INC 112 Basic Circuit Analysis
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Transcript of INC 112 Basic Circuit Analysis
INC 112 Basic Circuit Analysis
Week 4
Mesh Analysis
Mesh Analysis (Loop Analysis)
Mesh = A closed loop path which has no smaller loops inside
1Ω2Ω
5Ω
2A3V
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
3Ω
6Ω
42V
4Ω
10V
Use mesh analysis to find the power consumption in the resistor 3 Ω
I1 I2
Mesh current (loop current)
3Ω
6Ω
42V
4Ω
10VI1 I2
102713
01024)12(3
422319
0)21(31642
II
III
II
III
Equation 1
Equation 2
Loop 1
Loop 2
I1 = 6A, I2 = 4A, The current that pass through R 3Ω is 6-4 = 2A (downward)Power = 12 W
Example
7V
6V3Ω
1Ω 2Ω
1Ω
2Ω
+ Vx -
Use Mesh analysis to find Vx
I1
I2
I3
7V
6V3Ω
1Ω 2Ω
1Ω
2Ω
+ Vx -
I1
I2
I3
6362312
03)23(36)13(2
033261
0)32(322)12(1
132213
0)31(26)21(17
III
IIIII
III
IIIII
III
IIII
Equation 1
Equation 2
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.
7V3Ω
1Ω 2Ω
1Ω
2Ω
+ Vx -
7A
ExampleUse Mesh analysis to find Vx
I1
I2
I3
033261
0)32(322)12(1
III
IIIIIEquation from 2nd loop
7V3Ω
1Ω 2Ω
1Ω
2Ω
+ Vx -
7A
I1
I2
I3
7V3Ω
1Ω 2Ω
1Ω
2Ω
+ Vx -
7A
Equation 2
I1
I2
I3
Supermesh
731
734241
03)23(3)21(17
II
III
IIIII
Equation 3
I1 = 9A
I2 = 2.5A
I3 = 2A
Vx = 3(I3-I2) = -1.5V
How to choose betweenNode 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
7V3Ω
1Ω 2Ω
1Ω
2Ω
+ Vx -
7A
From the previous example, if we want to use Nodal Analysis
0V
7V
V1 V2
V3
Example: Dependent Source
3Ω
1Ω 2Ω
1Ω
2Ω
+ Vx -15A
1/9 Vx
Find Vx
I1
I2
I3
3Ω
1Ω 2Ω
1Ω
2Ω
+ Vx -15A
1/9 Vx
I1
I2
I3
)23(39
113
033261
0)32(322)12(1
151
IIVx
VxII
III
IIIII
I
Equation 1
Equation 2
Equation 3
Equation 4
I1=15A, I2=11A, I3=17A
Vx = 3(17-11) = 18V
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
2V
1V
1Ω
1V
1Ω
2V
1ΩI total I1 I2
I1 = 1A
I2 = 2A
I total = 1+2 = 3A
Example
2V
1Ω
2V
1ΩI total I1 I2
1A
1Ω
1A
I1 = 1A
I2 = 0A
I total = 1+0 = 1A
Example
3Ω
6Ω
42V
4Ω
10V+
Vx-
Find voltage Vx
3Ω
6Ω
42V
4Ω+
Vx-
V
Vx V
333.9
42)7/12(6
)7/12(42
)4||3(6
)4||3()42(
V
Vx V
333.3
1042
210
4)3||6(
)3||6()10(
3Ω
6Ω 4Ω
10V+
Vx-
3Ω
6Ω
42V
4Ω
10V+
Vx-
V
VxVxVx VV
6333.3333.9
)10()42(
Superposition andDependent Source
Dependent sources cannot be used with superposition.
They have to be active all the time.
Example
2Ω
10V
1Ω
3A +-2Ix
Ix
Use superposition to find Ix
2Ω
10V
1Ω
+-2Ix
Ix
Find Ix by eliminating the current source 3A
KVL
2
021210
)10(
VIx
IxIxIx
2Ω 1Ω
3A +-2Ix
Ix
Find Ix by eliminating the voltage source 10V
Ix+3
KVL outer loop
6.0
02)3(12
)3(
AIx
IxIxIx
2Ω
10V
1Ω
3A +-2Ix
Ix
A
IxIxIx AV
4.1)6.0(2
)3()10(