# Easy Tutorial Mesh and Supermesh for AC Circuit Analysis

Contents

Like nodal and supernode for ac circuit that have the same procedure as in the dc circuit, mesh and supermesh for ac circuit will do the same.

Learning Kichhoff’s laws for AC circuit will lead us to:

1. Node and supernode for ac circuit
2. Mesh and supermesh for ac circuit
3. Superposition for ac circuit
4. Source transformation for ac circuit
5. Thevenin and Norton for ac circuit

## Mesh and Supermesh for AC Circuit

Kirchhoff’s voltage law (KVL) forms the basis of mesh analysis. The validity of KVL for ac circuit can be seen in Mesh Analysis for dc circuit here.

Keep in mind that the very nature of using mesh analysis is that it is to be applied to the planar circuit.

Make sure to read what is ac circuit first.

## Mesh Analysis of AC Circuit

Like the nodal analysis of ac circuit, we will not cover the basic explanation of its principle. You have to read the link above in order to understand ‘how to use it’ properly.

Let us review the examples below to be able to analyze the ac circuit using mesh analysis.

Determine current Io in the circuit of Figure.(1) using mesh analysis. Figure 1. Example of mesh analysis for ac circuit

Applying KVL to mesh 1, we obtain (1.1)

For mesh 2, (1.2)

For mesh 3, I3 = 5. Substituting this to Equations.(1.1) and (1.2), we obtain (1.3) (1.4)

Equations.(1.3) and (1.4) can be rewritten in matrix form as

and its determinants are

The desired current is

Read also : energy in a coupled circuit

## Supermesh Analysis of AC Circuit

Next, we will analyze an ac circuit with supermesh within it.

Before doing that, make sure you have understood the principle of supermesh analysis first.

Solve for Vo in the circuit of Figure.(2) with mesh analysis. Figure 2. Example of supermesh analysis for ac circuit

As shown above, meshes 3 and 4 form a supermesh due to the current source between the meshes. For mesh 1, KVL results

or (2.1)

For mesh 2, (2.2)

For the supermesh, (2.3)

Due to the current source between meshes 3 and 4, at node A, (2.4)

We reduce the above four equations to two by elimination, instead of solving them directly.

Combining Equations.(2.1) and (2.2), (2.5)

Combining Equations.(2.2) to (2.4), (2.6) Figure 3. Analysis circuit in Figure.(2)

From Equations.(2.5) and (2.6), we obtain the matrix equation

We obtain the determinants

Current I1 is obtained as

The required voltage Vo is