Practice Questions: Network Theorems in AC Circuits | Network Theory (Electric Circuits) - Electrical Engineering (EE) PDF Download

 Page 1


 
Network Theorems - Alternating Current examples   
 
In the previous chapter, we have been dealing mainly with direct current resistive circuits in 
order to the principles of the various theorems clear.  As was mentioned, these theories are 
equally valid for a.c.  
 
Example 1 
 
 
 
 
 
 
For the circuit  shown in the figure, if the frequency of the supply is 50 Hz, determine using 
Ohm’s Law and Kirchoff’s Laws the current I in the 160 ? capacitor. 
Impedance of capacitor and inductor at 50 Hz are 
 j X
L
  = j 2p × 50 × 63.66 × 10
-3
 = j 20 ?  
and - j X
C
 = 1/( j 2p × 50 × 19.89 × 10
-6
) = - j 160 ? 
Using Kirchoff’s current law 
 I  =  I
1
  - I
2  
 
Using Kirchoff’s voltage law 
 100?0
o
 = 20 I
1
  – j160 (I
1
 – I
2
) ? 5 = (1-j8) I
1
 +j8 I
2
  ...... (1) 
and  70?-30
o
 =  – j 20 I
2
 – j 160 (I
1
 – I
2
) ? 7?-30
o
 =  – j 16 I
1
+ j 14 I
2
  ...... (2) 
multiplying equation (1) by 7 and equation (2) by 4 and subtracting gives 
35 - 28?-30
o
 = (7 – j 56 + j 64) I
1
 + 0  ? 10.751 + j 14 =  (7 + j 8)I
1
 
i.e. I
1
 = 
o
o
81 . 48 63 . 10
48 . 52 65 . 17
?
?
= 1.660?3.67
o  
A  
substituting in (1),  
 j8 I
2
 = 5 – (1-j8) × 1.660?3.67
o
 
i.e. 8?90
o
 I
2
 = 5 – 8.062?-82.87
o
 × 1.660?3.67
o
 = 5 – 13.383?-79.20
o
 = 2.492 + j13.146 
i.e. I
2
 = 
o
o
90 8
27 . 79 38 . 13
?
?
= 1.673?-10.73
o
 A 
Thus the required current I is  = 1.660?3.67
o
 – 1.673?-10.73
o
 
    = 1.657 + j 0.106 – 1.644 + j 0.311  =  0.013 +j 0.417 
    = 0.42?88.2
o
 A 
The problem could probably have been worked out with lesser steps, but I have done it in this 
manner so that  you can get more familiarised with the solution of problems using complex 
numbers.  
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
 
I
2 
Page 2


 
Network Theorems - Alternating Current examples   
 
In the previous chapter, we have been dealing mainly with direct current resistive circuits in 
order to the principles of the various theorems clear.  As was mentioned, these theories are 
equally valid for a.c.  
 
Example 1 
 
 
 
 
 
 
For the circuit  shown in the figure, if the frequency of the supply is 50 Hz, determine using 
Ohm’s Law and Kirchoff’s Laws the current I in the 160 ? capacitor. 
Impedance of capacitor and inductor at 50 Hz are 
 j X
L
  = j 2p × 50 × 63.66 × 10
-3
 = j 20 ?  
and - j X
C
 = 1/( j 2p × 50 × 19.89 × 10
-6
) = - j 160 ? 
Using Kirchoff’s current law 
 I  =  I
1
  - I
2  
 
Using Kirchoff’s voltage law 
 100?0
o
 = 20 I
1
  – j160 (I
1
 – I
2
) ? 5 = (1-j8) I
1
 +j8 I
2
  ...... (1) 
and  70?-30
o
 =  – j 20 I
2
 – j 160 (I
1
 – I
2
) ? 7?-30
o
 =  – j 16 I
1
+ j 14 I
2
  ...... (2) 
multiplying equation (1) by 7 and equation (2) by 4 and subtracting gives 
35 - 28?-30
o
 = (7 – j 56 + j 64) I
1
 + 0  ? 10.751 + j 14 =  (7 + j 8)I
1
 
i.e. I
1
 = 
o
o
81 . 48 63 . 10
48 . 52 65 . 17
?
?
= 1.660?3.67
o  
A  
substituting in (1),  
 j8 I
2
 = 5 – (1-j8) × 1.660?3.67
o
 
i.e. 8?90
o
 I
2
 = 5 – 8.062?-82.87
o
 × 1.660?3.67
o
 = 5 – 13.383?-79.20
o
 = 2.492 + j13.146 
i.e. I
2
 = 
o
o
90 8
27 . 79 38 . 13
?
?
= 1.673?-10.73
o
 A 
Thus the required current I is  = 1.660?3.67
o
 – 1.673?-10.73
o
 
    = 1.657 + j 0.106 – 1.644 + j 0.311  =  0.013 +j 0.417 
    = 0.42?88.2
o
 A 
The problem could probably have been worked out with lesser steps, but I have done it in this 
manner so that  you can get more familiarised with the solution of problems using complex 
numbers.  
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
 
I
2 
 
Example 2 
Let us solve the same problem as earlier, but using Superposition theorem. 
 
 
 
           
 
 
 
This circuit can  be broken into its two constituent components as shown. 
 
 
 
 
 
 
Using series parallel addition of impedances, we can obtain the supply currents as follows. 
Equivalent  Z
s1
 = 20 + (-j160)//j20, Z
s2
 = j20 + 20//(-j160) 
  = 20 + 
140
20 160
j
j j
-
× -
,  = j20 +
160 20
) 160 ( 20
j
j
-
- ×
 
  = 20 + j 22.857,  = j20 + 19.692 – j 2.462 = 19.692 + j17.538 
  = 30.372?48.81
o
  ?,  = 26.370?41.69
o
 ? 
source current I
1A
 = 
o
o
81 . 48 372 . 30
0 100
?
?
, -I
2B
  = 
o
o
69 . 41 370 . 26
30 70
?
- ?
 
  = 3.293?-48.81
o
,  = 2.655?-71.69
o
, 
Using the current division rule (note directions of currents and signs), 
 I
A
 = 
140
20
81 . 48 293 . 3
j
j
o
-
× - ? , I
B
 = 
160 20
20
69 . 71 655 . 2
j
o
-
× - ? 
  = 0.470?131.19
o
,  = 
o
o
87 . 82 25 . 161
69 . 71 10 . 53
- ?
- ?
= 0.329?11.18
o 
Using superposition theorem, the total current in  
 I  =  0.470?131.19
o 
 + 0.329?11.18
o
  =  -0.310 + j 0.354 + 0.323 + j0.064 
  = 0.013 + j 0.419 = 0.42?88.2
o
 A 
which is the same answer obtained in the  earlier example.   
 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
 
I
2 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
I
1A 
I
A 
I
2A 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1B 
I
B 
I
2B 
I
B 
+
Page 3


 
Network Theorems - Alternating Current examples   
 
In the previous chapter, we have been dealing mainly with direct current resistive circuits in 
order to the principles of the various theorems clear.  As was mentioned, these theories are 
equally valid for a.c.  
 
Example 1 
 
 
 
 
 
 
For the circuit  shown in the figure, if the frequency of the supply is 50 Hz, determine using 
Ohm’s Law and Kirchoff’s Laws the current I in the 160 ? capacitor. 
Impedance of capacitor and inductor at 50 Hz are 
 j X
L
  = j 2p × 50 × 63.66 × 10
-3
 = j 20 ?  
and - j X
C
 = 1/( j 2p × 50 × 19.89 × 10
-6
) = - j 160 ? 
Using Kirchoff’s current law 
 I  =  I
1
  - I
2  
 
Using Kirchoff’s voltage law 
 100?0
o
 = 20 I
1
  – j160 (I
1
 – I
2
) ? 5 = (1-j8) I
1
 +j8 I
2
  ...... (1) 
and  70?-30
o
 =  – j 20 I
2
 – j 160 (I
1
 – I
2
) ? 7?-30
o
 =  – j 16 I
1
+ j 14 I
2
  ...... (2) 
multiplying equation (1) by 7 and equation (2) by 4 and subtracting gives 
35 - 28?-30
o
 = (7 – j 56 + j 64) I
1
 + 0  ? 10.751 + j 14 =  (7 + j 8)I
1
 
i.e. I
1
 = 
o
o
81 . 48 63 . 10
48 . 52 65 . 17
?
?
= 1.660?3.67
o  
A  
substituting in (1),  
 j8 I
2
 = 5 – (1-j8) × 1.660?3.67
o
 
i.e. 8?90
o
 I
2
 = 5 – 8.062?-82.87
o
 × 1.660?3.67
o
 = 5 – 13.383?-79.20
o
 = 2.492 + j13.146 
i.e. I
2
 = 
o
o
90 8
27 . 79 38 . 13
?
?
= 1.673?-10.73
o
 A 
Thus the required current I is  = 1.660?3.67
o
 – 1.673?-10.73
o
 
    = 1.657 + j 0.106 – 1.644 + j 0.311  =  0.013 +j 0.417 
    = 0.42?88.2
o
 A 
The problem could probably have been worked out with lesser steps, but I have done it in this 
manner so that  you can get more familiarised with the solution of problems using complex 
numbers.  
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
 
I
2 
 
Example 2 
Let us solve the same problem as earlier, but using Superposition theorem. 
 
 
 
           
 
 
 
This circuit can  be broken into its two constituent components as shown. 
 
 
 
 
 
 
Using series parallel addition of impedances, we can obtain the supply currents as follows. 
Equivalent  Z
s1
 = 20 + (-j160)//j20, Z
s2
 = j20 + 20//(-j160) 
  = 20 + 
140
20 160
j
j j
-
× -
,  = j20 +
160 20
) 160 ( 20
j
j
-
- ×
 
  = 20 + j 22.857,  = j20 + 19.692 – j 2.462 = 19.692 + j17.538 
  = 30.372?48.81
o
  ?,  = 26.370?41.69
o
 ? 
source current I
1A
 = 
o
o
81 . 48 372 . 30
0 100
?
?
, -I
2B
  = 
o
o
69 . 41 370 . 26
30 70
?
- ?
 
  = 3.293?-48.81
o
,  = 2.655?-71.69
o
, 
Using the current division rule (note directions of currents and signs), 
 I
A
 = 
140
20
81 . 48 293 . 3
j
j
o
-
× - ? , I
B
 = 
160 20
20
69 . 71 655 . 2
j
o
-
× - ? 
  = 0.470?131.19
o
,  = 
o
o
87 . 82 25 . 161
69 . 71 10 . 53
- ?
- ?
= 0.329?11.18
o 
Using superposition theorem, the total current in  
 I  =  0.470?131.19
o 
 + 0.329?11.18
o
  =  -0.310 + j 0.354 + 0.323 + j0.064 
  = 0.013 + j 0.419 = 0.42?88.2
o
 A 
which is the same answer obtained in the  earlier example.   
 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
 
I
2 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
I
1A 
I
A 
I
2A 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1B 
I
B 
I
2B 
I
B 
+
 
Example 3 
Let us again consider the same example  to illustrate Thevenin’s Theorem. 
 
 
 
 
 
 
 
Consider the capacitor disconnected at P and Q. 
Current flowing in the circuit under this condition = 
20 20
30 70 0 100
j
o o
+
- ? - ?
 
 
o
o
o
j
j
37 . 3 863 . 1
45 28 . 28
63 . 41 69 . 52
20 20
35 62 . 60 100
- ? =
?
?
=
+
+ -
= 
?Thevenin’s voltage source = 100?0
o
 – 20 × 1.863?-3.37
o
 = 62.80 + j 2.19 = 62.84?2.00
o
 
Also, Thevenin’s impedance across Q = 20//j20 = 
20 20
20 20
j
j
+
×
  = 14.142?45
o
 = 10 + j 10 
?Thevenin’s equivalent circuit is 
From this circuit, it follows that 
I  = 
160 10 10
00 . 2 84 . 62
j j
o
- +
?
 = 
o
o
19 . 86 33 . 150
00 . 2 84 . 62
- ?
?
 
 = 0.418?-88.2
o
 A 
which is again the same result. 
 
Example 4 
Let us again consider the same example  to illustrate Norton’s Theorem. 
 
 
 
 
 
 
 
Consider the capacitor short-circuited at P and Q. 
The Norton’s current source is given as 
20
30 70
20
0 100
j
o o
- ?
+
?
= 5 –1.75 –j3.031  
= 3.25 – j 3.031 = 4.444?-43.00
o 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
2 
P
 
I
 
Q
 
P
 
I
 
Q
 
10 + j 10 ? 
19.89 µ F 
-j160 ? 
E
th
 = 62.84?2.00
o
 V 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
2 
P
 
I
 
Q
 
Page 4


 
Network Theorems - Alternating Current examples   
 
In the previous chapter, we have been dealing mainly with direct current resistive circuits in 
order to the principles of the various theorems clear.  As was mentioned, these theories are 
equally valid for a.c.  
 
Example 1 
 
 
 
 
 
 
For the circuit  shown in the figure, if the frequency of the supply is 50 Hz, determine using 
Ohm’s Law and Kirchoff’s Laws the current I in the 160 ? capacitor. 
Impedance of capacitor and inductor at 50 Hz are 
 j X
L
  = j 2p × 50 × 63.66 × 10
-3
 = j 20 ?  
and - j X
C
 = 1/( j 2p × 50 × 19.89 × 10
-6
) = - j 160 ? 
Using Kirchoff’s current law 
 I  =  I
1
  - I
2  
 
Using Kirchoff’s voltage law 
 100?0
o
 = 20 I
1
  – j160 (I
1
 – I
2
) ? 5 = (1-j8) I
1
 +j8 I
2
  ...... (1) 
and  70?-30
o
 =  – j 20 I
2
 – j 160 (I
1
 – I
2
) ? 7?-30
o
 =  – j 16 I
1
+ j 14 I
2
  ...... (2) 
multiplying equation (1) by 7 and equation (2) by 4 and subtracting gives 
35 - 28?-30
o
 = (7 – j 56 + j 64) I
1
 + 0  ? 10.751 + j 14 =  (7 + j 8)I
1
 
i.e. I
1
 = 
o
o
81 . 48 63 . 10
48 . 52 65 . 17
?
?
= 1.660?3.67
o  
A  
substituting in (1),  
 j8 I
2
 = 5 – (1-j8) × 1.660?3.67
o
 
i.e. 8?90
o
 I
2
 = 5 – 8.062?-82.87
o
 × 1.660?3.67
o
 = 5 – 13.383?-79.20
o
 = 2.492 + j13.146 
i.e. I
2
 = 
o
o
90 8
27 . 79 38 . 13
?
?
= 1.673?-10.73
o
 A 
Thus the required current I is  = 1.660?3.67
o
 – 1.673?-10.73
o
 
    = 1.657 + j 0.106 – 1.644 + j 0.311  =  0.013 +j 0.417 
    = 0.42?88.2
o
 A 
The problem could probably have been worked out with lesser steps, but I have done it in this 
manner so that  you can get more familiarised with the solution of problems using complex 
numbers.  
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
 
I
2 
 
Example 2 
Let us solve the same problem as earlier, but using Superposition theorem. 
 
 
 
           
 
 
 
This circuit can  be broken into its two constituent components as shown. 
 
 
 
 
 
 
Using series parallel addition of impedances, we can obtain the supply currents as follows. 
Equivalent  Z
s1
 = 20 + (-j160)//j20, Z
s2
 = j20 + 20//(-j160) 
  = 20 + 
140
20 160
j
j j
-
× -
,  = j20 +
160 20
) 160 ( 20
j
j
-
- ×
 
  = 20 + j 22.857,  = j20 + 19.692 – j 2.462 = 19.692 + j17.538 
  = 30.372?48.81
o
  ?,  = 26.370?41.69
o
 ? 
source current I
1A
 = 
o
o
81 . 48 372 . 30
0 100
?
?
, -I
2B
  = 
o
o
69 . 41 370 . 26
30 70
?
- ?
 
  = 3.293?-48.81
o
,  = 2.655?-71.69
o
, 
Using the current division rule (note directions of currents and signs), 
 I
A
 = 
140
20
81 . 48 293 . 3
j
j
o
-
× - ? , I
B
 = 
160 20
20
69 . 71 655 . 2
j
o
-
× - ? 
  = 0.470?131.19
o
,  = 
o
o
87 . 82 25 . 161
69 . 71 10 . 53
- ?
- ?
= 0.329?11.18
o 
Using superposition theorem, the total current in  
 I  =  0.470?131.19
o 
 + 0.329?11.18
o
  =  -0.310 + j 0.354 + 0.323 + j0.064 
  = 0.013 + j 0.419 = 0.42?88.2
o
 A 
which is the same answer obtained in the  earlier example.   
 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
 
I
2 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
I
1A 
I
A 
I
2A 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1B 
I
B 
I
2B 
I
B 
+
 
Example 3 
Let us again consider the same example  to illustrate Thevenin’s Theorem. 
 
 
 
 
 
 
 
Consider the capacitor disconnected at P and Q. 
Current flowing in the circuit under this condition = 
20 20
30 70 0 100
j
o o
+
- ? - ?
 
 
o
o
o
j
j
37 . 3 863 . 1
45 28 . 28
63 . 41 69 . 52
20 20
35 62 . 60 100
- ? =
?
?
=
+
+ -
= 
?Thevenin’s voltage source = 100?0
o
 – 20 × 1.863?-3.37
o
 = 62.80 + j 2.19 = 62.84?2.00
o
 
Also, Thevenin’s impedance across Q = 20//j20 = 
20 20
20 20
j
j
+
×
  = 14.142?45
o
 = 10 + j 10 
?Thevenin’s equivalent circuit is 
From this circuit, it follows that 
I  = 
160 10 10
00 . 2 84 . 62
j j
o
- +
?
 = 
o
o
19 . 86 33 . 150
00 . 2 84 . 62
- ?
?
 
 = 0.418?-88.2
o
 A 
which is again the same result. 
 
Example 4 
Let us again consider the same example  to illustrate Norton’s Theorem. 
 
 
 
 
 
 
 
Consider the capacitor short-circuited at P and Q. 
The Norton’s current source is given as 
20
30 70
20
0 100
j
o o
- ?
+
?
= 5 –1.75 –j3.031  
= 3.25 – j 3.031 = 4.444?-43.00
o 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
2 
P
 
I
 
Q
 
P
 
I
 
Q
 
10 + j 10 ? 
19.89 µ F 
-j160 ? 
E
th
 = 62.84?2.00
o
 V 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
2 
P
 
I
 
Q
 
 
Norton’s admittance = 05 . 0 05 . 0
20
1
20
1
j
j
- = + S 
or same as Thevenin’s impedance 10 + j10 ? 
The Norton’s equivalent circuit is as shown in the figure. 
The current through the capacitor can be determined using the current division rule. 
 I  =  4.444?-43.0
o
 × 
160 10 10
10 10
j j
j
- +
+
 = 4.444?-43.0
o
 ×
o
o
19 . 86 33 . 150
45 142 . 14
- ?
?
 
  =  0.418?-88.2
o
 A 
 
Example 5 
Using Millmann’s theorem find the current in  the capacitor. 
 
 
 
 
 
 
 
 V
SN
  = 
?
?
Y
V Y.
 = 
20
1
160
1
20
1
30 70
20
1
0
160
1
0 100
20
1
j j
j j
o o
+
-
+
- ? · + ·
-
+ ? ·
 
   =  
19 . 41 06643 . 0
00 . 43 444 . 4
04375 . 0 05 . 0
031 . 3 25 . 3
05 . 0 00625 . 0 05 . 0
031 . 3 75 . 1 0 5
- ?
- ?
=
-
-
=
- +
- - +
o
j
j
j j
j
 
   =  66.89?-1.81
o
 V 
? I  =  66.89?-1.81
o
 /(-j160) = 0.418?88.19
o
 A  
 
Example 6 
Determine the delta equivalent of the star connected network shown. 
 
 
 
 
 
 
 
I
 
-j160 
4.444?-43.0
o
 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
 
I
2 
S 
N 
20 ? 
j20 ? 
-j160 ? 
S 
C 
A B 
C 
A B 
Y
AB 
Y
BC 
Y
CA =
 
Page 5


 
Network Theorems - Alternating Current examples   
 
In the previous chapter, we have been dealing mainly with direct current resistive circuits in 
order to the principles of the various theorems clear.  As was mentioned, these theories are 
equally valid for a.c.  
 
Example 1 
 
 
 
 
 
 
For the circuit  shown in the figure, if the frequency of the supply is 50 Hz, determine using 
Ohm’s Law and Kirchoff’s Laws the current I in the 160 ? capacitor. 
Impedance of capacitor and inductor at 50 Hz are 
 j X
L
  = j 2p × 50 × 63.66 × 10
-3
 = j 20 ?  
and - j X
C
 = 1/( j 2p × 50 × 19.89 × 10
-6
) = - j 160 ? 
Using Kirchoff’s current law 
 I  =  I
1
  - I
2  
 
Using Kirchoff’s voltage law 
 100?0
o
 = 20 I
1
  – j160 (I
1
 – I
2
) ? 5 = (1-j8) I
1
 +j8 I
2
  ...... (1) 
and  70?-30
o
 =  – j 20 I
2
 – j 160 (I
1
 – I
2
) ? 7?-30
o
 =  – j 16 I
1
+ j 14 I
2
  ...... (2) 
multiplying equation (1) by 7 and equation (2) by 4 and subtracting gives 
35 - 28?-30
o
 = (7 – j 56 + j 64) I
1
 + 0  ? 10.751 + j 14 =  (7 + j 8)I
1
 
i.e. I
1
 = 
o
o
81 . 48 63 . 10
48 . 52 65 . 17
?
?
= 1.660?3.67
o  
A  
substituting in (1),  
 j8 I
2
 = 5 – (1-j8) × 1.660?3.67
o
 
i.e. 8?90
o
 I
2
 = 5 – 8.062?-82.87
o
 × 1.660?3.67
o
 = 5 – 13.383?-79.20
o
 = 2.492 + j13.146 
i.e. I
2
 = 
o
o
90 8
27 . 79 38 . 13
?
?
= 1.673?-10.73
o
 A 
Thus the required current I is  = 1.660?3.67
o
 – 1.673?-10.73
o
 
    = 1.657 + j 0.106 – 1.644 + j 0.311  =  0.013 +j 0.417 
    = 0.42?88.2
o
 A 
The problem could probably have been worked out with lesser steps, but I have done it in this 
manner so that  you can get more familiarised with the solution of problems using complex 
numbers.  
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
 
I
2 
 
Example 2 
Let us solve the same problem as earlier, but using Superposition theorem. 
 
 
 
           
 
 
 
This circuit can  be broken into its two constituent components as shown. 
 
 
 
 
 
 
Using series parallel addition of impedances, we can obtain the supply currents as follows. 
Equivalent  Z
s1
 = 20 + (-j160)//j20, Z
s2
 = j20 + 20//(-j160) 
  = 20 + 
140
20 160
j
j j
-
× -
,  = j20 +
160 20
) 160 ( 20
j
j
-
- ×
 
  = 20 + j 22.857,  = j20 + 19.692 – j 2.462 = 19.692 + j17.538 
  = 30.372?48.81
o
  ?,  = 26.370?41.69
o
 ? 
source current I
1A
 = 
o
o
81 . 48 372 . 30
0 100
?
?
, -I
2B
  = 
o
o
69 . 41 370 . 26
30 70
?
- ?
 
  = 3.293?-48.81
o
,  = 2.655?-71.69
o
, 
Using the current division rule (note directions of currents and signs), 
 I
A
 = 
140
20
81 . 48 293 . 3
j
j
o
-
× - ? , I
B
 = 
160 20
20
69 . 71 655 . 2
j
o
-
× - ? 
  = 0.470?131.19
o
,  = 
o
o
87 . 82 25 . 161
69 . 71 10 . 53
- ?
- ?
= 0.329?11.18
o 
Using superposition theorem, the total current in  
 I  =  0.470?131.19
o 
 + 0.329?11.18
o
  =  -0.310 + j 0.354 + 0.323 + j0.064 
  = 0.013 + j 0.419 = 0.42?88.2
o
 A 
which is the same answer obtained in the  earlier example.   
 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
 
I
2 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
I
1A 
I
A 
I
2A 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1B 
I
B 
I
2B 
I
B 
+
 
Example 3 
Let us again consider the same example  to illustrate Thevenin’s Theorem. 
 
 
 
 
 
 
 
Consider the capacitor disconnected at P and Q. 
Current flowing in the circuit under this condition = 
20 20
30 70 0 100
j
o o
+
- ? - ?
 
 
o
o
o
j
j
37 . 3 863 . 1
45 28 . 28
63 . 41 69 . 52
20 20
35 62 . 60 100
- ? =
?
?
=
+
+ -
= 
?Thevenin’s voltage source = 100?0
o
 – 20 × 1.863?-3.37
o
 = 62.80 + j 2.19 = 62.84?2.00
o
 
Also, Thevenin’s impedance across Q = 20//j20 = 
20 20
20 20
j
j
+
×
  = 14.142?45
o
 = 10 + j 10 
?Thevenin’s equivalent circuit is 
From this circuit, it follows that 
I  = 
160 10 10
00 . 2 84 . 62
j j
o
- +
?
 = 
o
o
19 . 86 33 . 150
00 . 2 84 . 62
- ?
?
 
 = 0.418?-88.2
o
 A 
which is again the same result. 
 
Example 4 
Let us again consider the same example  to illustrate Norton’s Theorem. 
 
 
 
 
 
 
 
Consider the capacitor short-circuited at P and Q. 
The Norton’s current source is given as 
20
30 70
20
0 100
j
o o
- ?
+
?
= 5 –1.75 –j3.031  
= 3.25 – j 3.031 = 4.444?-43.00
o 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
2 
P
 
I
 
Q
 
P
 
I
 
Q
 
10 + j 10 ? 
19.89 µ F 
-j160 ? 
E
th
 = 62.84?2.00
o
 V 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
2 
P
 
I
 
Q
 
 
Norton’s admittance = 05 . 0 05 . 0
20
1
20
1
j
j
- = + S 
or same as Thevenin’s impedance 10 + j10 ? 
The Norton’s equivalent circuit is as shown in the figure. 
The current through the capacitor can be determined using the current division rule. 
 I  =  4.444?-43.0
o
 × 
160 10 10
10 10
j j
j
- +
+
 = 4.444?-43.0
o
 ×
o
o
19 . 86 33 . 150
45 142 . 14
- ?
?
 
  =  0.418?-88.2
o
 A 
 
Example 5 
Using Millmann’s theorem find the current in  the capacitor. 
 
 
 
 
 
 
 
 V
SN
  = 
?
?
Y
V Y.
 = 
20
1
160
1
20
1
30 70
20
1
0
160
1
0 100
20
1
j j
j j
o o
+
-
+
- ? · + ·
-
+ ? ·
 
   =  
19 . 41 06643 . 0
00 . 43 444 . 4
04375 . 0 05 . 0
031 . 3 25 . 3
05 . 0 00625 . 0 05 . 0
031 . 3 75 . 1 0 5
- ?
- ?
=
-
-
=
- +
- - +
o
j
j
j j
j
 
   =  66.89?-1.81
o
 V 
? I  =  66.89?-1.81
o
 /(-j160) = 0.418?88.19
o
 A  
 
Example 6 
Determine the delta equivalent of the star connected network shown. 
 
 
 
 
 
 
 
I
 
-j160 
4.444?-43.0
o
 
E
1
 = 100?0
o
 V 
20 ? 
j20 ? 
63.66 m? 
19.89 µ F 
-j160 ? 
E
2
 = 70?-30
o
 V 
I
1 
I
 
I
2 
S 
N 
20 ? 
j20 ? 
-j160 ? 
S 
C 
A B 
C 
A B 
Y
AB 
Y
BC 
Y
CA =
 
 
 Y
AB
  =  
20
1
160
1
20
1
20
1
20
1
j j
j
+
-
+
×
=
20 5 . 17
1
20 5 . 2 20
1
j j +
=
+ -
,     ? Z
AB
 = 17.5 + j 20 ? 
 Y
BC
 = 
20
1
160
1
20
1
20
1
160
1
j j
j j
+
-
+
×
-
 = 
140 160
1
160 20 160
1
j j j -
=
- +
,  ? Z
BC
 = 160 – j 140 ? 
 Y
CA
 =
20
1
160
1
20
1
160
1
20
1
j j
j
+
-
+
-
×
=
160 140
1
160 20 160
1
j j - -
=
- + -
, ? Z
CA
 = –140 – j 160 ? 
Example 7 
Determine the star equivalent of the delta connected network shown. 
 
 
 
 
 
 
 
 Z
A
  = 
140 160 160 140 20 5 . 17
) 160 140 )( 20 5 . 17 (
j j j
j j
- + - - +
- - +
= 
280 5 . 37
19 . 131 603 . 212 81 . 48 575 . 26
j
o o
-
- ? × ?
 
   = 20 01 . 0 00 . 20
37 . 82 5 . 282
38 . 82 5650
= - ? =
- ?
- ?
o
o
o
?  (same as original value in Ex 6). 
 Z
B
  = 
140 160 160 140 20 5 . 17
) 140 160 )( 20 5 . 17 (
j j j
j j
- + - - +
- +
= 
280 5 . 37
19 . 41 603 . 212 81 . 48 575 . 26
j
o o
-
- ? × ?
 
   = 20 99 . 89 00 . 20
37 . 82 5 . 282
62 . 7 5650
j
o
o
o
= ? =
- ?
?
?  (same as original value in Ex 6). 
      Z
C
 =
140 160 160 140 20 5 . 17
) 160 140 )( 140 160 (
j j j
j j
- + - - +
- - -
=
280 5 . 37
19 . 131 603 . 212 19 . 41 603 . 212
j
o o
-
- ? × - ?
 
   = 160 01 . 90 00 . 160
37 . 82 5 . 282
38 . 172 45200
j
o
o
o
- = - ? =
- ?
- ?
?  (same as original value in Ex 6). 
In order to show that the working is correct, I have selected the reverse problem for this 
example and used the results of the previous example to find the original quantities. You can  
see that  the answers differ only due to the cumulative calculation errors. 
 
C 
A B 
17.5+j20 ? 
160-j140 ? - 140 - j160? =
 
Z
A 
Z
B 
Z
C 
S 
C 
A B