CHAPTER 16 SINGLE-PHASE PARALLEL A.C. CIRCUITS
Exercise 91, Page 256
1. A 30  resistor is connected in parallel with a pure inductance of 3 mH across a 110 V, 2 kHz
supply. Calculate (a) the current in each branch, (b) the circuit current, (c) the circuit phase
angle, (d) the circuit impedance, (e) the power consumed, and (f) the circuit power factor.
The circuit is shown below.
(a) Current in resistor, I R 
V 110

= 3.67 A
R 30
Inductive reactance, X L  2 f L  2  2000   3 103  = 37.70 
Current in inductance, I L 
(b) Circuit current, I =
(c) tan  
I
2
R
V
110

= 2.92 A
X L 37.70
 IL 2  
3.67
2
 2.922  = 4.69 A
IL
 2.92 
hence, circuit phase angle,   tan 1 
 = 38.51 lagging
IR
 3.67 
(d) Circuit impedance, Z =
V 110

= 23.45 
I 4.69
(e) Power consumed, P  VI cos   (110)(4.69) cos 38.51 = 404 W
or
P  I R 2 R   3.67   30  = 404 W
2
(f) Power factor = cos  = cos 38.51 = 0.782 lagging
© John Bird Published by Taylor and Francis
200
2. A 40  resistance is connected in parallel with a coil of inductance L and negligible resistance
across a 200 V, 50 Hz supply and the supply current is found to be 8 A. Sketch the phasor
diagram and determine the inductance of the coil.
The circuit diagram is shown in (a) below.
(a)
Current, I R 
(b)
V 200

=5A
R 40
From the phasor diagram in diagram (b) above,
I
current IL 
Inductive reactance, X L 
from which,
V
IL
2
 IR 2   82  52  = 6.245 
i.e. 2 f L 
inductance, L =
200
6.245
200
= 0.102 H or 102 mH
 6.245 250 
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Exercise 92, Page 257
1. A 1500 nF capacitor is connected in parallel with a 16  resistor across a 10 V, 10 kHz supply.
Calculate (a) the current in each branch, (b) the supply current, (c) the circuit phase angle, (d) the
circuit impedance, (e) the power consumed, (f) the apparent power, and (g) the circuit power
factor. Sketch the phasor diagram
The circuit diagram is shown in (i) below.
(i)
(ii)
(a) Current in resistor, I R 
V 10

= 0.625 A
R 16
Capacitive reactance, X C 
1
1

= 10.61 
3
2fC 2 10 10 1500 109 
Current in capacitor, IC 
V
10

= 0.943 A
X L 10.61
(b) Supply current, I =
I
2
R
 IC 2  
 0.625
2
 0.9432  = 1.131 A
IC
 0.943 
hence, circuit phase angle,   tan 1 
 = 56.46 leading, as shown in the
IR
 0.625 
phasor diagram in (ii) above
V
10

(d) Circuit impedance, Z =
= 8.84 
I 1.131
(c) tan  
(e) Power consumed, P  VI cos   (10)(1.131) cos 56.46 = 6.25 W
or
P  I R 2 R   0.625  16  = 6.25 W
2
(f) Apparent power, S = VI = (10)(1.131) = 11.31 VA
(g) Power factor = cos  = cos 56.46 = 0.553 leading
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2. A capacitor C is connected in parallel with a resistance R across a 60 V, 100 Hz supply. The
supply current is 0.6 A at a power factor of 0.8 leading. Calculate the values of R and C.
The circuit diagram is shown in (a) below.
(a)
(b)
Power factor = cos  = 0.8 hence,   cos 1 0.8 = 36.87 leading
From the phasor diagram in (b) above,
current in resistor, IR  I cos   0.6cos36.87 = 0.48 A
and
current in capacitor, IC  Isin   0.6sin 36.87 = 0.36 A
Resistance, R =
V
60

= 125 
I R 0.48
Capacitive reactance, X C 
from which,
1
V
60

= 166.67 =
2fC
IC 0.36
capacitance, C =
1
166.67  2100 
= 9.55 F
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Exercise 93, Page 259
1. An inductance of 80 mH is connected in parallel with a capacitance of 10 F across a 60 V,
100 Hz supply. Determine (a) the branch currents, (b) the supply current, (c) the circuit phase
angle, (d) the circuit impedance, and (e) the power consumed.
The circuit diagram is shown in (i) below.
(i)
(ii)
(a) Inductive reactance, X L  2 f L  2 100   80 103  = 50.265 
Hence, current in inductance, I L 
Capacitive reactance, X C 
V
60

= 1.194 A
X L 50.265
1
1

= 159.155 
2fC 2 100  10 106 
Hence, current in capacitor, IC 
V
60

= 0.377 A
X C 159.155
(b) Supply current, I = IL  IC = 1.194 – 0.377 = 0.817 A
(c) From the phasor diagram in (ii) above, circuit phase angle,  = 90 lagging
(d) Impedance, Z =
V
60

= 73.44 
I 0.817
(e) Power consumed, P = VI cos   (60)(0.817) cos 90 = 0 W
2. Repeat problem 1 for a supply frequency of 200 Hz.
(a) Inductive reactance, X L  2 f L  2  200   80 103  = 100.53 
© John Bird Published by Taylor and Francis
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Hence, current in inductance, I L 
Capacitive reactance, X C 
V
60

= 0.597 A
X L 100.53
1
1

= 79.577 
2fC 2  200  10 106 
Hence, current in capacitor, IC 
V
60

= 0.754 A
X C 79.577
(b) Supply current, I = IC  IL = 0.754 – 0.597 = 0.157 A
(c) From the phasor diagram in below, circuit phase angle,  = 90 leading
(d) Impedance, Z =
V
60

= 382.2 
I 0.157
(e) Power consumed, P = VI cos   (60)(0.157) cos 90 = 0 W
© John Bird Published by Taylor and Francis
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Exercise 94, Page 261
1. A coil of resistance 60  and inductance 318.4 mH is connected in parallel with a 15 F
capacitor across a 200 V, 50 Hz supply. Calculate (a) the current in the coil, (b) the current in the
capacitor, (c) the supply current and its phase angle, (d) the circuit impedance, (e) the power
consumed, (f) the apparent power and (g) the reactive power. Sketch the phasor diagram.
The circuit is shown in (i) below.
(i)
(ii)
(a) X L  2 f L  2  50   318.4 103   100 
Hence, ZLR  R 2  XL 2 
 60
2
 1002   116.62 
Current in coil, I LR 
V
200

= 1.715 A
ZLR 116.62
Phase angle,   tan 1
XL
 100 
 tan 1 
 = 59.04 lagging
R
 60 
(b) X C 
1
1

 212.2 
2fC 2  50  15 106 
Current in capacitor, IC 
V
200

= 0.943 A leading by 90
X C 212.2
The phasor diagram is shown in (ii) above.
(c) Horizontal component of ILR  ILR cos59.04  1.715cos59.04  0.8823 A
Horizontal component of IC  IC cos90 = 0
Total horizontal component = 0.8823 + 0 = 0.8823 A
© John Bird Published by Taylor and Francis
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Vertical component of ILR  ILR sin 59.04  1.715sin 59.04  1.4707 A
Vertical component of IC  IC sin 90  0.943sin 90  0.943 A
Total vertical component = -1.4707 + 0.943 = - 0.5277 A
From the diagram below, resultant current, I =
 0.8823
2
 0.52772  = 1.028 A
 0.5277 
Phase angle,   tan 1 
 = 30.88 lagging
 0.8823 
(d) Circuit impedance, Z =
V 200

= 194.6 
I 1.028
(e) Power consumed, P = V I cos  = (200)(1.028) cos 30.88 = 176.5 W
or
P = I LR 2 R  1.715   60  = 176.5 W
2
(f) Apparent power, S = V I = (200)(1.028) = 205.6 VA
(g) Reactive power, Q = (200)(1.028) sin 30.88 = 105.5 var
The phasor diagram is shown sketched in (ii) above.
2. A 25 nF capacitor is connected in parallel with a coil of resistance 2 k and inductance 0.20 H
across a 100 V, 4 kHz supply. Determine (a) the current in the coil, (b) the current in the
capacitor, (c) the supply current and its phase angle, (d) the circuit impedance, and (e) the power
consumed.
The circuit diagram is shown in (i) below.
(a) XL  2 f L  2  4000 0.2  5026.55 
Hence, ZLR  R 2  XL 2 
 2000
2
 5026.552   5410 
© John Bird Published by Taylor and Francis
207
Current in coil, I LR 
V
100

= 18.48 mA
ZLR 5410
Phase angle,   tan 1
XL
 5026.55 
 tan 1 
 = 68.30 lagging
R
 2000 
(i)
(b) X C 
(ii)
1
1

 1591.55 
2fC 2  4000   25 109 
Current in capacitor, IC 
V
100

= 62.83 mA leading by 90
X C 1591.55
The phasor diagram is shown in (ii) above.
(c) Total horizontal component = 18.48 cos 68.30 + 62.83 cos 90 = 6.833 mA
Total vertical component = - 18.48 sin 68.30 + 62.83 sin 90 = 45.660 mA
From the diagram below, resultant current, I =
 6.833
2
 45.6602  = 46.17 mA
 45.66 
Phase angle,   tan 1 
 = 81.49 leading
 6.833 
(d) Circuit impedance, Z =
V
100

= 2166 
I 46.17 103
or 2.166 k
(e) Power consumed, P = V I cos  = (100) (46.17 103 )2 cos 81.49 = 0.683 W
© John Bird Published by Taylor and Francis
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or
P = ILR 2 R  18.48 103   2000  = 0.683 W
2
© John Bird Published by Taylor and Francis
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Exercise 95, Page 265
1. A 0.15 μF capacitor and a pure inductance of 0.01 H are connected in parallel across a 10 V,
variable frequency supply. Determine (a) the resonant frequency of the circuit, and (b) the current
circulating in the capacitor and inductance.
(a) Parallel resonant frequency, fr =
1  1 R2  1  1 




 when R = 0
2  LC L2  2   LC 
=

1 
1

6 
2  (0.01)(0.15 10 ) 
=
1
(25819.0) = 4109 Hz = 4.11 kHz
2
(b) Current circulating in L and C at resonance,
ICIRC =
Hence,
V
V
=
= 2π fr C V
XC
 1 


 2f r C 
ICIRC = 2π(4109)(0.15 × 10-6)(10) = 0.03873 = 38.73 mA
(Alternatively, ICIRC =
10
V
V
=
=
= 38.73 mA)
XL
2f r L
2(4109)(0.01)
2. A 30 F capacitor is connected in parallel with a coil of inductance 50 mH and unknown
resistance R across a 120 V, 50 Hz supply. If the circuit has an overall power factor of 1 find
(a) the value of R, (b) the current in the coil, and (c) the supply current.
The circuit is shown below.
© John Bird Published by Taylor and Francis
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Since power factor = 1, then cos  = 1 and   cos 1 1  0 , hence the circuit is at resonance.
1  1 R2 
(a) Resonant frequency, f r 



2  LC L2 


1 
1
R2


50 =
2   50 103  30 106   50 103 2 


i.e.

R2 
2(50)   666.667 103 

0.0025 

and
 2  50    666.667 103  400R 2
2
i.e.
400R 2  666.667 103  2  50    567.97 103
2
Hence,
from which,
and
R2 
567.97 103
 1419.927
400
resistance, R = 1419.927 = 37.68 
(b) X L  2  50   50 103   15.71 
Hence,
ZLR  R 2  XL 2 
and current in the coil, I LR 
(c) Current at resonance, I r 
=
37.68
2
 15.712  = 40.824 
V
120

= 2.94 A
ZLR 40.824
V
V
VCR


L
RD
L
CR
120 37.7  30 10 6 
50 103
= 2.714 A
3. A coil of resistance 25  and inductance 150 mH inductance is connected in parallel with a
10 F capacitor across a 60 V, variable frequency supply. Calculate (a) the resonant frequency,
(b) the dynamic resistance, (c) the current at resonance and (d) the Q-factor at resonance.
© John Bird Published by Taylor and Francis
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The circuit is shown below.
(a) Resonant frequency, f r 
1  1 R2  1



2  LC L2  2
=
1
2


1
252



 150 103 10 106  150 103 2 


 666.667 10
(b) Dynamic resistance, R D 
L
150 103
= 600 

CR 10 106   25
(c) Current at resonance, I r 
V
60

= 0.10 A
R D 600
3
 27.778  103  = 127.2 Hz
3
2f r L 2 127.2 150  10 

(d) Q-factor =
= 4.80
R
25
4. A coil having resistance R and inductance 80 mH is connected in parallel with a 5 nF capacitor
across a 25 V, 3 kHz supply. Determine for the condition when the current is a minimum, (a) the
resistance R of the coil, (b) the dynamic resistance, (c) the supply current, and (d) the Q-factor.
(a) The supply current is a minimum when the parallel circuit is at resonance
1  1 R2 
and resonant frequency, fr =



2  LC L2 
Transposing for R gives: (2πfr)2 =
1
R2
- 2
LC L
1
R2
=
- (2πfr)2
2
LC
L
and
2
 1
R 2  L2 
  2f r  
 LC

© John Bird Published by Taylor and Francis
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  1
2 
R   L2 
  2f r   

  LC
from which,
When L = 80 mH, C = 5 109 F and fr = 3000 Hz,
then
resistance, R =


2 
1
2
 80 103  


2

3000


3
9
80

10
5

10


 





= 3704.9 Ω = 3.705 kΩ
(b) Dynamic resistance, R D 
L
80 103
= 4.318 k

CR  5 109  3.705 103 
(c) Supply current at resonance, I r 
V
25

= 5.79 mA
R D 4.318 103
2f r L 2  3000  80  10

(d) Q-factor =
R
3.705  103
3

= 0.41
5. A coil of resistance 1.5 k and 0.25 H inductance is connected in parallel with a variable
capacitance across a 10 V, 8 kHz supply. Calculate (a) the capacitance of the capacitor when the
supply current is a minimum, (b) the dynamic resistance, and (c) the supply current.
The circuit is shown below. Since the supply current is a minimum, the circuit is at resonance.
(a) Resonant frequency,
from which,
fr 
 2 f r  
2
1  1 R2 



2  LC L2 
1 R2

LC L2
© John Bird Published by Taylor and Francis
213
i.e.
 2 f r 
2

R2
1

2
L
LC
from which, capacitance, C =
1

R 
2
L  2 f r   2 
L 

2

1
2

15002 
0.25  2  8000   

0.252 

= 1561 pF
(b) Dynamic resistance, R D 
L
0.25

= 106.769 k or 106.8 k
CR 15611012 1.5 103 
(c) Current at resonance, I r 
V
10

= 93.66 A
R D 106.769 103
6. A parallel circuit as shown below is tuned to resonance by varying capacitance C.
Resistance, R = 30 , inductance, L = 400 H, and the supply voltage, V = 200 V, 5 MHz.
Calculate (a) the value of C to give resonance at 5 MHz, (b) the dynamic impedance, (c) the
Q-factor, (d) the bandwidth, (e) the current in each branch, (f) the supply current, and (g) the
power dissipated at resonance.
1  1 R2 
(a)Transposing f r 


 for C:
2  LC L2 
 2f r  
from which,
 1 R2 
 2 

 LC L 
 2f r 
2

R2
1

2
L
LC
and
and
 2f r 
2

1 R2

LC L2
1
 2f r 
2
R2
 2
L
 LC
© John Bird Published by Taylor and Francis
214
and
C=
i.e.
1

R 
2
L  2f r   2 
L 

2
1
2


30 



6 2
400 10  2 5 10  

2

 400 106  
6
capacitance, C = 2.533 pF
(b) Dynamic impedance, Z D =
(c) Q-factor =

L
400 106
= 5.264 M

CR  2.533 1012   30 
X L 2f r L 2(5 106 )(400 106 )


= 418.9
R
R
30
(d) Bandwidth,  f 2  f1  
f r 5 106
= 11.94 kHz

Q 418.9
(e) Capacitive current, IC =
V
2000

XC 

1

   90
 2  5 106  2.533 1012  


=
2000
12566.517  90
= 0.01591590 or 15.91590mA
Inductive branch current,
I LR =
=
V
V
2000


ZLR R  jX L 30  j  2  5 106  400 106  


2000
2000

30  j12566.3706 12566.4064 89.863 
= 15.915-89.863mA
(f) Supply current, I r =
Alternatively,
V
200

= 38 A
ZD 5.264 10 6
Ir  IC  ILR  15.91590  15.915  89.863 mA
= (0 + j15.915) + (0.038 – j15.915)
= 0.038 mA or 38 A
(g) Power, P = V I cos  = (200)(38 10 6 ) cos 0 = 7.60 mW
© John Bird Published by Taylor and Francis
215
Alternatively, P = IR 2 R  15.915 103  (30) = 7.60 mW
2
(Note that if the latter formula is used, the current must be the current flowing through the
resistor, i.e. I LR in the above circuit)
© John Bird Published by Taylor and Francis
216
Exercise 96, Page 270
1. A 415 V alternator is supplying a load of 55 kW at a power factor of 0.65lagging. Calculate (a) the
kVA loading and (b) the current taken from the alternator. (c) If the power factor is now raised to
unity find the new kVA loading.
(a) Power = VI cos  = (VI)(power factor)
Hence VI =
power
55 10 3
=
= 84615 VA = 84.6 kVA
p.f .
0.65
(b) VI = 84615 VA hence I =
84615 84615
=
= 203.9 A
V
415
(c) The kVA loading remains at 84.6 kVA irrespective of changes in power factor
2. A single phase motor takes 30 A at a power factor of 0.65 lagging from a 240 V, 50 Hz supply.
Determine (a) the current taken by the capacitor connected in parallel to correct the power factor
to unity, and (b) the value of the supply current after power factor correction.
The circuit diagram is shown below in (i) and the phasor diagram is shown in (ii).
(i)
(ii)
(a) Power factor = cos  = 0.65 from which, phase angle,  = cos 1 0.65 = 49.46
For unity power factor, current taken by capacitor, IC = ab in the phasor diagram
= 30 sin 49.46 = 22.80 A
(b) After power factor correction, new supply current = oa in the above phasor diagram
= 30 cos 49.46 = 19.50 A
© John Bird Published by Taylor and Francis
217
3. A 20  non-reactive resistor is connected in series with a coil of inductance 80 mH and
negligible resistance. The combined circuit is connected to a 200 V, 50 Hz supply. Calculate
(a) the reactance of the coil, (b) the impedance of the circuit, (c) the current in the circuit,
(d) the power factor of the circuit, (e) the power absorbed by the circuit, (f) the value of a power
factor correction capacitor to produce a power factor of unity, and (g) the value of a power
factor correction capacitor to produce a power factor of 0.9
(a) Inductive reactance, X L  2 f L  2  50  80  103  = 25.13 
(b) Circuit impedance, ZLR  R  jXL   20  j25.13  = 32.1251.49 
(c) Circuit current, I LR 
V
2000

= 6.227-51.49 A
ZLR 32.1251.49
(d) Power factor = cos  = cos 51.49 = 0.623
(e) Power, P = I LR 2 R   6.227   20  = 775.5 W
2
(f) The circuit including the capacitor is shown below in (i) and the phasor diagram in (ii).
(i)
(ii)
Power factor = cos  = 1 from which  = 0 Hence the new circuit current I is shown as oa in
diagram (ii).
Capacitor current, IC  ab  6.227sin 51.49 = 4.873 A. IC 
which,
capacitance, C =
V

XC
V
 2 f C V
1
2 f C
from
IC
4.873
= 77.56 F

2 f V 2 50  200 
© John Bird Published by Taylor and Francis
218
(g) Power factor = cos  = 0.9 from which,  = cos 1 0.9 = 25.84
The phasor diagram is shown in (iii) below where IC  bc  ab  ac
(iii)
ab = 4.873 A from part (f)
ac
ac

oa 3.877
tan 25.84 =
and oa = 6.227 cos 51.49 = 3.877 A
from which,
ac = 3.877 tan 25.84 = 1.878 A
Hence, bc = ab – ac = 4.873 – 1.878 = 2.995 A = IC
IC 
V

XC
V
 2 f C V
1
2 f C
capacitance, C =
from which,
IC
2.995
= 47.67 F

2 f V 2 50  200 
4. A motor has an output of 6 kW, an efficiency of 75% and a power factor of 0.64 lagging when
operated from a 250 V, 60 Hz supply. It is required to raise the power factor to 0.925 lagging by
connecting a capacitor in parallel with the motor. Determine (a) the current taken by the motor,
(b) the supply current after power factor correction, (c) the current taken by the capacitor, (d) the
capacitance of the capacitor, (d) the capacitance of the capacitor and (e) the kvar rating of the
capacitor.
(a) Efficiency =
from which,
Hence,
power output
power input
hence,
75
6000

100 power input
power input =
6000
= 8000 W
0.75
8000 = VIM cos    250 IM  0.64 
since cos  = power factor = 0.64
where   cos1 0.64  50.21
© John Bird Published by Taylor and Francis
219
Current taken by motor, IM 
8000
= 50 A
 250 0.64
(b) To improve the power factor to 0.925 lagging, the new phase angle = cos 1 0.925  22.33 as
shown in the phasor diagram below.
After power factor correction, the supply current is shown as I in the phasor diagram.
In triangle oab, oa = 50 cos 50.21 = 32 A and ab =
In triangle oac, cos 22.33 =
502  322 = 38.42 A by Pythagoras
oa 32
32

from which, current, I = oc =
= 34.59 A
oc oc
cos 22.33
(c) The current taken by the capacitor, IC , is given by length bc, where bc = ab – ac
ac = 34.59 sin 22.33 = 13.14
Hence, capacitor current, IC = bc = ab – ac = 38.42 – 13.14 = 25.28 A
(d) Capacitive reactance, X C 
i.e.
V
250

 9.89 
IC 25.28
1
1
1
 9.89 from which, capacitance, C =
= 268.2 F

2 f C
2 f  9.89  2  60  9.89 
(e) The kvar rating of the capacitor =
V IC  250  25.28 

= 6.32 kvar
1000
1000
5. A supply of 250 V, 80 Hz is connected across an inductive load and the power consumed is 2 kW,
when the supply current is 10 A. Determine the resistance and inductance of the circuit. What
value of capacitance connected in parallel with the load is needed to improve the overall power
factor to unity ?
© John Bird Published by Taylor and Francis
220
Power, P = I2 R hence, 2000 = (10)2 R from which, resistance, R =
Impedance, Z =
2000
= 20 Ω
10 2
V 250

= 25 Ω
I
10
Impedance, Z =
R 2  X L 2 from which, X L  Z2  R 2  252  202 = 15 Ω
X L  2 f L from which, inductance, L =
XL
15

= 29.84 mH
2f 2(80)
The circuit including the capacitor is shown below in (i) and the phasor diagram in (ii).
(i)
(ii)
Power factor = cos  = 1 from which  = 0 Hence the new circuit current I is shown as oa in
diagram (ii).
X 
 15 
  tan 1  L   tan 1   = 36.87º
 20 
 R 
Capacitor current, IC  ab  10sin 36.87 = 6 A. IC 
which,
capacitance, C =
V

XC
V
 2 f C V
1
2 f C
from
IC
6
= 47.75 F

2 f V 2 80  250 
6. A 200 V, 50 Hz single-phase supply feeds the following loads: (i) fluorescent lamps taking a
current of 8 A at a power factor of 0.9 leading, (ii) incandescent lamps taking a current of 6 A at
unity power factor, (iii) a motor taking a current of 12 A at a power factor of 0.65 lagging.
Determine the total current taken from the supply and the overall power factor. Find also the
value of a static capacitor connected in parallel with the loads to improve the overall power factor
to 0.98 lagging
© John Bird Published by Taylor and Francis
221
(i) Fluorescent lamps take a current of 8 A at a phase angle of cos 1 0.9 leading, i.e. 25.84 leading
(ii) Incandescent lamps take a current of 6 A at a phase angle of cos 1 1 , i.e. 0
(iii) Motor takes a current of 12 A at a phase angle of cos 1 0.65 lagging, i.e. 49.46 lagging
These currents are shown in the phasor diagram below.
Total horizontal component of currents = 8 cos 25.84 + 6 cos 0 + 12 cos 49.46
= 7.20 + 6.00 + 7.80 = 21.0 A
Total vertical component of currents = 8 sin 25.84 + 6 sin 0 - 12 sin 49.46
= 3.487 + 0 – 9.119 = - 5.632 A
Total supply current, I =
21.02  5.6322 = 21.74 A
 5.632 
at a phase angle of   tan 1 
 = 15.01 lagging
 21.0 
Hence, power factor = cos 15.01 = 0.966 lagging
If power factor is 0.98, then cos  = 0.98 and  = cos 1 0.98  11.48 
If a capacitor is connected in parallel with the loads, the capacitor current, IC , is as shown in the
phasor diagram below, where IC = length bc
From triangle oac, oa = horizontal component of I = 21.0 A
© John Bird Published by Taylor and Francis
222
and ac = vertical component of I = 21.72 sin 15.01 = 5.631 A
From triangle oab, tan 11.48 =
ab
ab

from which, ab = 21.0tan11.48 = 4.265 A
oa 21.0
Hence, capacitor current, IC = ac – ab = 5.631 – 4.265 = 1.366 A
IC 
V

XC
IC
1.366
V
= 21.74 F
 2 f C V from which, capacitance, C =

1
2 f V 2 50  200 
2 f C
© John Bird Published by Taylor and Francis
223