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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
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FAQs on Solution of Current in AC Parallel & Series Parallel Circuits - Basic Electrical Technology - Electrical Engineering (EE)
| 1. How do I calculate the total current in an AC parallel circuit? |  |
Ans. To calculate the total current in an AC parallel circuit, you need to add up the individual currents flowing through each branch. The total current is the sum of the currents in each branch. This can be calculated using Ohm's Law, where the total current (I) is equal to the sum of the currents in each branch (I1, I2, I3, ...), given by the equation I = I1 + I2 + I3 + ...
| 2. What happens to the total current in an AC series parallel circuit when a new branch is added? | |
Ans. When a new branch is added to an AC series parallel circuit, the total current will increase. This is because the total resistance of the circuit decreases when a new branch is added, which in turn decreases the total impedance. According to Ohm's Law, as the impedance decreases, the total current increases.
| 3. How does the total current behave in an AC series parallel circuit when the frequency of the AC source is changed? | |
Ans. In an AC series parallel circuit, when the frequency of the AC source is changed, the total current can be affected. At higher frequencies, the reactance of inductors increases, while the reactance of capacitors decreases. This can lead to changes in the total impedance of the circuit, resulting in a change in the total current. It is important to note that the behavior of the total current will depend on the specific values of the inductors, capacitors, and resistors in the circuit.
| 4. How can I determine the total current in a complex AC series parallel circuit? | |
Ans. To determine the total current in a complex AC series parallel circuit, you can use Kirchhoff's laws. First, analyze the circuit to identify series and parallel combinations of components. Then, apply Ohm's Law to calculate the currents in each individual branch. Finally, use Kirchhoff's current law at different nodes to determine the total current. By applying these principles, you can determine the total current in a complex AC series parallel circuit.
| 5. What factors can impact the total current in an AC series parallel circuit? | |
Ans. Several factors can impact the total current in an AC series parallel circuit. These include the values of resistors, inductors, and capacitors in the circuit, as well as the frequency of the AC source. The total current is also influenced by the arrangement of components in series and parallel combinations. Additionally, the presence of any other external circuit elements or loads connected to the circuit can also affect the total current.
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