INC 111 Basic Circuit Analysis

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INC 111 Basic Circuit Analysis Week 4 Mesh Analysis

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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. Mesh Analysis. Procedure Count the number of meshes. Let the number equal N. Define mesh current on each mesh. Let the values be I1, I2, I3, … - PowerPoint PPT Presentation

Transcript of INC 111 Basic Circuit Analysis

Page 1: INC 111 Basic Circuit Analysis

INC 111 Basic Circuit Analysis

Week 4

Mesh Analysis

Page 2: INC 111 Basic Circuit Analysis

Mesh Analysis (Loop Analysis)

Mesh = A closed loop path which has no smaller loops inside

1Ω2Ω

2A3V

Page 3: INC 111 Basic Circuit Analysis

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

Page 4: INC 111 Basic Circuit Analysis

Example

42V

10V

Use mesh analysis to find the power consumption in the resistor 3 Ω

I1 I2

Mesh current (loop current)

Page 5: INC 111 Basic Circuit Analysis

42V

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

Page 6: INC 111 Basic Circuit Analysis

Example

7V

6V3Ω

1Ω 2Ω

+ Vx -

Use Mesh analysis to find Vx

I1

I2

I3

Page 7: INC 111 Basic Circuit Analysis

7V

6V3Ω

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

Page 8: INC 111 Basic Circuit Analysis

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.

Page 9: INC 111 Basic Circuit Analysis

7V3Ω

1Ω 2Ω

+ Vx -

7A

ExampleUse Mesh analysis to find Vx

I1

I2

I3

Page 10: INC 111 Basic Circuit Analysis

033261

0)32(322)12(1

III

IIIIIEquation from 2nd loop

7V3Ω

1Ω 2Ω

+ Vx -

7A

I1

I2

I3

Page 11: INC 111 Basic Circuit Analysis

7V3Ω

1Ω 2Ω

+ Vx -

7A

Equation 2

I1

I2

I3

Supermesh

731

734241

03)23(3)21(17

II

III

IIIII

Equation 3

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I1 = 9A

I2 = 2.5A

I3 = 2A

Vx = 3(I3-I2) = -1.5V

Page 13: INC 111 Basic Circuit Analysis

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.

Page 14: INC 111 Basic Circuit Analysis

Example

7V3Ω

1Ω 2Ω

+ Vx -

7A

From the previous example, if we want to use Nodal Analysis

0V

7V

V1 V2

V3

Page 15: INC 111 Basic Circuit Analysis

Example: Dependent Source

1Ω 2Ω

+ Vx -15A

1/9 Vx

Find Vx

I1

I2

I3

Page 16: INC 111 Basic Circuit Analysis

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

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Special Techniques

• Superposition Theorem

• Thevenin’s Theorem

• Norton’s Theorem

• Source Transformation

Page 18: INC 111 Basic Circuit Analysis

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.

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Linearity

V and I have linear relationship

I

V

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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

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Example

2V

1V

1V

2V

1ΩI total I1 I2

I1 = 1A

I2 = 2A

I total = 1+2 = 3A

Page 22: INC 111 Basic Circuit Analysis

Example

2V

2V

1ΩI total I1 I2

1A

1A

I1 = 1A

I2 = 0A

I total = 1+0 = 1A

Page 23: INC 111 Basic Circuit Analysis

Example

42V

10V+

Vx-

Find voltage Vx

Page 24: INC 111 Basic Circuit Analysis

42V

4Ω+

Vx-

V

Vx V

333.9

42)7/12(6

)7/12(42

)4||3(6

)4||3()42(

Page 25: INC 111 Basic Circuit Analysis

V

Vx V

333.3

1042

210

4)3||6(

)3||6()10(

6Ω 4Ω

10V+

Vx-

Page 26: INC 111 Basic Circuit Analysis

42V

10V+

Vx-

V

VxVxVx VV

6333.3333.9

)10()42(

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Superposition andDependent Source

Dependent sources cannot be used with superposition.

They have to be active all the time.

Page 28: INC 111 Basic Circuit Analysis

Example

10V

3A +-2Ix

Ix

Use superposition to find Ix

Page 29: INC 111 Basic Circuit Analysis

10V

+-2Ix

Ix

Find Ix by eliminating the current source 3A

KVL

2

021210

)10(

VIx

IxIxIx

Page 30: INC 111 Basic Circuit Analysis

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

Page 31: INC 111 Basic Circuit Analysis

10V

3A +-2Ix

Ix

A

IxIxIx AV

4.1)6.0(2

)3()10(