Lecture 4

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INC 111 Basic Circuit Analysis
Week 4
Mesh Analysis
Mesh Analysis (Loop Analysis)
Mesh = A closed loop path which has no smaller loops inside
5Ω
3V
2Ω
1Ω
2A
Mesh Analysis
Procedure
1. Count the number of meshes. Let the number equal N.
2. Define mesh current on each mesh. Let the values be
I1, I2, I3, …
3. Use Kirchoff’s voltage law (KVL) on each mesh,
generating N equations
4. Solve the equations
Example
Use mesh analysis to find the power consumption in the resistor 3 Ω
6Ω
42V
I1
4Ω
3Ω
I2
Mesh current (loop current)
10V
6Ω
42V
Loop 1
I1
3Ω
I2
10V
 42  6 I1  3( I1  I 2)  0
9 I1  3I 2  42
Loop 2
4Ω
Equation 1
3( I 2  I1)  4 I 2  10  0
 3I1  7 I 2  10
Equation 2
I1 = 6A, I2 = 4A, The current that pass through R 3Ω is 6-4 = 2A (downward)
Power = 12 W
Example
Use Mesh analysis to find Vx
1Ω
7V
I2
2Ω
+ Vx -
I1
3Ω
6V
I3
2Ω
1Ω
1Ω
7V
I2
2Ω
 7  1( I 1  I 2)  6  2( I 1  I 3)  0
3I 1  I 2  2 I 3  1 Equation 1
+ Vx -
I1
3Ω
6V
I3
2Ω
1Ω
1( I 2  I 1)  2 I 2  3( I 2  I 3)  0
 I 1  6 I 2  3I 3  0 Equation 2
2( I 3  I 1)  6  3( I 3  I 2)  I 3  0
 2 I 1  3I 2  6 I 3  6 Equation 3
I1 = 3A, I2 = 2A, I3 = 3A
Vx = 3(I3-I2) = 3V
Supermesh
When there is a current source in the mesh path, we cannot use KVL
because we do not know the voltage across the current source.
We have to use supermesh, which is a combination of 2 meshes to be a
big mesh, and avoid the inclusion of the current source in the mesh path.
Example
Use Mesh analysis to find Vx
1Ω
7V
I2
2Ω
+ Vx -
I1
3Ω
7A
I3
2Ω
1Ω
1Ω
7V
I2
2Ω
+ Vx -
I1
3Ω
7A
I3
1Ω
2Ω
1( I 2  I1)  2 I 2  3( I 2  I 3)  0
 I 1  6 I 2  3I 3  0
Equation from 2nd loop
Supermesh
1Ω
7V
I2
2Ω
+ Vx -
I1
3Ω
7A
I3
1Ω
2Ω
 7  1( I1  I 2)  3( I 3  I 2)  I 3  0
Equation 2
I1  4 I 2  4 I 3  7
I1  I 3  7
Equation 3
I1 = 9A
I2 = 2.5A
I3 = 2A
Vx = 3(I3-I2) = -1.5V
How to choose between
Node and Mesh Analysis
The hardest part in analyzing circuits is solving
equations. Solving 3 or more equations can be time
consuming.
Normally, we will count the number of equations
according to each method and select the method
that have lesser equations.
Example
From the previous example, if we want to use Nodal Analysis
7V
1Ω
V1
7V
2Ω
+ Vx -
V2
3Ω
7A
V3
1Ω
2Ω
0V
Example: Dependent Source
Find Vx
1Ω
I2
2Ω
+ Vx 15A
I1
3Ω
1/9 Vx
I3
2Ω
1Ω
I1  15
1Ω
I2
2Ω
1( I 2  I1)  2 I 2  3( I 2  I 3)  0
+ Vx 15A
I1
 I 1  6 I 2  3I 3  0
3Ω
1/9 Vx
I3
Equation 1
1Ω
2Ω
1
I 3  I1  Vx
9
Vx  3( I 3  I 2)
I1=15A, I2=11A, I3=17A
Vx = 3(17-11) = 18V
Equation 2
Equation 3
Equation 4
Special Techniques
• Superposition Theorem
• Thevenin’s Theorem
• Norton’s Theorem
• Source Transformation
Superposition Theorem
In a linear circuit, we can calculate the value of
current (or voltage) that is the result from each
voltage source and current source independently.
Then, the real value is the sum of all current (or
voltage) from the sources.
Linearity
V and I have linear relationship
I
V
Implementation
When calculating the effect of a source, the other sources
will be set to zero.
• For voltage sources, when set as 0V, it will be similar to
short circuit
• For current sources, when set as 0A, it will be similar to
open circuit
Example
1V
1V
1Ω
I total
1Ω
1Ω
I1
2V
2V
I1 = 1A
I2 = 2A
I total = 1+2 = 3A
I2
Example
1A
1A
1Ω
I total
1Ω
1Ω
I1
2V
2V
I1 = 1A
I2 = 0A
I total = 1+0 = 1A
I2
Example
Find voltage Vx
6Ω
42V
4Ω
3Ω
+
Vx
-
10V
6Ω
42V
4Ω
3Ω
+
Vx
-
(3 || 4)
(12 / 7)
Vx ( 42V ) 
 42 
 42
6  (3 || 4)
6  (12 / 7)
 9.333V
6Ω
4Ω
3Ω
+
Vx
-
10V
(6 || 3)
2
Vx (10V )  
10  
10
(6 || 3)  4
24
 3.333V
6Ω
42V
4Ω
3Ω
+
Vx
-
Vx  Vx( 42V )  Vx(10V )
 9.333 3.333  6V
10V
Superposition and
Dependent Source
Dependent sources cannot be used with superposition.
They have to be active all the time.
Example
Use superposition to find Ix
Ix
2Ω
10V
1Ω
3A
2Ix
+
-
Find Ix by eliminating the current source 3A
Ix
2Ω
10V
KVL
1Ω
2Ix
+
-
 10  2Ix  1Ix  2Ix  0
Ix(10V )  2
Find Ix by eliminating the voltage source 10V
Ix+3
Ix
2Ω
1Ω
3A
KVL outer loop
2Ix
+
-
2 Ix  1( Ix  3)  2 Ix  0
Ix(3 A)  0.6
Ix
2Ω
10V
1Ω
3A
2Ix
Ix  Ix(10V )  Ix(3 A)
 2  (0.6)  1.4 A
+
-
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