**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 I_{1}
and I_{2}.

**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,

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

On solving this equation, we get,

`\I_2=8/7" A"`

Also, by putting this value of current I_{2}
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 I_{1}
and I_{2}.

**Solution**–

The current in mesh 3 is,

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,

On substituting the value of current I_{1}
from equation (6) into equation (5), we get,

On solving this equation, we get,

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

Here, the negative sign represents that
the actual direction of current I_{2} is opposite to the assumed
direction.

On substituting the value of current I_{2}
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 I_{1}
and I_{2}.

**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 I_{1}
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 I_{2}
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.