Dr. Sameir Abd Alkhalik Aziez
University of Technology ( Lecture (9))
-
٩٧
-
Network Theorems:-
1- Superposition Theorem:-
In any circuit network contain more than one sources ( voltage or
current ) to find the current ( or voltage ) in a certain part of a network , remove
the sources of the network and find the current ( or voltage ) in the existence of
only one source each time. The resultant current ( or voltage ) will be the
algebraic sum of current ( or voltage ) due to all sources when acting
independently once a time .
(Removing the sources means:- Short circuiting the voltage source and open
circuiting the current source) .
Example 1:- In the following circuit diagram, find all branch current's using
superposition theorem:-
3Ω
6Ω
7Ω
25V
3A
4Ω
I
5
I
2
I
1
I
4
I
3
Solution :-
1.) Effect of 25 V source :-
Dr. Sameir Abd Alkhalik Aziez
University of Technology ( Lecture (9))
-
٠٨
-
2.) Effect of 3 A source :-
A
I
I
I
I
I
9
.
0
2
.
1
1
.
2
4
5
2
1
3
3.) Superpose :-
A
I
I
I
6
.
4
1
.
2
5
.
2
1
1
1
A
I
I
I
3
.
1
5
.
2
2
.
1
2
2
2
A
I
I
I
3
.
4
8
.
1
5
.
2
5
2
5
A
I
I
I
6
.
1
9
.
0
5
.
2
4
1
4
A
I
I
I
9
.
5
9
.
0
5
3
3
3
A
I
5
.
2
3
7
25
1
A
I
5
.
2
6
4
25
2
A
I
I
I
5
2
1
3
A
I
1
.
2
3
7
7
*
3
1
A
I
9
.
0
3
7
3
*
3
4
A
I
2
.
1
6
4
4
*
3
2
A
I
8
.
1
6
4
6
*
3
5
Dr. Sameir Abd Alkhalik Aziez
University of Technology ( Lecture (9))
-
١٨
-
Example 2:- For the following circuit network, find the current in all branches,
using superposition theorem:-
Solution:-
1.) Effect of 150 V source:-
30Ω
60Ω
40Ω
150V
I'
1
I'
2
I'
3
60
30
60
30
*
60
40
T
R
A
I
5
.
2
60
150
1
A
I
83
.
0
30
60
30
*
5
.
2
2
A
I
67
.
1
30
60
60
*
5
.
2
3
Dr. Sameir Abd Alkhalik Aziez
University of Technology ( Lecture (9))
-
٢٨
-
2.) Effect of 270 V source:-
54
60
40
60
*
40
30
T
R
A
I
5
54
270
1
A
I
2
60
40
40
*
5
2
A
I
3
60
40
60
*
5
3
3.) Superpose :-
A
I
I
I
5
.
0
3
5
.
2
3
1
1
A
I
I
I
83
.
2
2
83
.
0
2
2
2
A
I
I
I
33
.
3
5
67
.
1
1
3
3
Dr. Sameir Abd Alkhalik Aziez
University of Technology ( Lecture (9))
-
٣٨
-
Example 3:- Find the current in all branch in the following circuit diagram:-
4Ω
15V
8Ω
5Ω
2A
12V
I
4
I
2
I
3
I
1
I
5
Solution:-
1.) Effect of 15 V source:-
A
I
12
.
2
8
5
8
*
5
4
15
1
A
I
3
.
1
8
5
8
*
12
.
2
2
A
I
82
.
0
8
5
5
*
12
.
2
3
2.) Effect of 12 V source:-
Dr. Sameir Abd Alkhalik Aziez
University of Technology ( Lecture (9))
-
٤٨
-
8Ω
5Ω
12V
I"
2
I"
1
I"
3
4Ω
A
I
57
.
1
4
8
4
*
8
5
12
1
A
I
52
.
0
4
8
4
*
57
.
1
2
A
I
04
.
1
4
8
8
*
57
.
1
3
3.) Effect of 2 A source:-
74
.
1
5
1
4
1
8
1
1
T
T
R
R
V
R
V
T
o
5
.
3
*
2
A
V
I
o
88
.
0
4
1
Dr. Sameir Abd Alkhalik Aziez
University of Technology ( Lecture (9))
-
٥٨
-
A
V
I
o
7
.
0
5
2
A
V
I
o
44
.
0
8
3
A
I
I
12
.
1
2
1
4
A
I
I
3
.
1
2
2
5
3.) Superpose :-
A
I
I
I
I
04
.
0
12
.
1
04
.
1
12
.
2
4
3
1
1
A
I
I
I
I
97
.
0
7
.
0
57
.
1
3
.
1
2
1
2
2
A
I
I
I
I
03
.
1
3
.
1
57
.
1
3
.
1
5
1
2
3
A
I
I
I
I
9
.
0
44
.
0
52
.
0
82
.
0
3
2
3
4
A
I
I
I
I
96
.
1
88
.
0
04
.
1
12
.
2
1
3
1
5