By

Hayder Hamzah

University of Babylon, Hillah, Iraq

haydermenkash@yahoo.com

+9647814471323

Mesh Analysis

1. Nodal method

2. Mesh method

3. Superposition method

4. Matrix method

1. Ohm's law

2. Kirchhoff's Voltage Law

Where

R: resistance

V: voltage

I: current

V4 = V1+ V2 + V3

V4 -V1- V2 - V3 = 0

Basic Circuits

Mesh Analysis: Basic Concepts:

In formulating mesh analysis we assign a mesh

current to each mesh.

Mesh currents are sort of fictitious in that a particular

mesh current does not define the current in each branch

of the mesh to which it is assigned.

I1 I2 I3

Basic Circuits

Mesh Analysis: Basic Concepts:

R1

Rx

R2

+

_ I1 I2

+

_VA VB

+ +

+

_

_

_V1

VL1

V2

Figure 7.2: A circuit for illustrating mesh analysis.

AXX

XL

AL

VIRIRRso

RIIVRIVwhere

VVV

211

211111

11

)(,

;

Eq 7.1

Around mesh 1:

Basic Circuits Mesh Analysis: Basic Concepts:

R1

Rx

R2

+

_ I1 I2

+

_VA VB

+ +

+

_

_

_V1

VL1

V2

BVIRXRIXRor

BVIRXRIXR

givesinEqngSubstituti

RIVXRIILVwith

BVVLV

havewemeshAround

2)2(1

2)2(1

,2.73.7

222;)12(1;

21

2

Eq 7.2

Eq 7.3

Eq 7.4

Basic Circuits

Mesh Analysis: Basic Concepts:

We are left with 2 equations: From (7.1) and (7.4)

we have,

AXX VIRIRR 211 )(

BXX VIRRIR 221 )(

Eq 7.5

Eq 7.6

We can easily solve these equations for I1 and I2.

Basic Circuits

Mesh Analysis: Basic Concepts:

The previous equations can be written in matrix form as:

B

A

XX

XX

B

A

XX

XX

V

V

RRR

RRR

I

I

or

V

V

I

I

RRR

RRR

1

2

1

2

1

2

1

2

1

(

)(

(

)(Eq (7.7)

Eq (7.8)

Basic Circuits Mesh Analysis: Example 7.1.

Write the mesh equations and solve for the currents I1, and I2.

+

_10V

4 2

6 7

2V20V

I1 I2+

+_

_

Figure 7.2: Circuit for Example 7.1.

Mesh 1 4I1 + 6(I1 – I2) = 10 - 2

Mesh 2 6(I2 – I1) + 2I2 + 7I2 = 2 + 20

Eq (7.9)

Eq (7.10)

Basic Circuits

Mesh Analysis: Example 7.1, continued.

Simplifying Eq (7.9) and (7.10) gives,

10I1 – 6I2 = 8

-6I1 + 15I2 = 22

Eq (7.11)

Eq (7.12)

» % A MATLAB Solution

»

» R = [10 -6;-6 15];

»

» V = [8;22];

»

» I = inv(R)*V

I =

2.2105

2.3509

I1 = 2.2105

I2 = 2.3509

Basic Circuits Mesh Analysis: Example 7.2

Solve for the mesh currents in the circuit below.

+

_

6

10

9

11

3

4

20V 10V

8V

12V

I1 I2

I3

+

+

__

_

_

+

+_

Figure 7.3: Circuit for Example 7.2.

The plan: Write KVL, clockwise, for each mesh. Look for a

pattern in the final equations.

Basic Circuits Mesh Analysis: Example 7.2

+

_

6

10

9

11

3

4

20V 10V

8V

12V

I1 I2

I3

+

+

__

_

_

+

+_

Mesh 1: 6I1 + 10(I1 – I3) + 4(I1 – I2) = 20 + 10

Mesh 2: 4(I2 – I1) + 11(I2 – I3) + 3I2 = - 10 - 8

Mesh 3: 9I3 + 11(I3 – I2) + 10(I3 – I1) = 12 + 8

Eq (7.13)

Eq (7.14)

Eq (7.15)

Basic Circuits Mesh Analysis: Example 7.2

Clearing Equations (7.13), (7.14) and (7.15) gives,

20I1 – 4I2 – 10I3 = 30

-4I1 + 18I2 – 11I3 = -18

-10I1 – 11I2 + 30I3 = 20

In matrix form:

20

18

30

3

2

1

301110

11184

10420

I

I

I

WE NOW MAKE AN IMPORTANT

OBSERVATION!!

Standard Equation form

Basic Circuits Mesh Analysis: Standard form for mesh equations

Consider the following:

R11 =

of resistance around mesh 1, common to mesh 1 current I1.

R22 = of resistance around mesh 2, common to mesh 2 current I2.

R33 = of resistance around mesh 3, common to mesh 3 current I3.

)3(

)2(

)1(

3

2

1

333231

232221

131211

emfs

emfs

emfs

I

I

I

RRR

RRR

RRR

Basic Circuits Mesh Analysis: Standard form for mesh equations

R12 = R21 = - resistance common between mesh 1 and 2

when I1 and I2 are opposite through R1,R2.

R13 = R31 = - resistance common between mesh 1 and 3

when I1 and I3 are opposite through R1,R3.

R23 = R32 = - resistance common between mesh 2 and 3

when I2 and I3 are opposite through R2,R3.

)1(emfs = sum of emf around mesh 1 in the direction of I1.

)2(emfs = sum of emf around mesh 2 in the direction of I2.

)3(emfs = sum of emf around mesh 3 in the direction of I3.

Basic Circuits Mesh Analysis: Example 7.3 - Direct method.

20V

10V

15V

30V

20

10

30

10

12

8

+_

I1 I2 I3+

+

+

_

__

Use the direct method to write the mesh equations for the following.

Figure 7.4: Circuit diagram for Example 7.3.

15

25

10

3

2

1

30100

105010

01030

I

I

I

Eq (7.13)

Basic Circuits Mesh Analysis: With current sources in the circuit

Example 7.4: Consider the following:

10V

20V

4A

10

5

20

2

+

_

15

+

_

I1 I2

I3

Figure 7.5: Circuit diagram for Example 7.4.

Use the direct method to write the mesh equations.

10V

20V

4A

10

5

20

2

+

_

15

+

_

I1 I2

I3

Basic Circuits Mesh Analysis: With current sources in the circuit

This case is explained by using an example.

Example 7.4: Find the three mesh currents in the circuit below.

Figure 7.5: Circuit for Example 7.4.

When a current source is present, it will be directly related to

one or more of the mesh current. In this case I2 = -4A.

Basic Circuits Mesh Analysis: With current sources in the circuit

Example 7.4: Continued. An easy way to handle this case is to

remove the current source as shown below. Next, write the mesh

equations for the remaining meshes.

Note that I 2 is retained for writing the equations through the

5 and 20 resistors.

10V

20V

10

5

20

2

+

_

15

+

_

I1 I2

I3

Basic Circuits Mesh Analysis: With current sources in the circuit

Example 7.4: Continued.

10V

20V

10

5

20

2

+

_

15

+

_

I1 I2

I3 Equation for mesh 1:

10I1 + (I1-I2)5 = 10

or

15I1 – 5I2 = 10

Equations for mesh 2:

2I3 + (I3-I2)20 = 20

or

- 20I2 + 22I3 = 20

Constraint Equation

I2 = - 4A

Basic Circuits Mesh Analysis: With current sources in the circuit

Example 7.4: Continued. Express the previous equations in

Matrix form:

1

2

3

15 5 0 10

0 20 22 20

0 1 0 4

I

I

I

I1 = -0.667 A

I2 = - 4 A

I3 = - 2.73 A

End of Lesson 7

circuits

Mesh Analysis