Mesh Analysis Example with Solution

In this article, we will discuss mesh analysis or mesh current method with the solved examples.

Mesh Analysis is a simple circuit analysis technique. It helps in solving electrical circuits with the help of linear equations. The mesh analysis method of circuit analysis is based on the KVL (Kirchhoff’s VoltageLaw).

In the case of mesh analysis, the solution of an electric circuit is obtained by assuming mesh currents that do not split at a junction but complete their paths around meshes.

There can be three types of cases of mesh analysis:

• Mesh analysis of circuits having only voltage sources and current sources which are convertible into voltage sources.
• Mesh analysis of circuits having voltage sources and current sources which are not convertible into voltage sources.
• Mesh analysis of circuits having supermesh.

Let us understand each of these cases of mesh analysis with the help of numerical examples in detail.

Case 1 – Mesh analysis of circuits having only voltage sources and current sources convertible to voltage sources:

Consider an electric circuit shown in figure-1. Using mesh analysis or the loop current method, we will find the electric currents I₁ and I₂.

Solution –

The KVL equation for mesh 1 is,

`\6-4I_1-8(I_1-I_2 )-10=0`

`\⇒-12I_1+8I_2=4" "…(1)`

The KVL equation for mesh 2 is,

`\-2+10-8(I_2-I_1 )-2I_2=0`

`\⇒8I_1-10I_2=-8" "…(2)`

By rearranging equation (2), we get,

`\I_1=(10I_2-8)/8" "…(3)`

On substituting the value of current I_1­ from equation (3) into equation (1), we get,

`\-12((10I_2-8)/8)+8I_2=4`

On solving this equation, we get,

`\I_2=8/7" A"`

Also, by putting this value of current I₂ into equation (3), we get,

`\I_1=((10×8/7)-8)/8`

`\∴I_1=3/7" A"`

In this way, we can use mesh analysis to find the solution of an electric circuit which contains voltage sources and current sources convertible into voltage sources.

Case 2 – Mesh analysis of circuits having voltage sources and current sources which are not convertible into voltage sources:

Consider an electric circuit shown in figure-2. Using mesh analysis or the loop current method, we will find the electric currents I₁ and I₂.

Solution –

The current in mesh 3 is,

`\I_3=5" A"`

The KVL equation for mesh 1 is,

`\10-5I_1-2(I1-I_2 )=0`

`\⇒7I_1-2I_2=10" "…(4)`

The KVL equation for mesh 2 is,

`\-6-2(I_2-I_1 )-5(I_2+I_3 )=0`

`\⇒2I_1-7I_2=31" "…(5)`

By rearranging equation (5), we get,

`\I_1=(31+7I_2)/2" "…(6)`

On substituting the value of current I₁ from equation (6) into equation (5), we get,

`\2((31+7I_2)/2)-7I_2=31`

On solving this equation, we get,

`\I_2=(-197)/45" A"`

Here, the negative sign represents that the actual direction of current I₂ is opposite to the assumed direction.

On substituting the value of current I₂ in equation (6), we have,

`\I_1=(31+7((-197)/45))/2`

`\∴I_1=8/45" A"`

In this way, we can apply mesh analysis or mesh current method to solve an electric circuit which is having voltage sources and current sources that are not convertible into voltage sources.

Case 3 – Mesh analysis of circuit having supermesh:

Consider an electric circuit shown in figure-3. Using mesh analysis or the loop current method, we will find the electric currents I₁ and I₂.

Solution –

The KVL equation for supermesh is,

`\12-2I_1-2I_2-2I_2=0`

`\⇒2I_1+4I_2=12`

`\⇒I_1+2I_2=6" "…(7)`

The KCL equation for the current source is,

`\I_1+2=I_2`

`\⇒I_1=I_2-2" "…(8)`

On substituting the value of current I₁ from equation (8) into equation (7), we get,

`\(I_2-2)+2I_2=6`

`\⇒3I_2=8`

`\∴I_2=8/3" A"`

Also, putting this value of current I₂ into equation (8), we get,

`\I_1=8/3-2`

`\∴I_1=2/3" A"`

In this way, we can apply the mesh analysis method to solve an electric circuit which has a current source in a common branch, i.e. form a supermesh.