Mesh Analysis - aast.edu

Electrical Circuits I

Emam Fathy

Department of Electrical and Control Engineering

email: emfmz@aast.edu

http://www.aast.edu/cv.php?disp_unit=346&ser=68525

Lecture 4

Mesh Analysis

Mesh Analysis

• Nodal analysis was developed by applying

KCL at each non-reference node

• Mesh analysis is developed by applying

KVL around loops in the circuit

• Mesh (loop) analysis results in a system of

linear equations which must be solved for

unknown currents

Steps of Mesh Analysis

1. Identify mesh (loops).

2. Assign a current to each mesh.

3. Apply KVL around each loop to get an

equation in terms of the loop currents.

4. Solve the resulting system of linear

equations for the mesh/loop currents.

Example

+

–

Vout

1kW

1kW

1kW

V1 V2

+

–

+

–

• Find Vout

1. Identifying the Meshes

Mesh 2

1kW

1kW

1kW

V1 V2Mesh 1+

–

+

–

2. Assigning Mesh Currents

1kW

1kW

1kW

V1 V2I1 I2

+

–

+

–

3. KVL Around Mesh 1

I1 1kW + (I1 – I2) 1kW = V1

Or I1 (1kW+ 1kW) – I2 1kW = V1

1kW

1kW

1kW

V1 V2I1 I2

+

–

+

–

3. KVL Around Mesh 2

1kW

1kW

1kW

V1 V2I1 I2

+

–

+

–

(I2 – I1) 1kW + I2 1kW = –V2

Or – I1 1kW + I2 (1kW + 1kW) = – V2

Solution

• The two equations can be combined into a

single matrix/vector equation

I1 (1kW+ 1kW) – I2 1kW = V1

– I1 1kW + I2 (1kW + 1kW) = – V2

W+WW

WW+W

2

1

2

1

k1k1k1

k1k1k1

V

V

I

I

4. Solving the Equations

Let: V1 = 7V and V2 = 4V

Results:

I1 = 3.33 mA

I2 = –0.33 mA

Finally

Vout = (I1 – I2) 1kW = 3.66V

Example

1kW

2kW

2kW

12V 4mA

2mA

I0

+

–

Mesh 2

Mesh 3

Mesh 1

1. Identify Meshes

1kW

2kW

2kW

12V 4mA

2mA

I0

+

–

2. Assign Mesh Currents

I1 I2

I31kW

2kW

2kW

12V 4mA

2mA

I0

+

–

The

Supermesh

surrounds

this source!

The

Supermesh

does not

include this

source!

12V 4mA

1kW

2kW

2kW

2mA

I0

I1 I2

I3

+

–

Supermesh

I2 = –4 mA

I1 – I3 = 2 mA

I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 12V KVL Around the Supermesh

A Supermesh results when 2 meshes have a current

source in common.

Solution

• The three equations can be combined into

a single matrix/vector equation

W+WWWW

V12

mA2

mA4

1k2k2k1k2k

101

010

3

2

1

I

I

I

I1 = 1.2 mA, I2 = – 4 mA, I3 = – 0.8 mA

I0 = I1 – I2 = 5.2 mA

Example

Find power dissipated in 12W-resistor and 3W-resistor

using mesh analysis.

12 V 8 V

2 W 9 W

12 W

4 W 3 W

Solution:

12 V 8 V

2 W 9 W

12 W

4 W 3 W

I1

I2

KVL I1:

18I1 – 12I2 = 12

KVL I2:

-12I1 + 24I2 = -8

Example

R1

R4

12 V

R2

R3

I1

I3

I2I

s

I3 = -Is

End of Lec