Chapter III

Circuit Analysis Techniques

• Circuit Reduction & Source Transformation

• Node-Voltage method

• Mesh- Current method.

• Superposition methods.

• Thevenin’s and Norton’s circuits.

• Maximum Power Transfer theorem

The Mesh- Current method

“LOOP Analysis”

Mesh-Current Method (Loop Analysis)

• Nodal analysis was developed by applying KCL at each non-

reference node.

• Mesh-Current method is developed by applying KVL around meshes

in the circuit.

• Loop (mesh) analysis results in a system of linear equations which

must be solved for unknown currents.

• Reduces the number of required equations to the number of meshes

• Can be done systematically with little thinking

• As usual, be careful writing mesh equations – follow sign convention.

Definitions

+

-

A

B C

Mesh: Loop that does not enclose other loops

Branch: Path between 2 essential nodes

How many mesh-currents?

No. of essential nodes Ne = 4

No. of branches

Be = 6

No. of Mesh-currents

M = Be –(Ne-1)

•Enough equations to get unknowns

Steps of Mesh Analysis

1. Identify the number of basic meshes.

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.

Steps of Mesh Analysis

2. Assigning Mesh Currents

Mesh 2

1kW

1kW

1kW

V1 V2Mesh 1

+

–

+

–

1kW

1kW

1kW

V1 V2I1 I2

+

–

+

–

1. Identifying the Meshes

R

I1

+ –VR

VR = I1 R

R

I1

+ –VR

I2

VR = (I1 - I2 ) R

3. Formulation of voltage drop in terms of Mesh Currents

Steps of Mesh Analysis

R2 W

R3 W

R1 W

V1 V2I1 I2

+

–

+

–

-V1 + I1 R1 + (I1 - I2) R3 = 0

I2 R2 + V2 + (I2-I1) R3 = 0

4. Writing Mesh-Current Equations

Example

Find the power supplied by the sources using Mesh current method

M= 3-(2-1) = 2

• Assign mesh currents

• Write mesh equations

i1(12) + (i1-i2)2+ i1(4) -12= 0

i2(9) +8+ i2(3) + (i2-i1) 2= 0

• Solve mesh equations:

• I1 = 0.6129 A & I2 = - 0.4838 A

• P12V = -12 I1 & P9V = 8 I2

Mesh current method Cases

Mesh 1

-10 + 4i1 + 6(i1-i2) = 0

Mesh 2

i2 = - 5A

No need to write a loop equation

Case I: When a current source exists only in one mesh

Case II: Super Mesh

Case II: When a current source exists between two meshes (super Mesh)

4i1 + 2 (i1 – i3) + 2(i2 – i3) + 8i2 = 0

i2 – i1 = 5 A

-10 + 2(i3 – i2 ) + 2(i3 –i1) = 0

𝑖2 − 𝑖1 = 4A

2 𝑖1 − 𝑖3 + 4𝑖1 + 8𝑖2 + 6 𝑖2 − 𝑖4 = 0

6 𝑖4 − 𝑖2 + 𝑖4 + 3 𝑖4 − 𝑖3 = 0

−30 + 2 𝑖3 − 𝑖1 + 3 𝑖3 − 𝑖4 = 0

Example

Source current

Super Mesh equation

Find the power supplied by the voltage source using the Mesh current method

Solving i3 = 8.561 A and the P30V = 30x8.561 = 256.83 W

Case III: Mesh with Dependent Sources

−20 + 4 𝑖1

− 𝑖3

+ 2 𝑖1

− 𝑖2

= 0

4 𝑖3

− 𝑖1

+ 6𝑖3

+ 8 𝑖3

− 𝑖2

= 0

−10𝑖0

+ 2 𝑖2

− 𝑖1

+ 8 𝑖2

− 𝑖3

= 0

In addition to the 3 Mesh equations, an extra equation defining the control parameter

of the dependent source has to be formulated 𝒊𝟎

= 𝒊𝟑

Example

Use Mesh current method to find io

𝑖1 = 4A

3 𝑖2 − 𝑖1 + 2𝑖2 − 5𝑖𝑜 + 𝑖2 − 𝑖3 = 0

5𝑖𝑜 + 𝑖3 − 𝑖2 + 4𝑖3 − 20 + 5 𝑖3 − 𝑖1 = 0

𝑖𝑜 = 𝑖1 − 𝑖3

Example

Use nodal analysis and Mesh current method to find VD and if

Solution (Nodal Analysis)

Solution (Loop Analysis)

35 i1 -10i2 – 5i3 – 20i4 -175 if = 0

-10i1 + 110 i2 + 60 = 0

-5i1 + 205i3 – 200i4 – 60 = 0

-200i3 – 20i1 + 620i4 – 400i5 = 0

I5 = -0.625 VD

VD = 5 (i1 – i 3)

If = (i4 – i 3)