0.1.

Millman’s Theorem

0.1 Millman’s Theorem

Statement:
In a network if it contains a several voltage sources E1 , E2 , E3 ... with an internal impedances
Z1 , Z2 , Z3 ... , which are connected in parallel may be replaced by a single voltage source E
with internal impedance Z, where E and Z are

E1 Y1 + E2 Y2 + E3 Y3
E=
Y1 + Y2 + Y3
and
1 1
Z= =
Y Y1 + Y2 + Y3
Consider a network containing source E1 , E2 , E3 ... with an internal impedances Z1 , Z2 , Z3 ... ,
A

Z1 Z2 Z3

E1 E2 E3

B
which are connected in parallel is as shown in Figure 1.

Figure 1
Replace the each voltage source by current source with its internal resistance connected in parallel,
which is as shown in Figure 2

I1 Z1 I2 Z2 I3 Z3

B
Figure 2
All the current source are added to form a single current source I where I

I = I1 + I2 + I3
Replace the impedances by a single impedance Z where Z is
1 1 1 1
Y = = + +
Z Z1 Z2 Z3

Z
A A
I Z
E
B B

Figure 3
Next the current source by voltage source E where E is
E1 E2 E3
I1 + I2 + I3 Z1 + Z2 + Z3
E = IZ = 1 = 1 1 1
Z Z1 + Z2 + Z3

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0.1. Millman’s Theorem

E1 Y1 + E2 Y2 + E3 Y3
E=
Y1 + Y2 + Y3
and
1 1 1 1
Y = = + +
Z Z1 Z2 Z3
Y = Y1 + Y2 + Y3

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0.1. Millman’s Theorem

Figure 5
2019-JULLY State and explain Millman’s theorem. The current through (6 + j8)Ω, impedance is
2019-JAN For the circuit shown in Figure 4 find E 1135
the current through (6 + j8)Ω, impedance using I = =
Z + ZL 10 + 6 + j8
Millman’s theorem. 1135 1135
= =
16 + j8 17.88∠26.56
6 = 63.47∠ − 26.56
10 415120 0 -j10
———————
j8 2018-JULY Using Millman’s theorem find current
j10 through RL for the circuit shown in Figure 6.
41500
4152400

Figure 4
2Ω 4Ω 5Ω IL
Solution:

+- 20 V +- 40 V +- 50 V
R L =9.4Ω
1 1
Y1 = = = 0.1
Z1 10
1 1
Y2 = = = −j0.1 Figure 6
Z2 j10
1 1 Solution:
Y3 = = = j0.1
Z3 −j10
1 1
Y1 = = = 0.5
Y = Y1 + Y2 + Y3 = 0.1 − j0.1 + j0.1 Z1 2
1 1
= 0.1 Y2 = = = 0.25
1 1 Z2 4
Z = = = 10Ω 1 1
Y 0.1 Y3 = = = 0.2
Z3 5

E1 Y1 = 415 × 0.1 = 41.5

E2 Y2 = 415∠120 × −j0.1 = 415∠120 × 0.1∠ − 90 Y = Y1 + Y2 + Y3 = 0.5 + 0.25 + 0.2
= 41.5∠30 = 36 + j20.75 = 0.95
E3 Y3 = 415∠240 × j0.1 = 41.5∠330 1 1
Z= = = 1.052Ω
= 36 − j20.75 Y 0.95

E1 Y1 + E2 Y2 + E3 Y3
E =
E1 Y1 + E2 Y2 + E3 Y3 Y1 + Y2 + Y3
E =
Y1 + Y2 + Y3 20 × 0.5 + 40 × 0.25 + 50 × 0.2
=
41.5 + 36 + j20.75 + 36 − j20.75 0.95
= 30
0.1 =
113.5 0.95
= = 31.57
0.1
= 1135
Millman’s equivalent circuit is is as shown in Figure
Millman’s equivalent circuit is is as shown in 7.
Figure 5.

1.052Ω IL
6Ω
10Ω
+- 31.57 V
R L =9.4Ω
j8Ω

1136∠00
Figure 7

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0.1. Millman’s Theorem

The current IL through RL is —————

31.57 31.57 2014-JULLY Using Millman’s theorem find the
IL = = current IL through RL for the network shown in
Z + ZL 1.052 + 9.4
= 3.02A Figure 10.

———————
2017-JAN Apply Millman’s theorem to find VO and
IO for the circuit shown in Figure 8. 2Ω 3Ω 4Ω IL

+- 10 V +- 20 V +- 30 V
R L =10Ω
-j5Ω j5Ω
10Ω
2Ω Vo
+- +- +- -j100 V Figure 10
100 V j100 V Io Solution:

Figure 8 1 1
Y1 = = = 0.5
Solution: Z1 2
1 1
Y2 = = = 0.33
1 1 Z2 3
Y1 = = = 0.1
Z1 10 1 1
Y3 = = = 0.25
1 1 Z3 4
Y2 = = = j0.2
Z2 −j5
1 1
Y3 = = = −j0.2
Z3 j5 Y = Y1 + Y2 + Y3 = 0.5 + 0.33 + 0.25
= 1.0833
Y = Y1 + Y2 + Y3 = 0.1 + j0.2 − j0.2
1 1
= 0.1 Z= = = 0.923Ω
Y 1.0833
1 1
Z= = = 10Ω
Y 0.1 E1 Y1 + E2 Y2 + E3 Y3
E =
E1 Y1 + E2 Y2 + E3 Y3 Y1 + Y2 + Y3
E = 10 × 0.5 + 20 × 0.33 + 30 × 0.25
Y1 + Y2 + Y3 =
100 × 0.1 + j100 × j0.2 + j100 × −j0.2 1.0833
= 19.16
0.1 =
−30 1.0833
= = 17.689
0.1
= −300
Millman’s equivalent circuit is is as shown in Figure
Millman’s equivalent circuit is is as shown in Figure 11.
7.

0.923Ω IL
10Ω
2Ω Vo +- 17.686 V
R L =10Ω
-
-300 + Io
Figure 11
Figure 9 The current IL through RL is
The current IL through RL is
E 17.689
−300 −300 IL = =
IL = = Z + ZL 17.689 + 10
Z + ZL 10 + 2 = 1.6191A
= 25A
VO = 25 × 2 = 50V ———————–

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0.1. Millman’s Theorem

2013-JAN Using Millman’s theorem find the current

IL through RL for the network shown in Figure 12. IL
1Ω 2Ω 3Ω

+- 1 V +- 2 V +- 3 V
R L =10Ω
4Ω 4Ω 4Ω IL

- - +- 10 V
R L =10Ω Figure 14
+ 4V + 2V
Solution:
1
Y1 =
Figure 12 Z1
Solution:
1
Y1 = 1 1
Z1 Y1 = = =1
Z1 1
1 1 1 1
Y1 = = = 0.25 Y2 = = = 0.5
Z1 4 Z2 2
1 1 1 1
Y2 = = = 0.25 Y3 = = = 0.333
Z2 4 Z3 3
1 1
Y3 = = = 0.25
Z3 4
Y = Y1 + Y2 + Y3 = 1 + 0.5 + 0.333
= 1.833
Y = Y1 + Y2 + Y3 = 0.25 + 0.25 + 0.25
= 0.75 1 1
Z= = = 0.5454Ω
1 1 Y 1.833
Z= = = 1.333Ω
Y 0.75
E1 Y1 + E2 Y2 + E3 Y3
E =
E1 Y1 + E2 Y2 + E3 Y3 Y1 + Y2 + Y3
E =
Y1 + Y2 + Y3 1 × 1 + 2 × 0.5 + 3 × 0.333
=
−4 × 0.25 − 2 × 0.25 + 10 × 0.25 0.333
= 1
0.75 =
1 1.833
= = 1.636V
0.75
= 1.333
Millman’s equivalent circuit is is as shown in Figure
Millman’s equivalent circuit is is as shown in Figure 15.
13.

0.5454Ω IL
1.333Ω IL

+- 1.63 V
R L =10Ω
+- 1.333 V
R L =10Ω

Figure 15
Figure 13
The current IL through RL is
The current IL through RL is
E 1.333 E 1.636
IL = = IL = =
Z + ZL 1.333 + 10 Z + ZL 0.5454 + 10
= 0.1176A = 0.1552A

———————– ———————–
2013-JULY Using Millman’s theorem find the 2012-JULY Using Millman’s theorem find the
current IL through RL for the network shown in current IL through RL for the network shown in
Figure 14. Figure 16.

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0.1. Millman’s Theorem

j20Ω
2Ω 4Ω 5Ω IL ER
-+
+- 20 V - +- 50 V
R L =9.4Ω
+ 40 V
EY -j20Ω
-+ S
Figure 16
Solution:
EB 20Ω
-+
1 1
Y1 = = = 0.5 Figure 18
Z1 2
1 1 Solution:
Y2 = = = 0.25
Z2 4
1 1 1 1
Y3 = = = 0.2 Y1 = = = −j0.05
Z3 5 Z1 j20
1 1
Y2 = = = j0.05
Z2 −j20
1 1
Y = Y1 + Y2 + Y3 = 0.5 + 0.25 + 0.2 Y3 = = = 0.05
Z3 20
= 0.95

1 1
Z= = = 1.052Ω Y = Y1 + Y2 + Y3 = −j0.05 + j0.05 + 0.05
Y 0.95
= 0.05
1 1
E1 Y1 + E2 Y2 + E3 Y3 Z= = = 20Ω
E = Y 0.05
Y1 + Y2 + Y3
E1 Y1 = 230 × −j0.05 = −j11.5
20 × 0.5 + 40 × 0.25 + 50 × 0.2
= E2 Y2 = 230∠ − 120 × j0.05
0.95
30 E2 Y2 = 230∠ − 120 × 0.05∠90 = 11.5∠ − 30
=
0.95
= 31.57V E3 Y3 = 230∠120 × 0.05 = −j11.5
E3 Y3 = 230∠120 × 0.05 = 11.5∠120
Millman’s equivalent circuit is is as shown in Figure
17. E1 Y1 + E2 Y2 + E3 Y3
E =
Y1 + Y2 + Y3
−j11.5 + 11.5∠ − 30 + 11.5∠120
IL
=
1.052Ω 0.05
−j11.5 + 9.95 − j5.75 − 5.75 + j9.95
=
+- 31.57 V
R L 9.4Ω 0.05
4.2 − j7.3 8.42∠ − 60
= =
0.05 0.05
Figure 17 = 168.4∠ − 60V
The current IL through RL is Millman’s equivalent circuit is is as shown in Figure
19.
E 31.57
IL = =
Z + ZL 1.052 + 9.4
= 3.02A
168.4 −60° V
20Ω
-+
———————–
2012-JAN Using Millman’s theorem determine
voltage VS of the network shown in Figure 18 given
that ER = 230∠0 V, EY = 230∠ − 120 V EB =
230∠120 V .

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0.1. Millman’s Theorem

Figure 19 Y = Y1 + Y2 + Y3 0.12 + j0.16

————————- = 0.37 + j0.11
Using Millman’s theorem determine current flowing = 0.386∠16.56
through (4 + j3) Ω of the network shown in Figure
20

5 30° 1 1
10Ω 4Ω 3Ω Z = = = 2.6∠ − 16.56Ω
Y 0.386∠16.56
4 −30°
10 30° 5Ω
j10Ω j3Ω -j4Ω

B E1 Y1 = 141.14∠75 × 0.0707∠ − 45◦ =

Figure 20 = 9.978∠30 = 8.641 + j4.989

Solution: E2 Y2 = 5∠30 × 0.2 = 1∠30 = 0.8666 + j0.5
Replace the current source and parallel resis- E3 Y3 = 20∠ − 83.13 × 0.2∠53.13
tance by a voltage source = 4∠ − 30 = 3.464 − j2
Z1 = 10 + j10 = 14.14∠45◦ Ω
E1 = 10∠30 × 14.14∠45◦ = 141.4∠75◦
Similarly
Z3 = 3 − j14 = 5∠ − 53.13◦ Ω E1 Y1 + E2 Y2 + E3 Y3
E =
Y1 + Y2 + Y3
E3 = 4∠ − 30 × 5∠ − 53.13◦ = 20∠ − 83.13◦
8.641 + j4.989 + 0.8666 + j0.5 + 3.464 − j2
The modified network as shown in Figure 21 =
0.386∠16.56
A 12.971 + j3.489 13.432∠15
= =
0.386∠16.56 0.386∠16.56
141.4 75° 5 30° 20 −83.13°
4Ω = 168.4∠ − 60V
14.14 45°Ω 5 −53.13 Ω
5Ω
j3Ω Millman’s equivalent circuit is is as shown in Figure
19.
B

A
Figure 21

2.591 −16.56°Ω 4Ω
1 1
Y1 = = = 0.0707∠ − 45◦
Z1 14.14∠45◦
Y1 = 0.05 − j0.05 25 34.45°
j3Ω
1 1
Y2 = = = 0.2
Z2 5
1 1 B
Y3 = = = 0.2∠53.13◦
Z3 5∠ − 53.13◦
= 0.12 + j0.16 Figure 22

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0.2. Thevenin’s and Norton’s Theorems

0.2 Thevenin’s and Norton’s Theorems

0.2.1 Thevenin’s Theorem
Statement:
Any linear, bilateral network with two terminals can be replaced by a single voltage source
ET H in series with an impedance ZT H , where the ET H is an open circuit voltage at the
terminals and an impedance ZT H is the equivalent impedance as viewed from the terminals
into the network.

ZTH A
A
Any Linear, IL
IL
bilateral
two ZL E TH ZL
terminal
network
B B

Figure 23

Proof: • Step 2: Determine the Tehevenin’s

Consider a network as shown in Figure 24. The Impedance
current in the terminal AB is

2Ω 8Ω
A
2Ω 8Ω
IL A
4Ω
10 V +-
R L =10Ω
4Ω ZTH =9.333Ω
B

Figure 24 B

V − 10 V V Figure 26
+ + = 0
2 4 18
V [0.5 + 0.25 + 0.0556] = 5 2×4
5 ZT H = 8 + = 9.333Ω
V = = 6.026 2+4
0.8056
6.026
IL = = 0.3448A • Step 3: Tehevenin’s Equivalent Circuit
18
• Step 1: Determine the Open Circuit
Voltage
ZTH =9.333Ω
A
2Ω 8Ω
A
+ E =6.667V
- TH
4Ω
10 V +- i1
B
B
Figure 27
Figure 25

• Step 4: Current through IL is

6i1 − 10 = 0
i1 = 1.666A
ET H 6.667V
ET H = 1.666A × 4 = 6.667V IL = = = 0.344 A
ZT H + ZL 9.333 + 10

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0.2. Thevenin’s and Norton’s Theorems

ZTH =9.333Ω
A
IL
ETH =6.667V +- R L =10Ω

Figure 28

0.2.2 Norton’s Theorem

Statement:
Any linear, bilateral network with two terminals can be replaced by a single current source
IN in parallel with an impedance ZN , where the IN is an short circuit current through the
terminals and an impedance ZN is the equivalent impedance as viewed from the terminals
into the network.

A A
Any Linear, IL
IL
bilateral
two ZL IN ZN
terminal
network B
B

Figure 29

0.2.3 Maximum Power Transfer Theorem

Statement:
In Any linear, bilateral network maximum power is delivered to the load RL by the source
when the load resistance RL is equal to the Thevenin’s resistance RT H .

ZTH A
A
Any Linear, IL
IL
bilateral
two ZL E TH ZL
terminal
network
B B

Figure 30

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0.2. Thevenin’s and Norton’s Theorems

Q 1) Find the Thevenin and Norton equivalent for 6Ω 4Ω 4Ω A

the circuit shown in Figure 31 with respect terminals
A-B
96 V + x 12Ω y 12Ω z
6Ω 4Ω 4Ω A
- ISC
IL

20Ω B
96 V +
-
12Ω 12Ω

Figure 34
B

Figure 31 18x − 12y + 0z = 96

Solution: −12x + 28y − 12z = 0
Determine the Thevenin voltage VT H . Apply KVL 0x − 12y + 16z = 0
for the circuit shown in Figure 32.
6Ω 4Ω 4Ω A
By Solving

96 V + 12Ω y 12Ω VOC x = 9.212A y = 5.818A z = 4.36A

- x
ISC = IN = z = 4.36A
B
8.8Ω A A
Figure 32

38.4 +
- 4.36 A
8.8Ω

18x − 12y = 96
−12x + 28y = 0 B B
Thevenin’s Equivalent Norton’s Equivalent
By Solving
Figure 35
x = 7.467A y = 3.2A Current through RL is
VOC 38.4
VOC = 7.467 × 12A = 38.4V IL = = = 1.33A
RL 20 + 8.8
Power through is
ZT H = [(6||12) + 4]||12 + 4
PL = IL2 RL = (1.33)2 × 20 = 35.56W
= [4 + 4]||12 + 4
= 8||12 + 4 = 4.8 + 4 Maximum power is RL is
= 8.8Ω VOC 38.4
IL = = = 2.18A
RN + RL 8.8 + 8.8
6Ω 4Ω 4Ω A PL = IL2 RL = (2.18)2 × 8.8 = 41.856W
—————–
12Ω 12Ω
RTH

Figure 33
To determine the short circuit current Apply KVL
for the circuit shown in Figure 34

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0.2. Thevenin’s and Norton’s Theorems

Q 2) Find the Thevenin and Norton equivalent for Figure 39

the circuit shown in Figure 36 with respect terminals Q 3) Find the Thevenin and Norton equivalent for
A-B the circuit shown in Figure 40 with respect terminals
4Ω 1Ω A A-B
IL 10 2 A
RL IL
32 V +
-
12Ω
2A 6Ω
RL
50 V +
-
5 3
10
B

B
Figure 36
Solution: Figure 40
Determine the Thevenin voltage VT H . Apply KVL Solution:
for the for the circuit shown in Figure 37. Determine the Thevenin voltage VT H . Apply the
KVL for the circuit shown in Figure 41.
y = −2A 15x − 5y+ = 50
16x − 12y = 32 −5x + 10y = 0
16x − 24 = 32
32 − 24 By Solving
x = = 0.5A x = 4A y = 2A
16
VOC = 12[0.5A − (−2)].5A × 3 = 30V VOC = 2A × 3 = 6V

4Ω 1Ω 10 2 A
A

y
x
32 V +
-
12Ω VOC 50 V +
- x 5 y 3 VOC
2A

B B

Figure 37 Figure 41

ZT H = (4||12) + 1 ZT H = [(10||5) + 2]||3

= 3 + 1 = 4Ω = [3.33 + 2]||3
= 5.33||3 = 1.92Ω
4Ω 1Ω
A
10 2 A
12Ω
RTH 5 3
RTH
B

Figure 38 B

4Ω Figure 42
A A
Determine the short circuit current by Applying
KVL for the circuit 43
30 V +
- 7.5 A 4Ω
15x − 5y = 50
−5x + 7y = 0

B B By Solving
Thevenin’s Equivalent Norton’s Equivalent
x = 4.375A y = 3.125A

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0.2. Thevenin’s and Norton’s Theorems

ISC = IN = y = 3.125A Voltage across AB VOC = VT H is

10 2 A VOC = 6i = 6 × 2 = 12V

ISC 6Ω 2i1
50 V + x 5 y 3 10Ω
- -+ A
i1
6Ω
B 20 V +
-
VOC
i
Figure 43
B
1.92 A A
Figure 46
When dependant voltage sources are present
6 + 3.125 1.92 then Thevenin Resistance RT H is calculated
- A by determining the short circuit current at
terminals AB:
6Ω 2i1 10Ω
B B
Thevenin’s Equivalent Norton’s Equivalent
-+ A
i1
6Ω y
Figure 44 20 V +
-
ISC
x
Current through RL is
VOC 6 B
IL = = = 0.503A
RL 1.92 + 10
Figure 47
Power through is
PL = IL2 RL = (0.503)2 × 10 = 2.53W x − y = i1
Maximum power is RL is KVL for loop x
VOC 6
IL = = = 1.5625A 12x − 2i1 − 6y − 20 = 0
RN + RL 1.92 + 1.92
12x − 2(x − y) − 6y = 20
PL = IL2 RL = (1.5625)2 × 1.92 = 4.68W
10x − 4y = 20
—————–
Q 4) Find the Thevenin and Norton equivalent for KVL for loop y
the circuit shown in Figure ?? with respect terminals
a-b −6x + 16y = 0
6Ω 2i1 10Ω 6x − 16y = 0
-+ A
i1 Solving the following simultaneous equations
6Ω
20 V +
- 10x − 4y = 20
6x − 16y = 0
B
x = 2.353 y = 0.882
Figure 45
Solution: ISC = y = 0.882A
Determine the Thevenin voltage VT H . Apply KVL
for the circuit shown in Figure 46. Thevenin’s resistance is
By KVL around the loop VT H 12
RT H = = = 13.6Ω
6i − 2i + 6i − 20 = 0 I SC 0.882
10i = 20 Thevenin and Norton equivalent circuits as shown
i = 2A in Figure 48

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0.2. Thevenin’s and Norton’s Theorems

13.6Ω A A terminals AB. Apply KVL for loop x

6x + 2ia − 12 = 0
12 V +
- 0.882 A 13.6Ω 6x − 2x = 12
12
x = = 3A
4
B B
Thevenin’s Equivalent Norton’s Equivalent ia = −x = −3A
Apply KVL for loop y
Figure 48
————————— 3y − 2ia = 0
Q 5) Find the Thevenin and Norton equivalent for 3y − 2(−3) = 0
the circuit shown in Figure 49 with respect terminals −6
y = = −2A
a-b 3
6Ω Short circuit current ISC
3Ω
A ISC = y = −2A
Thevenin resistance is RT H
12 V +
- + 2ia
- −6
ia RT H = = 3Ω
−2
B
6Ω 3Ω
A
Figure 49
Solution: 12 V + x
- + 2ia y ISC
Determine the Thevenin voltage VT H . Apply the ia
-
KVL for the circuit shown in Figure 50. From the
B
figure it is observed that
Figure 51
ia = −x
Thevenin’s and Norton’s Circuits are as shown in
By KVL for the loop x Figure 52
3Ω A A
6x + 2ia − 12 = 0
6x − 2x = 12 -
12 6V + 2A 3Ω
x = = 3A
4

ia = −x = −3A B B
Thevenin’s Equivalent Norton’s Equivalent
Open Circuit voltage VOC is
Figure 52
VOC = 2ia = 2 × (−3) = −6V —————————
Q 6) Find the Thevenin and Norton equivalent for
the circuit shown in Figure 53 with respect terminals
6Ω 3Ω a-b
A
3ia
x
12 V +
- + 2ia VOC
ia
- 2Ω
A
B
ia
Figure 50 24 V +
-
4A 8Ω
When dependant voltage sources are present
then Thevenin Resistance RT H is calculated
B
by determining the short circuit current at

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0.2. Thevenin’s and Norton’s Theorems

Figure 53 2Ω V1
A
Solution:
Determine the Thevenin voltage VT H for circuit
shown in Figure 50. It is observed that 24 V +
-
4A ISC

V2 = 24 B
V1
ia = Figure 56
8
8
RT H = = 1Ω
3ia 8

1Ω A A
2Ω V1
V2 A
ia 8 + 8A 1Ω
-
24 V +
-
4A 8Ω
VOC

B B B
Thevenin’s Equivalent Norton’s Equivalent
Figure 54
Apply (KCL) node voltage for the circuit shown in Figure 57
Figure 54 for node 1 —————————
Q 7) Find the Thevenin and Norton equivalent for
V1 − V2 V1 the circuit shown in Figure 58 with respect terminals
+4+ + 3ia = 0 a-b
2 8
V1 − 24 V1 V1 5Va
+4+ +3 = 0
2 8 8
V1 V1 V1 +- A
+ +3 = 8 +
2 8 8
V1 = 8 3 mA Va 2kΩ 40kΩ
-
To find the short current Isc short the output
terminals a and b. B

3ia Figure 58
Solution:
2Ω Determine the Thevenin voltage VT H by apply node
V1
V2 A voltage method for the circuit shown in Figure 59
ia
24 V +
-
4A 8Ω ISC V1 V2
+ − 3 × 10−3 = 0
2kΩ 40kΩ
B 0.5 × 10−3 V1 + 0.025 × 10−3 V2 = 3 × 10−3
0.5V1 + 0.025V2 = 3
Figure 55
When it is short circuited the current Ia = 0 It is observed that
then dependent source becomes zero. The modified
Va = V1
circuit is as shown in Figure 56.

V1 − 24 V1 − V2 = 5Va
+ 4 + ISC = 0
2 V1 − V2 − 5Va = 0
0 − 24
+ 4 + ISC = 0 V1 − V2 − 5V1 = 0
2
ISC = −4 + 12 = 8A 4V1 + V2 = 0

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0.2. Thevenin’s and Norton’s Theorems

Solving following equations 20Ω 2Ω

A
0.5V1 + 0.025V2 = 3 i
4V1 + V2 = 0 8Ω
20 V +
-
5i

V1 = 7.5V, V2 = −30V B

5Va Figure 62
V1 V2
+- A Solution:
+ Determine the Thevenin voltage VT H by applying
3 mA Va 2kΩ 40kΩ KVL for the circuit shown in Figure 63. It is
VOC observed that there is a current source between two
-
loops, hence apply supermesh analysis.
B

Figure 59 y − x = 5i = 5x
Determine the short circuit current ISC for the 6x − y = 0
circuit shown in Figure 60. It is observed that 40 20x + 10y = 20
kΩ is also shorted hence entire current is flowing
through shortened terminals AB. Solving the above equations

x = 3mA x = 0.25A y = 1.5A

y = −x = −3mA
VOC = VT H = 1.5A × 8Ω = 6V
20 2
VOC = −30V A
VOC −30V i
ZT H = = = 10kΩ 8
ISC −3mA 20 V + x 5i y VOC
-
5Va
+- A B
+
40kΩ
Figure 63
3 mA Va 2kΩ ISC
Determine the ISC for the circuit shown in Figure
-
64.
B
y − x = 5i = 5x
Figure 60
6x − y = 0
Thevenin and Norton circuits are as shown in Figure
20x + 2y = 20
61
10kΩ A A Solving the above equations

x = 0.625A y = 3.75A
- 10kΩ
30V + 3 mA 20 2
A
i
8
B B 20 V +
- x 5i y ISC
Thevenin’s Equivalent Norton’s Equivalent

Figure 61 B
————— Figure 64
Q 8) Find the Thevenin and Norton equivalent
for the circuit shown in Figure refThevenin- VOC 6
dependent8-1 with respect terminals a-b ZT H = = = 1.6Ω
ISC 3.75

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0.2. Thevenin’s and Norton’s Theorems

1.6Ω Applying KCL for the circuit shown in Figure 68.

A A
VA = 0V

6V +
- 3.75 A 1.6Ω VC VC − VA
−2+ = 0
5 1
0.2VC + 1VC = 2
2
B B VC = = 1.67V
Thevenin’s Equivalent Norton’s Equivalent
1.2

VA − VC v1
Figure 65 − + ISC = 0
1 2
————— −VC 0.2VC
Q 9) Find the Thevenin and Norton equivalent for − = −ISC
1 2
the circuit shown in Figure 66 with respect terminals 1.67 0.2 × 1.67
a-b + = ISC
1 2
1Ω 1Ω ISC = 1.67 + 0.167 = 1.837A
C
A
- v1 + 1Ω C 1Ω
A
4Ω 2A v1 - v1 +
2
4Ω 2A v1 ISC
B 2

Figure 66 B
Solution:
Figure 68
Determine the Thevenin voltage VT H for circuit
shown in Figure 67. It is observed that VOC 22
ZT H = = = 12Ω
ISC 1.837A
VC
v1 = × 1 = 0.2VC
5 12Ω A A
Apply KCL

VC v1 22 V +
- 1.83 A 12Ω
−2− = 0
5 2
VC VC
− 2 − 0.2 × = 0
5 2
0.2VC − 0.1VC = 2 B B
Thevenin’s Equivalent Norton’s Equivalent
2
VC = = 20
0.1 Figure 69
By Test voltage method. Apply a test voltage
v1 of 1 V at the output terminals AB and
VOC = VT H = VC + 1 ×
2 determine the applied source current. The
0.2 × 20 modified circuit is as shown in Figure 70.
= 20 + 1 ×
2
= 22 1Ω 1Ω VA
A
1Ω
- v1 + Io
C 1Ω
A
- v1 + 4Ω x v1 y +- V =1V
S
v1 VOC
2
4Ω 2A
2
B
B
Figure 70
Figure 67 It is observed that

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0.2. Thevenin’s and Norton’s Theorems

3Ω C 5Ω
A
v1 = −1x = −x +
v1 −x Vx
y−x = = 3Ω 10 A Vx VOC
2 2 - 4
0.5x − y = 0
B
By supermesh analysis method
Figure 72
6x + 1 = 0 Applying KCL for the circuit shown in Figure 73.
1
x = − = −0167A
6 VA = 0V
y = 0.5x = 0.5 × (−0167A) = −0.0833A
VC VC − VA
The circuit impedance is − 10 + = 0
6 5
0.166VC + 0.2VC = 10
1V
ZT H = = 12Ω 0.366VC = 10
0.0833A
10
VC = = 27.27V
————— 0.366
Q 9-1) Find the Thevenin and Norton equivalent for
the circuit shown in Figure 71 with respect terminals VA − VC Vx
a-b − + ISC = 0
5 4
3Ω 5Ω −VC 0.5VC
− = −ISC
A 5 4
+ 27.27 0.5 × 27.27
Vx + = ISC
3Ω Vx 5 4
10 A ISC = 5.454 + 3.408 = 8.862A
- 4

B 3Ω C 5Ω
A
Figure 71 +
Vx
Solution: 3Ω 10 A Vx ISC
- 4
Determine the Thevenin voltage VT H for circuit
shown in Figure 72. It is observed that
B
VC
vx = × 3 = 0.5VC Figure 73
6
Apply KCL VOC 390.38
ZT H = = = 44Ω
ISC 8.862A
VC Vx
− 10 − = 0
6 4 44Ω A A
VC 0.5VC
− 10 − = 0
6 4
VC (0.166 − 0.125) = 10
398.38 V +
- 8.86 A 44Ω
0.0416VC = 10
VC = 240.38

B Norton’s B
Vx Thevenin’s
VOC = VT H = VC + 5 × Equivalent Equivalent
4
0.5VC
= 240.38 + 5 × Figure 74
4
0.5 × 240.38 —————
= 240.38 + 5 × Q 10) Find the Thevenin equivalent for the circuit
4
= 240.38 + 150 = 390.38 shown in Figure 75 with respect terminals a-b

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0.2. Thevenin’s and Norton’s Theorems

V1 2vx
A -+
ix
2ix 4Ω 2Ω
2Ω 2Ω
A

B +
5A 4Ω vx 6Ω
Figure 75 -
Solution: B
For this circuit it does not have any
Figure 78
independent sources. Apply a test voltage
of 1 V at the output terminals AB and Solution:
determine the applied source current. The Determine the Thevenin voltage VT H for circuit
modified circuit is as shown in Figure 76. shown in Figure 79.
2vx
V1 -+
A
ix z
2ix 4Ω 2Ω io 2Ω 2Ω
A
+ y
B 5A x 4Ω vx 6Ω
-
Figure 76
B
It is observed that
Figure 79
V1
ix = −
2
= −0.5V1 x=5
It is observed that
Apply KCL for the node V1
vx = 4(x − y)
V1 V1 vx = 20 − 4y
+ + + 2ix − io = 0
4 2 KVL for the mesh z
V1 [0.25 + .5] + 2(−0.5V1 ) − io = 0
V1 [0.25 + .5 − 1] − io = 0 −2vx + 2(z − y) = 0
−0.25V1 − io = 0 −2y + 2z − 2(20 − 4y) =
−io = 0.25V1 6y + 2z = 40
KVL for the mesh y
V1 V1
RT H = = = 4Ω 4(y − x) + 2(y − z) + 6y = 0
io 0.25V1
−4x + 12y − 2z = 0
4Ω A
−4 × 5 + 12y − 2z = 0
−20 + 12y − 2z = 0
12y − 2z = 20
Solving the following linear equations
12y − 2z = 20
B
Thevenin’s Equivalent 6y + 2z = 40

Figure 77 y = 3.33A x = 10A

————— VOC = 6y = 6 × 3.33A = 20V
Q 11) Find the Thevenin equivalent for the circuit Short circuit the output terminals AB and determine
shown in Figure 78 with respect terminals a-b the short circuit current ISC

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0.2. Thevenin’s and Norton’s Theorems

2vx 2vx
-+ -+
z
x
2Ω 2Ω
A 2Ω 2Ω
+ y A
5A x 4Ω vx 6Ω k ISC io
+ y
-
4Ω vx 6Ω z + Vo=1V
-
B
-
Figure 80 B

x=5 Figure 81
It is observed that
It is observed that
vx = −4y
vx = 4(x − y)
Apply KVL for the mesh x
vx = 20 − 4y
−2vx + 2(x − y) = 0
KVL for the mesh y
vx = x − y
4(y − x) + 2(y − z) + 6y − 6k = 0 −4y = x − y
−4x + 12y − 2z − 6k = 0 x = −3y
−4 × 5 + 12y − 2z − 6k = 0
Apply KVL for the mesh y
−20 + 12y − 2z − 6k = 0
−2x + 12y − 6z = 0
12y − 2z − 6k = 20
−2(−3y) + 12y − 6z = 0
KVL for the mesh z 18y − 6z = 0
2z − 2y − 2vx = 0 Apply KVL for the mesh z
−2y + 2z − 2(20 − 4y) = 0 −6y + 8z = −1
−2y + 2z − 40 + 8y = 0
Solving the following simultaneous equations
−2y + 2z − 2(20 − 4y) = 0
6y + 2z + 0k = 40 18y − 6z = 0
−6y + 8z = −1
KVL for the mesh k
y = −0.055, z = −0.166
−6y + 0z + 8k = 0
iO = −z = 0.166A
Solving the following equations VO 1V
RT H = = = 6Ω
iO 0.166A
12y − 2z − 6k = 20 The Thevenin and Norton circuits are as shown in
6y + 2z + 0k = 40 Figure 82
−6y + 0z + 8k = 0 3.33Ω A A
y = 4.44, z = 6.667, k = 3.333
ISC = k = 3.333 20 V +
- 0.882 A 6Ω
VT H 20
ZT H = = = 6Ω
ISC 3.333
Alternative Method: To determine the short B B
circuit current ISC , apply a test voltage of 1 Thevenin’s Equivalent Norton’s Equivalent
V at the output terminals AB and determine
the applied source current io . The modified Figure 82
circuit is as shown in Figure 81. —————

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0.2. Thevenin’s and Norton’s Theorems

Q 13) Find the Thevenin equivalent for the circuit

shown in Figure 83 with respect terminals a-b VOC 30
ZN = = = 6Ω
ISC 5
6I x
-+ A 6Ω A A

4Ω Ix 6Ω
30 V +
- 5A 6Ω

-+ B
20 V
B B
Thevenin’s Norton’s
Figure 83
Equivalent Equivalent
Solution:
Figure 86
10Ix − 6Ix + 20 = 0 —————
Q 14) Find the Thevenin equivalent for the circuit
4Ix = −20
−20 shown in Figure 83 with respect terminals a-b
Ix = = −5A
4 8Ω
VOC = −5A × 6 = −30V +-
12 V
20 V +
-
6I x
2Ω
-+ A 5Ω

4Ω x Ix 6Ω VOC
Figure 87
Solution:
-+ B
20 V
7x − 20 = 0
Figure 84 20
x = = 2.857A
7
Ix = 0 VOC = x × 2 − 12
= 2.857 × 2 − 12
4Ix + 20 = 0 = 6.29V
20
Ix = = 5A
4 8Ω
ISC = 5A +-
12 V
20 V +
-
x
6I x 2Ω VOC
5Ω
-+ A

4Ω Ix 6Ω ISC Figure 88

2×5
-+ B ZT H = 8+
2×5
20 V 10
= 8+ = 2.857A
7
Figure 85 = 9.43Ω

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0.2. Thevenin’s and Norton’s Theorems

5 × j2 2 × −j2
8Ω ZAB = +
5 + j2 2 − j2
= (0.689 + j1.724) + (1 − j1) = 1.689 + j0.724
5Ω 2Ω RTH
VAB
IAB =
ZAB + 4
16.69∠0−156
Figure 89 =
1.689 + j0.724 + 4
9.43Ω = 2.91∠ − 163.25◦
A A
Q 16) Find the Thevenin and Norton
equivalent circuit between terminals A-B for
+ 9.43Ω
6.29 V - 0.66 A the circuit shown in Figure 93.
j3 
5 2
A
B B
Thevenin’s Norton’s
Equivalent Equivalent 500o V j5 
6

Figure 90 B
Q 15) For the circuit shown in Figure 91
Figure 93
Find the Thevenin and Norton equivalent
circuit with respect terminals A-B Solution:
j3 
Solution: 5  VA 2
A
5 A 4 B 2
500o V j5  ISC
Ix 6
500o V j2  -j2 
5090o V B

Figure 94

Figure 91
VA − 50 VA VA
+ + = 0
5 A B 2 5 j5 2 + j3
VA [0.2 − j0.2 + 0.153 − j0.23] = 10
I1 0.353 − j0.43 = 10
500o V j2  -j2  I 2 10
5090 V
o
VA =
0.353 − j0.43
= 18∠50.6◦

Figure 92 j3 
5 2
A

50∠0◦ j5 
VA = × j2 6
5 + j2
= 18.57∠68.2◦ B

Figure 95
50∠90◦
VB = × −j2
2 − j2   
5 × j5
= 35.35∠45◦ ZAB = + (2 + j3) ||6
5 + j5
= [(2.5 + j2.5) + (2 + j3)] ||6
VAB = VA − VB = 18.57∠68.2◦ − 35.35∠45◦ = [(4.5 + j5.5)] ||6
◦
= 19.69∠ − 156 = 3.31 + j1.4 = 3.6∠23◦

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0.2. Thevenin’s and Norton’s Theorems

VA 18∠50.6◦ Figure 97
IN = =
2 + j3 2 + j3
= 14.99∠ − 5.71◦
50∠0◦ − 25∠90◦
Q 17) Find the Thevenin and Norton I = = 6.933∠ − 33.7◦
8 + j1
equivalent circuit between terminals A-B for
the circuit shown in Figure 96.
j5 
VAB = 50∠0◦ − 6.933∠ − 33.7◦ × (5 + j5)
5 A 3 -j4 
= 9.79∠ − 78.65◦

500o V ZL 2590o V j5 
5 A 3 -j4 

B ZAB

Figure 96
B
Solution:
j5  Figure 98
5 A 3 -j4 

500o V i 2590o V (5 + j5) × (3 − j4)

ZAB =
8 + j1
B
= 4.23 − j1.15 Ω

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