Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (7))
-
٠٦
-
Network Solution :-
To solve a circuit is to find the current and voltage in all branches.
1) Loop ( Mesh ) current method :-
Example( 1 ):-
Find the current through the 10 Ω resistor of the network
shown:
5Ω
2Ω
3Ω
8Ω
10Ω
15V
I
1
I
3
I
2
Solution :-
-
ﻣﺟﻣوع ا
ﻟﻣﻘﺎوﻣﺎت
*
ـﺗﯾﺎر اﻟ
Loop
+
اﻟﻣﻘﺎوﻣﺔ اﻟﻣﺷﺗرﻛﺔ
*
ـﺗﯾﺎر اﻟ
Loop
اﻟﻣﺷﺗرك
اﻟﻔوﻟﺗﯾﺔ
)
ﺣﺳب اﺗﺟﺎﻫﻬﺎ ﻣﻊ
ـﺗﯾﺎر اﻟ
Loop
= (
ﺻﻔر
.
The loop equations are :-
Loop 1 :-
- ( 8+3 )I
1
+ 3I
2
+8I
3
+ 15 = 0
Loop 2 :-
- ( 3+5+2 )I
2
+ 3I
1
+5I
3
= 0
Loop 3 :-
- ( 10+8+5 )I
3
+ 8I
1
+5I
2
= 0
Rearrange the equations , then :-
-11I
1
+ 3I
2
+8I
3
= -15
3I
1
- 10I
2
+5I
3
= 0
8I
1
+ 5I
2
- 23I
3
= 0
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (7))
-
١٦
-
A
I
I
A
D
D
I
22
.
1
22
.
1
23
5
8
5
10
3
8
3
11
0
5
8
0
10
3
15
3
11
10
3
3
3
Example( 2 ):- Solve following circuit diagram;
Solution :-
-I
1
( 5+7 ) + 7I
2
+ 20
– 5 = 0
-I
2
( 7+2+6 ) + 7I
1
+ 6I
3
+ 5 + 5 + 5 = 0
-I
3
( 6+8 ) + 6I
2
– 5 – 30 = 0
Rearrange;
-12I
1
+ 7I
2
+0 = -15
7I
1
- 15I
2
+6I
3
= -15
0 + 6I
2
- 14I
3
= 35
-------------------
( 1 )
-------------------
( 2 )
-------------------
( 3 )
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (7))
-
٢٦
-
A
D
D
I
862
.
1
1402
2610
14
6
0
6
15
7
0
7
12
14
6
35
6
15
15
0
7
15
1
1
A
D
D
I
049
.
1
1402
1470
1402
14
35
0
6
15
7
0
15
12
2
2
A
D
D
I
05
.
2
1402
2875
1402
35
6
0
15
15
7
15
7
12
3
3
Example( 3 ):- Find the current in the 10V source , for the following network;
Solution :-
I
2
= -5 A
Hence , we need only one equation to solve this circuit
-I
1
( 4+6 ) + 6 * ( -5 ) + 10 = 0
-10I
1
– 20 = 0
-10I
1
= 20
A
I
2
10
20
1
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (7))
-
٣٦
-
Example( 4 ):- Solve the following circuit diagram, also find the voltage across
15
Ω resistance?
Solution:-
I
4
= 2 A
-I
1
( 4+6+5 ) + 4I
2
+ 12
– 18 = 0
-I
2
( 8+3+4+7 ) + 4I
1
+ 8I
3
- 12 = 0
-I
3
( 15+2+8+9 ) + 8I
2
+ 15 * 2 = 0
Rearrange:-
-15I
1
+ 4I
2
+0 = 6
4I
1
- 22I
2
+8I
3
= 12
0 + 8I
2
- 34I
3
= -30
-------------------
( 1 )
-------------------
( 2 )
-------------------
( 3 )
1 8 V
1 2 V
7 O
9 O
2 A
I
1
I
2
I
3
I
4
15 Ω
2 Ω
9 Ω
3 Ω
7 Ω
6 Ω
4 Ω
5 Ω
8 Ω
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (7))
-
٤٦
-
D
D
I
1
1
D
D
I
2
2
D
D
I
3
3
V
15
= I
15
* R
15
= ( I
3
– I
4
) * 15
= ( I
3
– 2 ) * 15
Example( 5 ):- Solve the following circuit diagram .
3Ω
15Ω
40Ω
5Ω
60Ω
20A
10A
5A
Solution:- The above diagram can be reduced to the following diagram;
أﻛﻣل اﻟﺣل
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (7))
-
٥٦
-
-I
1
( 5+15+60 ) + 60I
2
- 50
– 75 = 0
-I
2
( 3+60+40 ) + 60I
1
+ 60 = 0
Rearrange:-
-80I
1
+ 60I
2
= 125
60I
1
- 103I
2
= -60
D
D
I
1
1
D
D
I
2
2
Example( 6 ):- Solve the following circuit diagram:
Solution:-
-------------------
( 1 )
-------------------
( 2 )
أﻛﻣل اﻟﺣل
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (7))
-
٦٦
-
-( 6+2 )I
1
+ 2I
2
+20 - V
o
= 0
-( 10+2+4 )I
2
+ 2I
1
+ V
o
= 0
I
2
- I
1
= 6
Add eq. 1 & eq. 2
-8I
1
+ 2I
2
+20 -16 I
2
+ 2I
1
= 0
-6I
1
- 14I
2
= -20
From eq. 3
I
1
– I
2
= -6
A
D
D
I
2
.
3
14
6
84
20
1
1
14
6
1
6
14
20
1
1
A
D
D
I
8
.
2
20
20
36
20
6
1
20
6
2
2
-------------------
( 1 )
-------------------
( 2 )
-------------------
( 3 )
-------------------
( 1 )
-------------------
( 2 )