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CHAPTER 3
NETWORK THEOREM
2
NETWORK THEOREM
Superposition Theorem Source Transformation Thevenin & Norton Equivalent
3
SUPERPOSITION THEOREM
4
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.
5
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
6
Practice Problem 10.5Find current in the circuit using the superposition theorem
7
Solution
0j4Ij2)I(8 21
03010j4Ij4)I(6 o12
12 j2)I(0.5I
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 circuitFor mesh 1
For mesh 2
…(1)
…(2)
8
Solution
j0.5560.08I'I 1o
o11 3010j4Ij2)Ij4)(0.5(6
j1411
3010I
o
1
Substitute equation (1) into equation (2)
9
Solution
2j81Z
53.0j4164.077.0j846.9
)769.2j846.1)(2(
4j||62 Z
For Io” consider circuit beside where the voltage source is short circuitLet
769.2j846.14j6
24j
)0(2ZZ
Z"I o
21
2o
10
Solution
A45.651939.1 o
"I'II ooo
086.1j4961.0
Therefore
11
Practice Problem 10.6Calculate Vo in the circuit using superposition theorem
12
Solution
o03030sin(5t)
)81.12t4.631sin(5'v oo
oo 81.12-4.631(30)
j1.258
j1.25'V
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)
By voltage division
j)(0.2F
5jH1
13
Solution
ot 02)10cos(2
(2)ZZ
ZI
21
2
For vo” we remove voltage source and replace with short circuitTransfer the circuit to frequency domain ( = 10)
By current division
j0.5)(F2.0 10jH1
Let
j0.5)(Z1 j10||8Z2
j3.94.878j108
j80
j0.1382.096
14
Solution
)86.24t1.05cos(10)81.12t4.631sin(5 oo "v'vv ooo
)86.23t1.05cos(10"v oo
Therefore
Thus
1o IZ"V j0.5)j0.138)((2.096
o23.86050.1
15
SOURCE TRANSFORMATION
16
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.
17
Example 10.7Calculate Vx in the circuit by using source transformation
18
Solution
j4)A(9045
9020I o
o
S
If we transform the voltage source to a current source, we obtain the circuit as shown.
The parallel combination of 5 resistance and (3+j4) impedance gives a new equivalent impedance Z1 j1.25)Ω(2.5
j48
j4)5(3Z1
19
Solution
j1.25)j4)(2.5(ZIV 1SS
V285.519V oX
Convert the current source to a voltage source yields the circuit as below
We then could solve for VX by using voltage division
j10)V(5
j10)(5j134j1.252.510
10VX
20
Practice Problem 10.7Find Io in the circuit by using the concept of source transformation
21
Solution
16j123j)(j4)(4ZIV SSS
If we transform the current source to a voltage source, we obtain the circuit as shown.
22
Solution
j2j34Z
We transform the voltage source to a current source as shown below
j2-6j31
j26
j1612
S
SS Z
VI
j)(13
10
j36
j2)(j5)(6j5||Z
Let Then
Note that
23
Solution
By current division
j3)(1.5j2)(1j)(1
310
j)(13
10
Io
o
o
j
j
1.17602.13
56.11672.44
413
4020
A46.99288.3 o
24
THEVENIN & NORTON EQUIVALENT CIRCUIT
25
Equivalent Circuit
Thevenin Equivalent Circuit
Norton Equivalent Circuit
26
Relationship
NNth IZV Keep in mind that the two equivalent circuit are related as
Nth ZZ and
Vth = VOC = open circuit voltage
IN = ISC = short circuit current
27
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
NoteWhen there is dependent source or sources with difference frequencies, the step to find the equivalent is not straight forward.
28
Practice Problem 10.8Find the Thevenin equivalent at terminals a-b of the circuit
29
Solution
j26
j2)j4)(6(10
j3.22.410
j2)(6||j4)(10Z th
To find Zth, set voltage source to zero
j3.212.4
30
Solution
o
oo
43.18324.6
)2030)(904(
o57.5197.18
j26
)20j4)(30()20(30
j4j26
j4V
oo
th
To find Vth, open circuit at terminals a-b
31
Example 10.9Find Thevenin equivalent of the circuit as seen from terminals a-b
32
Solution
oo 0.5II15
To find Vth, we apply KCL at node 1
Applying KVL to the loop on the right-hand side
0Vj3)(40.5Ij4)(2I thoo
j55)(j3)5(4j4)10(2Vth
V9055V oth or
10AIo
Thus the Thevenin voltage is
33
Solution
oo 0.5II3 At the node apply KCL
Applying KVL to the outer loop
j)2(6j4)2j3(4IV oS
j0.6667)Ω(4
3
j62
I
VZ
S
Sth
o
th 46.9055.4Z
The Thevenin impedance is
2AIo
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
34
Practice Problem 10.9Determine the Thevenin equivalent of the circuit as seen from the terminals a-b
35
Solution
j48
VV5
j24
V 211
j0.53
50VV 21
0j48
VV)V0.2(V5 21
21
To find Vth, consider circuit besideAt node 1,
Thus node 2 become
)Vj0.5)(V(150j)V(2 211
12 j0.5)V(3j0.5)V(150
0j48
VV0.2V5 21
O
At node 2,
21O VVV but
…(1)
…(2)
36
Solution
V72.97.35VV o2th
Substitute equation (2) into equation (1)
j0.53
j0.53(50)j0.5)V(3j0.5)V(150 22
j12)(3537
50j)V(2500 2
o2 72.97.35
j2
j16.222.702V
37
Solution
j24j48
V0.2VI S
OS
SO Vj24j48
j48V
j212
j0.82.6
j212
1
j212
j48(0.2)IS
To find Zth, we remove the independent source and insert 1V voltage source between terminals a-b
At node a,
So
1VS But
And
38
Solution
oth 7.64473.4Z
Therefore
SS
Sth I
1
I
VZ
o17.102.72
46.9166.12
j0.82.6
j212
o
39
Example 10.10Obtain current Io by using Norton’s Theorem
40
Solution
5ZN
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.
41
Solution
0j4)I(10j2)I(8j2)I(18j40 321
To find IN,
i) Short circuit terminal a-b
ii) Apply mesh analysis
Notice that mesh 2 and 3 form a supermesh.
Mesh 1
0j2)I(18j4)I(10j2)I(13 132 Supermesh
…(1)
…(2)
42
Solution
j833II 23
At node a, due to the current source between mesh 2 and 3
3II 23
j8I05Ij40 22
j8)A3(II 3N
Adding equation (1) and (2) gives
The Norton current is
From equation (3)
…(3)
43
Solution
NO Ij15205
5I
A38.481.465j35
j83 o
By using Norton’s equivalent circuit along with the impedance at terminal a-b, we could solve for Io.
By using current division
44
Practice Problem 10.10Determine the Norton equivalent circuit as seen from terminal a-b. Use the equivalent to find Io
45
Solution
j3)(9||j2)(4
j13
j3)j2)(9(4
To find ZN,
i) Short circuit voltage source
ii) Open circuit source
ZN =
j0.706)Ω(3.176
=
=
46
Solution
To find IN,
i) Short circuit terminal a-b
ii) Solve for IN using mesh
analysis
47
Solution
03j)I(9j3)I(18I20 321
j4II 21
0j3)I(18Ij)I(13 213
Supermesh
Mesh 3
Solving for IN
oo
o
2N 32.688.39618.439.487
51.1179.65
j39
j6250II
…(1)
…(2)
…(3)
48
Solution
Using Norton equivalent, we
could find Io
o
oo
o 18.0513.858
)32.68)(8.39612.53(3.254I
oo 2.101.971I
)32.68(8.396j4.29413.176
j0.7063.176I
j510Z
ZI o
NN
No
49
Problem 11.14
a) Determine Thevenin equivalent circuit looking from the load, Z.
b) Determine load, Z that will produce maximum power transfer.
c) Value of the maximum power.
(is = 5 cos 40t A)
50
Problem 10.36 (Buku Electric Circuit by Nilsson & Riedel)
a) Determine Thevenin equivalent circuit looking from the load, Z.
b) Determine load, Z that will produce maximum power transfer.
c) Value of the maximum power.
51
Problem 11.15
a) Determine Thevenin equivalent circuit looking from the load, ZL.
b) Determine load, ZL that will produce maximum power transfer.
c) Value of the maximum power.