CHAPTER 3
NETWORK THEOREM
NETWORK THEOREM



Superposition Theorem
Source Transformation
Thevenin & Norton Equivalent
2
SUPERPOSITION THEOREM
3
Principle
If a circuit has two or more independent sources, the
voltage across or current through an element in a
linear circuit, is the algebraic sum of voltages across
or current through that element due to each
independent source acting alone.
4
Steps to apply superposition principle

Turn off all independent sources except one sources.





For voltage source replace by short circuit
For current source replace by open circuit.
Find voltage or current due to that active source
using any technique.
Repeat the procedure for each of the other source
Find total contribution by adding algebraically all the
contribution due to the independent source
5
Practice Problem 10.5
Find current in the circuit using the superposition theorem
6
Solution
Let Io = Io’ + Io” where Io’ and Io” are
due to voltage source and current
source respectively.
For Io’ consider circuit beside where
the current source is open circuit
For mesh 1
(8  j2)I 1  j4I 2  0
I 2  (0.5  j2)I 1
…(1)
For mesh 2
(6  j4)I 2  j4I 1  1030 o  0
…(2)
7
Solution
Substitute equation (1) into equation (2)
(6  j4)(0.5  j2)I 1  j4I 1  1030 o
10 30 o
I1 
11  j14
I o '  I1  0.08  j0.556
8
Solution
For Io” consider circuit beside where
the voltage source is short circuit
Let
Z1  8  j2
Z 2  6 || j4
j24

 1.846  j2.769
6  j4
Io " 
Z2
(20o )
Z1  Z2
(2)(1.846  j2.769)

 0.4164  j0.53
9.846  j0.77
9
Solution
Therefore
I o  I o ' I o "
 0.4961  j1.086
 1.193965.45o A
10
Practice Problem 10.6
Calculate Vo in the circuit using superposition theorem
11
Solution
Let vo = vo’ + vo” where vo’ is due to
the voltage source and vo” is due to
the current source.
For vo’ we remove current source
which is now open circuit
Transfer the circuit to
frequency domain ( = 5)
30sin(5t)  300o
0.2F  ( j)
1H  j5
By voltage division
Vo ' 
 j1.25
(30)  4.631 - 81.12o
8  j1.25
v o '  4.631sin(5 t  81.12o )
12
Solution
For vo” we remove voltage source and
replace with short circuit
Transfer the circuit to
frequency domain ( = 10)
2 cos(10t )  20o
0.2F  ( j0.5)
1H  j10
Let
Z1  ( j0.5)
Z2  8 || j10
j80

 4.878  j3.9
8  j10
By current division
I
Z2
(2)
Z1  Z 2
 2.096  j0.138
13
Solution
Thus
Vo "  IZ1
 (2.096  j0.138)(  j0.5)
 1.050  86.23o
v o "  1.05cos(10 t  86.23o )
Therefore
vo  vo ' vo "
 4.631sin(5 t  81.12o )  1.05cos(10t  86.24o )
14
SOURCE TRANSFORMATION
15
Source Transformation
Source transformation in the frequency domain
involves transforming a voltage source in series with
an impedance to a current source in parallel with an
impedance, or vice versa.
16
Example 10.7
Calculate Vx in the circuit by using source transformation
17
Solution
If we transform the voltage source to a current source, we obtain
the circuit as shown.
20  90o
IS 
 4  90o  ( j4)A
5
The parallel combination
of 5 resistance and
(3+j4) impedance
gives a new equivalent
impedance Z1
Z1 
5(3  j4)
 (2.5  j1.25)Ω
8  j4
18
Solution
Convert the current source to a voltage source
yields the circuit as below
VS  IS Z1  ( j4)(2.5  j1.25)
 (5  j10)V
We then could solve for VX
by using voltage division
VX 
10
(5  j10)
10  2.5  j1.25  4  j13
VX  5.519  28o V
19
Practice Problem 10.7
Find Io in the circuit by using the concept of source transformation
20
Solution
If we transform the current source to a voltage source, we obtain
the circuit as shown.
VS  IS ZS  (j4)(4  3j)  12  16j
21
Solution
We transform the voltage source to a current source as shown below
Note that
Z || j5 
Let
Then
Z  4  j3  2  j
 6 - j2
IS 

(6  j2)(j5)
6  j3
10
(1  j)
3
VS
ZS

12  j16
 1  j3
6  j2
22
Solution
By current division
Io 
10
(1  j)
3
10
(1  j)  (1  j2)
3
(1.5  j3)
 20  j 40 44.72116.56o


13  j 4
13.60217.1o
 3.28899.46o A
23
THEVENIN & NORTON
EQUIVALENT CIRCUIT
24
Equivalent Circuit
Thevenin Equivalent
Circuit
Norton Equivalent
Circuit
25
Relationship
Keep in mind that the two equivalent circuit are related as
Vth  Z N I N
and
Zth  Z N
Vth = VOC = open circuit voltage
IN = ISC = short circuit current
26
Steps to determine equivalent circuit
Zth or ZN
Equivalent impedance looking from the terminals when the
independence sources are turn off. For voltage source replace by short
circuit and current source replace by open circuit.
Vth
Voltage across terminals when the terminals is open circuit
IN
Current through the terminals when the terminals is short circuit
Note
When there is dependent source or sources with difference frequencies, the
step to find the equivalent is not straight forward.
27
Practice Problem 10.8
Find the Thevenin equivalent at terminals a-b of the circuit
28
Solution
To find Zth, set voltage source to zero
Z th  10  ( j4) || (6  j2)
( j4)(6  j2)
 10 
6  j2
 10  2.4  j3.2
 12.4  j3.2
29
Solution
To find Vth, open circuit at terminals a-b
 j4
( j4)(30 20 o )
o
Vth 
(3020 ) 
6  j2  j4
6  j2
(4  90o )(3020o )

6.324  18.43o
 18.97  51.57 o
30
Example 10.9
Find Thevenin equivalent of the circuit as seen from terminals a-b
31
Solution
To find Vth, we apply KCL at
node 1
15  Io  0.5I o
I o  10A
Applying KVL to the loop on
the right-hand side
 Io (2  j4)  0.5Io (4  j3)  Vth  0
or
Thus the Thevenin voltage is
Vth  55  90o V
Vth  10(2  j4)  5(4  j3)  ( j55)
32
Solution
To find Zth, we remove the independent source. Due to the presence of the
dependent current source, connect 3A current source to terminals a-b
At the node apply KCL
3  Io  0.5Io
Io  2A
Applying KVL to the outer
loop
VS  Io (4  j3  2  j4)  2(6  j)
The Thevenin impedance is
Z th 
VS 26  j

 (4  j0.6667)Ω
IS
3
Z th  4.055  9.46o 
33
Practice Problem 10.9
Determine the Thevenin equivalent of the circuit as seen from the
terminals a-b
34
Solution
To find Vth, consider circuit beside
At node 1,
V1
V  V2
 5 1
4  j2
8  j4
 (2  j)V1  50  (1  j0.5)(V1  V2 )
50  (1  j0.5)V2  (3  j0.5)V1
At node 2,
5  0.2VO 
but
…(1)
Thus node 2 become
V1  V2
0
8  j4
VO  V1  V2
5  0.2(V1  V2 ) 
V1  V2
0
8  j4
50
V1  V2 
3  j0.5
…(2)
35
Solution
Substitute equation (2) into equation (1)
3  j0.5
50  (1  j0.5)V2  (3  j0.5)V2  (50)
3  j0.5
50
0  50  (2  j)V2  (35  j12)
37
 2.702  j16.22
V2 
 7.3572.9o
2 j
Vth  V2  7.3572.9o V
36
Solution
To find Zth, we remove the independent
source and insert 1V voltage source
between terminals a-b
At node a,
IS  0.2VO 
But
And
So
VS
8  j4  4  j2
VS  1
8  j4
 VO 
VS
8  j4  4  j2
8  j4
1
2.6  j0.8
IS  (0.2)


12  j2 12  j2
12  j2
37
Solution
Therefore
VS 1
Z th 

IS IS
12  j2
12.1669.46o


2.6  j0.8 2.7217.10o
Z th  4.473  7.64o 
38
Example 10.10
Obtain current Io by using Norton’s Theorem
39
Solution
To find ZN,
i) Short circuit voltage source
ii) Open circuit source
As a result, the (8-j2) and
(10+j4) impedances are short
circuit.
 Z N  5
40
Solution
To find IN,
i) Short circuit terminal a-b
ii) Apply mesh analysis
Notice that mesh 2 and 3 form a
supermesh.
Mesh 1
 j40  (18  j2)I1  (8  j2)I 2  (10  j4)I 3  0 …(1)
Supermesh
(13  j2)I 2  (10  j4)I 3  (18  j2)I1  0
…(2)
41
Solution
At node a, due to the current source between mesh 2 and 3
I3  I 2  3
…(3)
Adding equation (1) and (2) gives
 j40  5I2  0  I2  j8
From equation (3)
I3  I 2  3  3  j8
The Norton current is
I N  I3  (3  j8)A
42
Solution
By using Norton’s equivalent circuit
along with the impedance at terminal
a-b, we could solve for Io.
By using current division
5
IO 
IN
5  20  j15
3  j8

 1.46538.48o A
5  j3
43
Practice Problem 10.10
Determine the Norton equivalent circuit as seen from terminal a-b. Use the
equivalent to find Io
44
Solution
To find ZN,
i) Short circuit voltage source
ii) Open circuit source
ZN =
(4  j2) || (9  j3)
=
(4  j2)(9  j3)
13  j
=
(3.176  j0.706)Ω
45
Solution
To find IN,
i) Short circuit terminal a-b
ii) Solve for IN using mesh
analysis
46
Solution
Supermesh
 20  8I1  (1  j3)I 2  (9  3j)I 3  0
I1  I2  j4
…(1)
…(2)
Mesh 3
(13  j)I 3  8I1  (1  j3)I 2  0
…(3)
Solving for IN
50  j62 79.65  51.11o
o
I N  I2 


8.396


32.68
9  j3
9.487  18.43o
47
Solution
Using Norton equivalent, we
could find Io
ZN
3.176  j0.706
Io 
IN 
(8.396  32.68o )
Z N  10  j5
13.176  j4.294
(3.25412.53o )(8.396  32.68o )
Io 
13.858  18.05o
I o  1.971  2.10 o
48