John Bird Published by Taylor and Francis
191
(c) Q-factor =
1 L 1 100 10 3

= 36.51
R C 25 0.12 10 6
4. A coil of 0.5 H inductance and 8  resistance is connected in series with a capacitor across a 200 V, 50 Hz
supply. If the current is in phase with the supply voltage, determine the capacitance of the capacitor and the
p.d. across its terminals.
If the current is in phase with the supply voltage, then the circuit is resonant.
At resonance, X L  XC i.e. 2 f L 
capacitance, C =
1
2 f C
1
 2f 
2
from which,

L
1
 2 50   0.5 
2
= 20.26 μF
1
 V   1   200 

P.d. across the capacitor terminals, VC  IXC    


6 
 R   2 f C   8  2 50  20.26 10 
= 3928 V = 3.928 kV
5. Calculate the inductance which must be connected in series with a 1000 pF capacitor to give a resonant
frequency of 400 kHz.
Resonant frequency, f r 
from which, 2 LC 
and
 1 
LC = 

 2 f r 
1
2 LC
1
fr
and
2
LC 
1
2 f r
2
and
1 1 
1
1


inductance, L = 
 
12 
3 
C  2 f r  1000 10  2  400 10 
= 158 H
2
or 0.158 mH
6. A series circuit comprises a coil of resistance 20  and inductance 2 mH and a 500 pF capacitor.
Determine the Q-factor of the circuit at resonance. If the supply voltage is 1.5 V, what is the
voltage across the capacitor?
© John Bird Published by Taylor and Francis
192
Q-factor =
Q=
VC
V
1 L 1
2 103

= 100
R C 20 500 1012
hence, voltage across the capacitor, VC  VQ  1.5100 = 150 V
© John Bird Published by Taylor and Francis
193
Exercise 88, Page 251
1. A voltage v = 200 sin t volts is applied across a pure resistance of 1.5 k. Find the power dissipated in the
resistor.
Power dissipated in the resistor, P = I2 R
Current, I =


200 / 2
V
= 0.09428 A

R
1500
(note that in the formula for power I has to be the r.m.s. value)
Hence, power dissipated = I 2 R   0.09428  (1500) = 13.33 W
2
2. A 50 F capacitor is connected to a 100 V, 200 Hz supply. Determine the true power and the
apparent power.
Capacitive reactance, X C 
Current, I =
1
1

= 15.915 
2fC 2  200   50 106 
V
100

= 6.283 A
X C 15.915
True power, P = V I cos  = (100)(6.283) cos 90 = 0
Apparent power, S = V I = (100)(6.283) = 628.3 VA
3. A motor takes a current of 10 A when supplied from a 250 V a.c. supply. Assuming a power
factor of 0.75 lagging find the power consumed. Find also the cost of running the motor for 1
week continuously if 1 kWh of electricity costs 12.20 p.
P = V I cos  = (250)(10)(0.75)
since power factor = cos 
= 1875 W
Energy = power  time = (1.875 kW)(7  24) = 315 kWh
Hence, cost of running motor for 1 week = 315  12.20 = 3843 p = £38.43
© John Bird Published by Taylor and Francis
194
4. A motor takes a current of 12 A when supplied from a 240 V a.c. supply. Assuming a power
factor of 0.70 lagging find the power consumed.
Power consumed, P = V I cos  = (240)(12)(0.70)
since power factor = cos 
= 2016 W or 2.016 kW
5. A transformer has a rated output of 100 kVA at a power factor of 0.6. Determine the rated power output and
the corresponding reactive power.
VI = 100 kVA = 100 × 103 and p.f. = 0.6 = cos 
Power output, P = VI cos  = (100 × 103)(0.6) = 60 kW
Reactive power, Q = VI sin 
If cos  = 0.6, then  = cos 1 0.6 = 53.13
Hence sin  = sin 53.13o = 0.8
Hence reactive power, Q = (100 × 103)(0.8) = 80 kvar
6. A substation is supplying 200 kVA and 150 kvar. Calculate the corresponding power and power
factor.
Apparent power, S = V I = 200 103 VA
Hence,
and
200 103 sin  = 150 10 3
and
reactive power, Q = V I sin  = 150 10 3 var
from which,
sin  =
150 103
 0.75
200 103
 = sin 1 0.75 = 48.59
Thus, power, P = V I cos  = 200 103 cos 48.59 = 132 kW
and
power factor = cos  = cos 48.59 = 0.66
7. A load takes 50 kW at a power factor of 0.8 lagging. Calculate the apparent power and the reactive power.
True power P = 50 kW = VI cos  and power factor = 0.8 = cos 
© John Bird Published by Taylor and Francis
195
Apparent power, S = VI =
P
50
=
= 62.5 kVA
cos  0.8
Angle  = cos 1 0.8 = 36.87o hence sin  = sin 36.87o = 0.6
Hence, reactive power, Q = VI sin  = 62.5 × 103 × 0.6 = 37.5 kvar
8. A coil of resistance 400  and inductance 0.20 H is connected to a 75 V, 400 Hz supply.
Calculate the power dissipated in the coil.
Inductive reactance, XL  2 f L  2  400 0.20  = 502.65 
Impedance, Z =
Current, I =
R 2  XL 2  4002  502.652 = 642.39 
V
75

= 0.11675 A
Z 642.39
From the impedance triangle, tan  
XL
R
and   tan 1
XL
 502.65 
 tan 1 
 = 51.49
R
 400 
Hence, power dissipated in coil, P = V I cos  = (75)(0.11675) cos 51.49 = 5.452 W
Alternatively, P = I 2 R   0.11675   400  = 5.452 W
2
9. An 80  resistor and a 6 μF capacitor are connected in series across a 150 V, 200 Hz supply. Calculate
(a) the circuit impedance, (b) the current flowing and (c) the power dissipated in the circuit.
(a) Capacitive reactance, X C 
Impedance, Z =
(b) Current, I =
1
1

= 132.63 
2 f C 2(200)(6 10 6 )
R 2  XL2  802  132.632 = 154.9 
V 150

= 0.968 A
Z 154.9
(c) From the impedance triangle, tan  
XC
R
and   tan 1
XC
 132.63 
 tan 1 
 = 58.90
R
 80 
Hence, power dissipated in coil, P = V I cos  = (150)(0.968) cos 58.90 = 75 W
© John Bird Published by Taylor and Francis
196
Alternatively, P = I 2 R   0.968   80  = 75 W
2
10. The power taken by a series circuit containing resistance and inductance is 240 W when
connected to a 200 V, 50 Hz supply. If the current flowing is 2 A find the values of the
resistance and inductance.
Power, P = I2 R
Impedance, Z =
240 =  2  R
2
i.e.
from which, resistance, R =
240
 2
2
= 60 
V 200

= 100 
I
2
From the impedance triangle, Z2  R 2  X L 2
from which,
i.e.
X L  Z2  R 2  1002  602 = 80 
2 f L = 80
from which, inductance, L =
80
= 0.255 H or 255 mH
2  50 
11. The power taken by a C-R series circuit, when connected to a 105 V, 2.5 kHz supply, is 0.9 kW and the
current is 15 A. Calculate (a) the resistance, (b) the impedance, (c) the reactance, (d) the capacitance,
(e) the power factor, and (f) the phase angle between voltage and current.
0.9 103  15  R
2
(a) Power, P = I2 R
i.e.
(b) Impedance, Z =
V 105

=7
I
15
from which, resistance, R =
900
15 
2
=4
(c) From the impedance triangle, Z2  R 2  X L 2
from which, capacitive reactance, X C  Z2  R 2  7 2  42 = 5.745 
(d) Capacitive reactance, X C 
1
1
i.e. 5.745 =
2 f C
2 (2500) C
from which, capacitance, C =
(e) Power factor, p.f. =
1
= 11.08 μF
2(2500)(5.745)
R 4
 = 0.571
Z 7
© John Bird Published by Taylor and Francis
197
(f) Tan ϕ =
X C 5.745

R
4
 5.745 
and the phase angle between voltage and current, ϕ = tan 1 
 = 55.15º leading
 4 
12. A circuit consisting of a resistor in series with an inductance takes 210 W at a power factor of
0.6 from a 50 V, 100 Hz supply. Find (a) the current flowing, (b) the circuit phase angle, (c) the
resistance, (d) the impedance and (e) the inductance.
(a) Power, P = V I cos 
i.e.
Hence,
210 = (50) I (0.6)
current, I =
(b) If cos  = 0.6 then
(c) Power, P = I2 R
210
=7A
 50 0.6
circuit phase angle,  = cos 1 0.6 = 53.13 lagging
i.e.
(d) Impedance, Z =
since p.f. = cos 
210 =  7  R
2
from which, resistance, R =
210
7
2
= 4.286 
V 50

= 7.143 
I
7
(e) From the impedance triangle, Z2  R 2  X L 2
from which,
i.e.
X L  Z2  R 2  7.1432  4.2862 = 5.71425 
2 f L = 5.71425
from which, inductance, L =
5.71425
= 9.095 mH
2 100 
13. A 200 V, 60 Hz supply is applied to a capacitive circuit. The current flowing is 2 A and the
power dissipated is 150 W. Calculate the values of the resistance and capacitance.
Power, P = I2 R
Impedance, Z =
i.e.
150 =  2  R
2
from which, resistance, R =
150
 2
2
= 37.5 
V 200

= 100 
I
2
From the impedance triangle, Z2  R 2  X C 2
© John Bird Published by Taylor and Francis
198
from which,
i.e.
X C  Z2  R 2  1002  37.52 = 92.702 
92.702 =
1
2 f C
from which, capacitance, C =
1
= 28.61 F
2  60  92.702 
© John Bird Published by Taylor and Francis
199