INC 111 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(