Solution of Current in AC Parallel & Series Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) PDF Download

 Page 1


In the last lesson, the following points were described: 
1. How to compute the total impedance/admittance in series/parallel circuits? 
2. How to solve for the current(s) in series/parallel circuits, fed from single phase ac 
supply, and then draw complete phasor diagram? 
3. How to find the power consumed in the circuit and also the different components, and 
the power factor (lag/lead)? 
 In this lesson, the computation of impedance/admittance in parallel and series-parallel 
circuits, fed from single phase ac supply, is presented. Then, the currents, both in 
magnitude and phase, are calculated. The process of drawing complete phasor diagram is 
described. The computation of total power and also power consumed in the different 
components, along with power factor, is explained. Some examples, of both parallel and 
series-parallel circuits, are presented in detail.  
Keywords: Parallel and series-parallel circuits, impedance, admittance, power, power 
factor. 
After going through this lesson, the students will be able to answer the following 
questions; 
1. How to compute the impedance/admittance, of the parallel and series-parallel circuits, 
fed from single phase ac supply?  
2. How to compute the different currents and also voltage drops in the components, both 
in magnitude and phase, of the circuit?  
3. How to draw the complete phasor diagram, showing the currents and voltage drops?  
4. How to compute the total power and also power consumed in the different 
components, along with power factor? 
This lesson starts with two examples of parallel circuits fed from single phase ac 
supply. The first example is presented in detail. The students are advised to study the two 
cases of parallel circuits given in the previous lesson. 
Example 16.1 
The circuit, having two impedances of O + = ) 15 8 (
1
j Z and O - = ) 8 6 (
2
j Z in 
parallel, is connected to a single phase ac supply (Fig. 16.1a), and the current drawn is 10 
A. Find each branch current, both in magnitude and phase, and also the supply voltage. 
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur 
Page 2


In the last lesson, the following points were described: 
1. How to compute the total impedance/admittance in series/parallel circuits? 
2. How to solve for the current(s) in series/parallel circuits, fed from single phase ac 
supply, and then draw complete phasor diagram? 
3. How to find the power consumed in the circuit and also the different components, and 
the power factor (lag/lead)? 
 In this lesson, the computation of impedance/admittance in parallel and series-parallel 
circuits, fed from single phase ac supply, is presented. Then, the currents, both in 
magnitude and phase, are calculated. The process of drawing complete phasor diagram is 
described. The computation of total power and also power consumed in the different 
components, along with power factor, is explained. Some examples, of both parallel and 
series-parallel circuits, are presented in detail.  
Keywords: Parallel and series-parallel circuits, impedance, admittance, power, power 
factor. 
After going through this lesson, the students will be able to answer the following 
questions; 
1. How to compute the impedance/admittance, of the parallel and series-parallel circuits, 
fed from single phase ac supply?  
2. How to compute the different currents and also voltage drops in the components, both 
in magnitude and phase, of the circuit?  
3. How to draw the complete phasor diagram, showing the currents and voltage drops?  
4. How to compute the total power and also power consumed in the different 
components, along with power factor? 
This lesson starts with two examples of parallel circuits fed from single phase ac 
supply. The first example is presented in detail. The students are advised to study the two 
cases of parallel circuits given in the previous lesson. 
Example 16.1 
The circuit, having two impedances of O + = ) 15 8 (
1
j Z and O - = ) 8 6 (
2
j Z in 
parallel, is connected to a single phase ac supply (Fig. 16.1a), and the current drawn is 10 
A. Find each branch current, both in magnitude and phase, and also the supply voltage. 
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur 
B 
I = 10A
Fig. 16.1 (a) Circuit diagram
A 
Z
2
 = (6 – j8) ? 
 Z
1
 = (8 + j15) ? 
I
1 
I
2 
 
 
 
 
 
 
 
 
 
Solution 
O ° ? = + = ? 93 . 61 17 ) 15 8 (
1 1
j Z f O ° - ? = - = - ? 13 . 53 10 ) 8 6 (
2 2
j Z f  
A j OC I ) 0 10 ( 0 10 ) ( 0 + = ° ? = ° ?   
The admittances, using impedances in rectangular form, are,  
1 3
2 2
1 1
1 1
10 ) 9 . 51 68 . 27 (
289
15 8
15 8
15 8
15 8
1 1
- -
O · - =
-
=
+
-
=
+
=
?
= - ? j
j j
j Z
Y
f
f 
1 3
2 2
2 2
2 2
10 ) 0 . 80 0 . 60 (
100
8 6
8 6
8 6
8 6
1 1
- -
O · + =
+
=
+
+
=
-
=
- ?
= ? j
j j
j Z
Y
f
f 
Alternatively, using impedances in polar form, the admittances are,  
1 3
1 1
1 1
10 ) 9 . 51 68 . 27 (
93 . 61 05882 . 0
93 . 61 0 . 17
1 1
- -
O · - =
° - ? =
° ?
=
?
= - ?
j
Z
Y
f
f
 
1 3
2 2
2 2
10 ) 0 . 80 0 . 60 ( 13 . 53 1 . 0
13 . 53 0 . 10
1 1
- -
O · + = ° ? =
° - ?
=
- ?
= ? j
Z
Y
f
f 
The total admittance is, 
3 3
2 1
10 ) 1 . 28 68 . 87 ( 10 )] 0 . 80 0 . 60 ( ) 9 . 51 68 . 27 [(
- -
· + = · + + - = + = ? j j j Y Y Y f 
1 3
77 . 17 10 07 . 92
- -
O ° ? · = 
The total impedance is, 
O - = ° - ? =
° ? ·
=
?
= - ?
-
) 315 . 3 343 . 10 ( 77 . 17 86 . 10
77 . 17 10 07 . 92
1 1
3
j
Y
Z
f
f 
The supply voltage is  
V Z I V V
AB
° - ? = ° - ? × = - ? · ° ? = - ? 77 . 17 6 . 108 77 . 17 ) 86 . 10 10 ( 0 ) ( f f 
V j ) 15 . 33 43 . 103 ( - = 
 
The branch currents are, 
A
Z
V
OD I ° - ? = ° + ° - ?
?
?
?
?
?
?
=
?
- ?
= - ? 7 . 79 39 . 6 ) 93 . 61 77 . 17 (
0 . 17
6 . 108
) (
1 1
1 1
f
f
? 
Version 2 EE IIT, Kharagpur 
Page 3


In the last lesson, the following points were described: 
1. How to compute the total impedance/admittance in series/parallel circuits? 
2. How to solve for the current(s) in series/parallel circuits, fed from single phase ac 
supply, and then draw complete phasor diagram? 
3. How to find the power consumed in the circuit and also the different components, and 
the power factor (lag/lead)? 
 In this lesson, the computation of impedance/admittance in parallel and series-parallel 
circuits, fed from single phase ac supply, is presented. Then, the currents, both in 
magnitude and phase, are calculated. The process of drawing complete phasor diagram is 
described. The computation of total power and also power consumed in the different 
components, along with power factor, is explained. Some examples, of both parallel and 
series-parallel circuits, are presented in detail.  
Keywords: Parallel and series-parallel circuits, impedance, admittance, power, power 
factor. 
After going through this lesson, the students will be able to answer the following 
questions; 
1. How to compute the impedance/admittance, of the parallel and series-parallel circuits, 
fed from single phase ac supply?  
2. How to compute the different currents and also voltage drops in the components, both 
in magnitude and phase, of the circuit?  
3. How to draw the complete phasor diagram, showing the currents and voltage drops?  
4. How to compute the total power and also power consumed in the different 
components, along with power factor? 
This lesson starts with two examples of parallel circuits fed from single phase ac 
supply. The first example is presented in detail. The students are advised to study the two 
cases of parallel circuits given in the previous lesson. 
Example 16.1 
The circuit, having two impedances of O + = ) 15 8 (
1
j Z and O - = ) 8 6 (
2
j Z in 
parallel, is connected to a single phase ac supply (Fig. 16.1a), and the current drawn is 10 
A. Find each branch current, both in magnitude and phase, and also the supply voltage. 
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur 
B 
I = 10A
Fig. 16.1 (a) Circuit diagram
A 
Z
2
 = (6 – j8) ? 
 Z
1
 = (8 + j15) ? 
I
1 
I
2 
 
 
 
 
 
 
 
 
 
Solution 
O ° ? = + = ? 93 . 61 17 ) 15 8 (
1 1
j Z f O ° - ? = - = - ? 13 . 53 10 ) 8 6 (
2 2
j Z f  
A j OC I ) 0 10 ( 0 10 ) ( 0 + = ° ? = ° ?   
The admittances, using impedances in rectangular form, are,  
1 3
2 2
1 1
1 1
10 ) 9 . 51 68 . 27 (
289
15 8
15 8
15 8
15 8
1 1
- -
O · - =
-
=
+
-
=
+
=
?
= - ? j
j j
j Z
Y
f
f 
1 3
2 2
2 2
2 2
10 ) 0 . 80 0 . 60 (
100
8 6
8 6
8 6
8 6
1 1
- -
O · + =
+
=
+
+
=
-
=
- ?
= ? j
j j
j Z
Y
f
f 
Alternatively, using impedances in polar form, the admittances are,  
1 3
1 1
1 1
10 ) 9 . 51 68 . 27 (
93 . 61 05882 . 0
93 . 61 0 . 17
1 1
- -
O · - =
° - ? =
° ?
=
?
= - ?
j
Z
Y
f
f
 
1 3
2 2
2 2
10 ) 0 . 80 0 . 60 ( 13 . 53 1 . 0
13 . 53 0 . 10
1 1
- -
O · + = ° ? =
° - ?
=
- ?
= ? j
Z
Y
f
f 
The total admittance is, 
3 3
2 1
10 ) 1 . 28 68 . 87 ( 10 )] 0 . 80 0 . 60 ( ) 9 . 51 68 . 27 [(
- -
· + = · + + - = + = ? j j j Y Y Y f 
1 3
77 . 17 10 07 . 92
- -
O ° ? · = 
The total impedance is, 
O - = ° - ? =
° ? ·
=
?
= - ?
-
) 315 . 3 343 . 10 ( 77 . 17 86 . 10
77 . 17 10 07 . 92
1 1
3
j
Y
Z
f
f 
The supply voltage is  
V Z I V V
AB
° - ? = ° - ? × = - ? · ° ? = - ? 77 . 17 6 . 108 77 . 17 ) 86 . 10 10 ( 0 ) ( f f 
V j ) 15 . 33 43 . 103 ( - = 
 
The branch currents are, 
A
Z
V
OD I ° - ? = ° + ° - ?
?
?
?
?
?
?
=
?
- ?
= - ? 7 . 79 39 . 6 ) 93 . 61 77 . 17 (
0 . 17
6 . 108
) (
1 1
1 1
f
f
? 
Version 2 EE IIT, Kharagpur 
A j ) 286 . 6 143 . 1 ( - =  
22 1 1
() 0 ( )
(10.0 0.0) (1.143 6.286) (8.857 6.286) 10.86 35.36
I OE I I OCOD OCCE
j jjA
?? ?=?°-?- - = -
=+ - - = + = ? °A
  
Alternatively, the current  is, 
2
I
22
22
108.6
( ) ( 17.77 53.13 ) 10.86 35.36
10.0
V
I OE A
Z
f
?
f
?- ??
?= = ?- °+ °= ?
??
?-
??
°  
A j ) 285 . 6 857 . 8 ( + = 
The phasor diagram with the total (input) current as reference is shown in Fig. 16.1b. 
 
Fig. 16.1 (b) Phasor diagram
53.13° 
I
2
= 10.86
61.90 
108.63 V 
? = 17.8° 
    I
1
 = 6.4 A 
D 
?
1
 = 79.7° 
?
1
 = 35.35 
O 
E 
V
AB
C 
I = 10A
 
 
 
 
 
 
 
 
 
 
 
 
Alternative Method 
O ° ? = + = - + + = - ? + ? = ' ? ' 565 . 26 65 . 15 ) 7 14 ( ) 8 6 ( ) 15 8 (
2 2 1 1
j j j Z Z Z f f f 
O - = ° - ? =
° - ° - ° ? ?
?
?
?
?
? ×
=
' ? '
- ? · ?
=
+
·
= - ?
) 315 . 3 343 . 10 ( 77 . 17 86 . 10
) 565 . 26 13 . 53 93 . 61 (
65 . 15
0 . 10 0 . 17
2 2 1 1
2 1
2 1
j
Z
Z Z
Z Z
Z Z
Z
f
f f
f
 
The supply voltage is  
( ) (10 10.86) 17.77 108.6 17.77
(103.43 33.15)
AB
VV IZ
jV
f ?- = · = × ?- ° = ?- °
=-
V
 
 
The branch currents are, 
2
11
12
10.0 10.0
( ) ( 53.13 26.565 ) 6.39 79.7
15.65
(1.142 6.286)
× ??
?- = = ? - °- ° = ?- °
??
+
??
=-
Z
I OD I A
ZZ
jA
?
 
Version 2 EE IIT, Kharagpur 
Page 4


In the last lesson, the following points were described: 
1. How to compute the total impedance/admittance in series/parallel circuits? 
2. How to solve for the current(s) in series/parallel circuits, fed from single phase ac 
supply, and then draw complete phasor diagram? 
3. How to find the power consumed in the circuit and also the different components, and 
the power factor (lag/lead)? 
 In this lesson, the computation of impedance/admittance in parallel and series-parallel 
circuits, fed from single phase ac supply, is presented. Then, the currents, both in 
magnitude and phase, are calculated. The process of drawing complete phasor diagram is 
described. The computation of total power and also power consumed in the different 
components, along with power factor, is explained. Some examples, of both parallel and 
series-parallel circuits, are presented in detail.  
Keywords: Parallel and series-parallel circuits, impedance, admittance, power, power 
factor. 
After going through this lesson, the students will be able to answer the following 
questions; 
1. How to compute the impedance/admittance, of the parallel and series-parallel circuits, 
fed from single phase ac supply?  
2. How to compute the different currents and also voltage drops in the components, both 
in magnitude and phase, of the circuit?  
3. How to draw the complete phasor diagram, showing the currents and voltage drops?  
4. How to compute the total power and also power consumed in the different 
components, along with power factor? 
This lesson starts with two examples of parallel circuits fed from single phase ac 
supply. The first example is presented in detail. The students are advised to study the two 
cases of parallel circuits given in the previous lesson. 
Example 16.1 
The circuit, having two impedances of O + = ) 15 8 (
1
j Z and O - = ) 8 6 (
2
j Z in 
parallel, is connected to a single phase ac supply (Fig. 16.1a), and the current drawn is 10 
A. Find each branch current, both in magnitude and phase, and also the supply voltage. 
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur 
B 
I = 10A
Fig. 16.1 (a) Circuit diagram
A 
Z
2
 = (6 – j8) ? 
 Z
1
 = (8 + j15) ? 
I
1 
I
2 
 
 
 
 
 
 
 
 
 
Solution 
O ° ? = + = ? 93 . 61 17 ) 15 8 (
1 1
j Z f O ° - ? = - = - ? 13 . 53 10 ) 8 6 (
2 2
j Z f  
A j OC I ) 0 10 ( 0 10 ) ( 0 + = ° ? = ° ?   
The admittances, using impedances in rectangular form, are,  
1 3
2 2
1 1
1 1
10 ) 9 . 51 68 . 27 (
289
15 8
15 8
15 8
15 8
1 1
- -
O · - =
-
=
+
-
=
+
=
?
= - ? j
j j
j Z
Y
f
f 
1 3
2 2
2 2
2 2
10 ) 0 . 80 0 . 60 (
100
8 6
8 6
8 6
8 6
1 1
- -
O · + =
+
=
+
+
=
-
=
- ?
= ? j
j j
j Z
Y
f
f 
Alternatively, using impedances in polar form, the admittances are,  
1 3
1 1
1 1
10 ) 9 . 51 68 . 27 (
93 . 61 05882 . 0
93 . 61 0 . 17
1 1
- -
O · - =
° - ? =
° ?
=
?
= - ?
j
Z
Y
f
f
 
1 3
2 2
2 2
10 ) 0 . 80 0 . 60 ( 13 . 53 1 . 0
13 . 53 0 . 10
1 1
- -
O · + = ° ? =
° - ?
=
- ?
= ? j
Z
Y
f
f 
The total admittance is, 
3 3
2 1
10 ) 1 . 28 68 . 87 ( 10 )] 0 . 80 0 . 60 ( ) 9 . 51 68 . 27 [(
- -
· + = · + + - = + = ? j j j Y Y Y f 
1 3
77 . 17 10 07 . 92
- -
O ° ? · = 
The total impedance is, 
O - = ° - ? =
° ? ·
=
?
= - ?
-
) 315 . 3 343 . 10 ( 77 . 17 86 . 10
77 . 17 10 07 . 92
1 1
3
j
Y
Z
f
f 
The supply voltage is  
V Z I V V
AB
° - ? = ° - ? × = - ? · ° ? = - ? 77 . 17 6 . 108 77 . 17 ) 86 . 10 10 ( 0 ) ( f f 
V j ) 15 . 33 43 . 103 ( - = 
 
The branch currents are, 
A
Z
V
OD I ° - ? = ° + ° - ?
?
?
?
?
?
?
=
?
- ?
= - ? 7 . 79 39 . 6 ) 93 . 61 77 . 17 (
0 . 17
6 . 108
) (
1 1
1 1
f
f
? 
Version 2 EE IIT, Kharagpur 
A j ) 286 . 6 143 . 1 ( - =  
22 1 1
() 0 ( )
(10.0 0.0) (1.143 6.286) (8.857 6.286) 10.86 35.36
I OE I I OCOD OCCE
j jjA
?? ?=?°-?- - = -
=+ - - = + = ? °A
  
Alternatively, the current  is, 
2
I
22
22
108.6
( ) ( 17.77 53.13 ) 10.86 35.36
10.0
V
I OE A
Z
f
?
f
?- ??
?= = ?- °+ °= ?
??
?-
??
°  
A j ) 285 . 6 857 . 8 ( + = 
The phasor diagram with the total (input) current as reference is shown in Fig. 16.1b. 
 
Fig. 16.1 (b) Phasor diagram
53.13° 
I
2
= 10.86
61.90 
108.63 V 
? = 17.8° 
    I
1
 = 6.4 A 
D 
?
1
 = 79.7° 
?
1
 = 35.35 
O 
E 
V
AB
C 
I = 10A
 
 
 
 
 
 
 
 
 
 
 
 
Alternative Method 
O ° ? = + = - + + = - ? + ? = ' ? ' 565 . 26 65 . 15 ) 7 14 ( ) 8 6 ( ) 15 8 (
2 2 1 1
j j j Z Z Z f f f 
O - = ° - ? =
° - ° - ° ? ?
?
?
?
?
? ×
=
' ? '
- ? · ?
=
+
·
= - ?
) 315 . 3 343 . 10 ( 77 . 17 86 . 10
) 565 . 26 13 . 53 93 . 61 (
65 . 15
0 . 10 0 . 17
2 2 1 1
2 1
2 1
j
Z
Z Z
Z Z
Z Z
Z
f
f f
f
 
The supply voltage is  
( ) (10 10.86) 17.77 108.6 17.77
(103.43 33.15)
AB
VV IZ
jV
f ?- = · = × ?- ° = ?- °
=-
V
 
 
The branch currents are, 
2
11
12
10.0 10.0
( ) ( 53.13 26.565 ) 6.39 79.7
15.65
(1.142 6.286)
× ??
?- = = ? - °- ° = ?- °
??
+
??
=-
Z
I OD I A
ZZ
jA
?
 
Version 2 EE IIT, Kharagpur 
22 1
( ) ( ) (10.0 0.0) (1.142 6.286)
(8.858 6.286) 10.862 35.36
I OE I I OCOD OCCE j j
jA A
??=- - = - = + - -
=+ = ? °
 
Alternatively, the current  is, 
2
I
1
22
12
10.0 17.0
( ) (61.93 26.565 ) 10.86 35..36
15.65
(8.858 6.286)
Z
I OE I A
ZZ
jA
?
× ??
?= = ? °- °= ?
??
+
??
=+
°
 
Example 16.2 
The power consumed in the inductive load (Fig. 16.2a) is 2.5 kW at 0.71 lagging 
power factor (pf). The input voltage is 230 V, 50 Hz. Find the value of the capacitor C, 
such that the resultant power factor of the input current is 0.866 lagging.  
 
 
+ 
- 
C 
Fig. 16.2 (a) Circuit diagram
I 
I
L
I
C
230 V 
L 
O 
A 
D 
 
 
 
 
 
 
 
Solution 
W KW P 2500 10 5 . 2 5 . 2
3
= · = = V = 230 V   = 50 Hz f
The power factor in the inductive branch is ) ( 71 . 0 cos lag
L
= f 
The phase angle is  ° ˜ ° = =
-
45 77 . 44 ) 71 . 0 ( cos
1
L
f
() 2500 cos 230 cos = · = · =
L L L L
I I V P f f 
A
V
P
I
L
L
31 . 15
71 . 0 230
2500
cos
=
×
= =
f
 
A I A I
L L L L
87 . 10 45 sin 32 . 15 sin ; 87 . 10 71 . 0 31 . 15 cos = ° × = = × = f f 
The current  is, 
L
I A j I
L L
) 87 . 10 87 . 10 ( 45 31 . 15 - = ° - ? = - ? f 
The power consumed in the circuit remains same, as the capacitor does not consume 
any power, but the reactive power in the circuit changes. The active component of the 
total current remains same as computed earlier. 
A I I
L L
87 . 10 cos cos = = f f 
The power factor of the current is cos ) ( 866 . 0 lag = f 
The phase angle is  ° = =
-
30 ) 866 . 0 ( cos
1
f
The magnitude of the current is A I 55 . 12 866 . 0 / 87 . 10 = = 
The current is A j I ) 276 . 6 87 . 10 ( 30 55 . 12 - = ° - ? = - ? f 
Version 2 EE IIT, Kharagpur 
Page 5


In the last lesson, the following points were described: 
1. How to compute the total impedance/admittance in series/parallel circuits? 
2. How to solve for the current(s) in series/parallel circuits, fed from single phase ac 
supply, and then draw complete phasor diagram? 
3. How to find the power consumed in the circuit and also the different components, and 
the power factor (lag/lead)? 
 In this lesson, the computation of impedance/admittance in parallel and series-parallel 
circuits, fed from single phase ac supply, is presented. Then, the currents, both in 
magnitude and phase, are calculated. The process of drawing complete phasor diagram is 
described. The computation of total power and also power consumed in the different 
components, along with power factor, is explained. Some examples, of both parallel and 
series-parallel circuits, are presented in detail.  
Keywords: Parallel and series-parallel circuits, impedance, admittance, power, power 
factor. 
After going through this lesson, the students will be able to answer the following 
questions; 
1. How to compute the impedance/admittance, of the parallel and series-parallel circuits, 
fed from single phase ac supply?  
2. How to compute the different currents and also voltage drops in the components, both 
in magnitude and phase, of the circuit?  
3. How to draw the complete phasor diagram, showing the currents and voltage drops?  
4. How to compute the total power and also power consumed in the different 
components, along with power factor? 
This lesson starts with two examples of parallel circuits fed from single phase ac 
supply. The first example is presented in detail. The students are advised to study the two 
cases of parallel circuits given in the previous lesson. 
Example 16.1 
The circuit, having two impedances of O + = ) 15 8 (
1
j Z and O - = ) 8 6 (
2
j Z in 
parallel, is connected to a single phase ac supply (Fig. 16.1a), and the current drawn is 10 
A. Find each branch current, both in magnitude and phase, and also the supply voltage. 
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur 
B 
I = 10A
Fig. 16.1 (a) Circuit diagram
A 
Z
2
 = (6 – j8) ? 
 Z
1
 = (8 + j15) ? 
I
1 
I
2 
 
 
 
 
 
 
 
 
 
Solution 
O ° ? = + = ? 93 . 61 17 ) 15 8 (
1 1
j Z f O ° - ? = - = - ? 13 . 53 10 ) 8 6 (
2 2
j Z f  
A j OC I ) 0 10 ( 0 10 ) ( 0 + = ° ? = ° ?   
The admittances, using impedances in rectangular form, are,  
1 3
2 2
1 1
1 1
10 ) 9 . 51 68 . 27 (
289
15 8
15 8
15 8
15 8
1 1
- -
O · - =
-
=
+
-
=
+
=
?
= - ? j
j j
j Z
Y
f
f 
1 3
2 2
2 2
2 2
10 ) 0 . 80 0 . 60 (
100
8 6
8 6
8 6
8 6
1 1
- -
O · + =
+
=
+
+
=
-
=
- ?
= ? j
j j
j Z
Y
f
f 
Alternatively, using impedances in polar form, the admittances are,  
1 3
1 1
1 1
10 ) 9 . 51 68 . 27 (
93 . 61 05882 . 0
93 . 61 0 . 17
1 1
- -
O · - =
° - ? =
° ?
=
?
= - ?
j
Z
Y
f
f
 
1 3
2 2
2 2
10 ) 0 . 80 0 . 60 ( 13 . 53 1 . 0
13 . 53 0 . 10
1 1
- -
O · + = ° ? =
° - ?
=
- ?
= ? j
Z
Y
f
f 
The total admittance is, 
3 3
2 1
10 ) 1 . 28 68 . 87 ( 10 )] 0 . 80 0 . 60 ( ) 9 . 51 68 . 27 [(
- -
· + = · + + - = + = ? j j j Y Y Y f 
1 3
77 . 17 10 07 . 92
- -
O ° ? · = 
The total impedance is, 
O - = ° - ? =
° ? ·
=
?
= - ?
-
) 315 . 3 343 . 10 ( 77 . 17 86 . 10
77 . 17 10 07 . 92
1 1
3
j
Y
Z
f
f 
The supply voltage is  
V Z I V V
AB
° - ? = ° - ? × = - ? · ° ? = - ? 77 . 17 6 . 108 77 . 17 ) 86 . 10 10 ( 0 ) ( f f 
V j ) 15 . 33 43 . 103 ( - = 
 
The branch currents are, 
A
Z
V
OD I ° - ? = ° + ° - ?
?
?
?
?
?
?
=
?
- ?
= - ? 7 . 79 39 . 6 ) 93 . 61 77 . 17 (
0 . 17
6 . 108
) (
1 1
1 1
f
f
? 
Version 2 EE IIT, Kharagpur 
A j ) 286 . 6 143 . 1 ( - =  
22 1 1
() 0 ( )
(10.0 0.0) (1.143 6.286) (8.857 6.286) 10.86 35.36
I OE I I OCOD OCCE
j jjA
?? ?=?°-?- - = -
=+ - - = + = ? °A
  
Alternatively, the current  is, 
2
I
22
22
108.6
( ) ( 17.77 53.13 ) 10.86 35.36
10.0
V
I OE A
Z
f
?
f
?- ??
?= = ?- °+ °= ?
??
?-
??
°  
A j ) 285 . 6 857 . 8 ( + = 
The phasor diagram with the total (input) current as reference is shown in Fig. 16.1b. 
 
Fig. 16.1 (b) Phasor diagram
53.13° 
I
2
= 10.86
61.90 
108.63 V 
? = 17.8° 
    I
1
 = 6.4 A 
D 
?
1
 = 79.7° 
?
1
 = 35.35 
O 
E 
V
AB
C 
I = 10A
 
 
 
 
 
 
 
 
 
 
 
 
Alternative Method 
O ° ? = + = - + + = - ? + ? = ' ? ' 565 . 26 65 . 15 ) 7 14 ( ) 8 6 ( ) 15 8 (
2 2 1 1
j j j Z Z Z f f f 
O - = ° - ? =
° - ° - ° ? ?
?
?
?
?
? ×
=
' ? '
- ? · ?
=
+
·
= - ?
) 315 . 3 343 . 10 ( 77 . 17 86 . 10
) 565 . 26 13 . 53 93 . 61 (
65 . 15
0 . 10 0 . 17
2 2 1 1
2 1
2 1
j
Z
Z Z
Z Z
Z Z
Z
f
f f
f
 
The supply voltage is  
( ) (10 10.86) 17.77 108.6 17.77
(103.43 33.15)
AB
VV IZ
jV
f ?- = · = × ?- ° = ?- °
=-
V
 
 
The branch currents are, 
2
11
12
10.0 10.0
( ) ( 53.13 26.565 ) 6.39 79.7
15.65
(1.142 6.286)
× ??
?- = = ? - °- ° = ?- °
??
+
??
=-
Z
I OD I A
ZZ
jA
?
 
Version 2 EE IIT, Kharagpur 
22 1
( ) ( ) (10.0 0.0) (1.142 6.286)
(8.858 6.286) 10.862 35.36
I OE I I OCOD OCCE j j
jA A
??=- - = - = + - -
=+ = ? °
 
Alternatively, the current  is, 
2
I
1
22
12
10.0 17.0
( ) (61.93 26.565 ) 10.86 35..36
15.65
(8.858 6.286)
Z
I OE I A
ZZ
jA
?
× ??
?= = ? °- °= ?
??
+
??
=+
°
 
Example 16.2 
The power consumed in the inductive load (Fig. 16.2a) is 2.5 kW at 0.71 lagging 
power factor (pf). The input voltage is 230 V, 50 Hz. Find the value of the capacitor C, 
such that the resultant power factor of the input current is 0.866 lagging.  
 
 
+ 
- 
C 
Fig. 16.2 (a) Circuit diagram
I 
I
L
I
C
230 V 
L 
O 
A 
D 
 
 
 
 
 
 
 
Solution 
W KW P 2500 10 5 . 2 5 . 2
3
= · = = V = 230 V   = 50 Hz f
The power factor in the inductive branch is ) ( 71 . 0 cos lag
L
= f 
The phase angle is  ° ˜ ° = =
-
45 77 . 44 ) 71 . 0 ( cos
1
L
f
() 2500 cos 230 cos = · = · =
L L L L
I I V P f f 
A
V
P
I
L
L
31 . 15
71 . 0 230
2500
cos
=
×
= =
f
 
A I A I
L L L L
87 . 10 45 sin 32 . 15 sin ; 87 . 10 71 . 0 31 . 15 cos = ° × = = × = f f 
The current  is, 
L
I A j I
L L
) 87 . 10 87 . 10 ( 45 31 . 15 - = ° - ? = - ? f 
The power consumed in the circuit remains same, as the capacitor does not consume 
any power, but the reactive power in the circuit changes. The active component of the 
total current remains same as computed earlier. 
A I I
L L
87 . 10 cos cos = = f f 
The power factor of the current is cos ) ( 866 . 0 lag = f 
The phase angle is  ° = =
-
30 ) 866 . 0 ( cos
1
f
The magnitude of the current is A I 55 . 12 866 . 0 / 87 . 10 = = 
The current is A j I ) 276 . 6 87 . 10 ( 30 55 . 12 - = ° - ? = - ? f 
Version 2 EE IIT, Kharagpur 
The current in the capacitor is  
A j
j j I I I
L L C
° ? = =
- - - = - ? - - ? = ° ?
90 504 . 4 504 . 4
) 87 . 10 87 . 10 ( ) 276 . 6 87 . 10 ( 90 f f
 
This current is the difference of two reactive currents, 
 A I I I
L L C
504 . 4 87 . 10 276 . 6 sin sin - = - = - = - f f  
The reactance of the capacitor, C is O = = = = 066 . 51
504 . 4
230
2
1
C
C
I
V
C f
X
p
 
The capacitor, C is F
X f
C
C
µ
p p
33 . 62 10 33 . 62
066 . 51 50 2
1
2
1
6
= · =
× ×
= =
-
 
The phasor diagram with the input voltage as reference is shown in Fig. 16.2b. 
 
230 V
C 
15.3  
A 
I
C
4.5 A 
f
45° 
Fig. 16.2 (b) Phasor diagram 
B 
I
C
 
 
 
 
 
 
 
 
 
 
Example 16.3 
An inductive load (R in series with L) is connected in parallel with a capacitance C of 
12.5 F µ (Fig. 16.3a). The input voltage to the circuit is 100 V at 31.8 Hz. The phase 
angle between the two branch currents, (
L
I I =
1
) and (
C
I I =
2
) is , and the current 
in the first branch is  . Find the total current, and also the values of R & L.  
° 120
A I I
L
5 . 0
1
= =
 
A 
B 
C = 12.5 µF 
I
2
I 
R 
D 
L 
+ 
- 
100 V 
Fig. 16.3 (a) Circuit diagram 
1
=0.5A I
 
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur