Mesh analysis is a circuit solving procedure using mesh current. In nodal analysis we apply KCL to find unknown voltages. But here using KVL we will find unknown currents using this method. We will use mesh currents instead of element current. Mesh analysis reduces the equations getting current. Nodal analysis can be applied in each general circuit but mesh analysis cannot be applied in each general circuit. Mesh analysis only be applied if the circuit is planer.

What is planer or non-planer circuit? A planner circuit is one that can be drawn in a plane with no branches crossing one another, otherwise it is non-planer.

As example see figure 1 (a), the circuit has two crossing branches. It can be redrawn as figure 1(b) having no change. But if you want to redraw the figure 2 circuit it is impossible. Hence figure 1 (a) is planer and figure 2 is non-planer. In planer circuit mesh analysis can be used. In figure 2 mesh analysis cannot be used but we can use nodal analysis here.

What is meant by mesh?

**A mesh is a loop which does not contain any other loops within it. **

Consider circuit in figure 3. We will find mesh current for this circuit.

There are two loops so we have two mesh paths are abefa and bcdeb.

The current through a mesh is called mesh current. Here two mesh currents are i_{1} and i_{2}. We are finding mesh current with voltage source. Mesh analysis with current source is also possible. But we are doing with voltage source here in figure 3 circuit.

**Three steps to determine mesh current**

- Assigning mesh currents as i
_{1}, i_{2},………i_{n}to the n meshes. - Applying KVL in each n meshes and using ohm’s law.
- Solving the equation in n meshes to get mesh currents.

**First step:** In figure 2 circuit there are two mesh currents are i_{1} and i_{2} are assigned meshes 1 and meshes 2.

**Second step: **Applying KVL in each mesh. The direction of the mesh current is arbitrary clockwise or counter clock-wise does not affect the validity of the solution.

In mesh 1, – V_{1} + R_{1}i_{1} + R_{3}i_{3} = 0

Or – V_{1} + R_{1}i_{1} + R_{3 }(i_{1} – i_{2}) = 0

Or (R_{1} + R_{3}) i_{1} – R_{3}i_{2} = V_{1 }—————— (i)

In mesh 2, R_{2}i_{2} + V_{2} + R_{3 }(i_{2 }– i_{1}) = 0

Or – R_{3}i_{1} + (R_{2} + R_{3}) i_{2} = – V_{2} ———————–(ii)

**Third step: **From equation (i) and (ii) we will get mesh current. You can use method of substitution or matrix.

Note that here mesh current and branch currents are different unless the mesh is isolated. To differ between the two types of currents we use I for a branch current and i for a mesh current. It is clear from figure 3

I_{1} = i_{1}, I_{2} = i_{2}, I_{3} = i_{1} – i_{2}

**Problem solution of mesh analysis **

Calculate the mesh current i_{1} and i_{2} for figure 4.

For mesh 1 applying KVL,

-15 + 5i_{1} + 10 (i_{1} – i_{2}) + 10 = 0

Or 3i_{1} – 2i_{2} = 1 ————– (a)

For mesh 2, 6i_{2} + 4i_{2} + 10 (i_{2 }– i_{1}) – 10 = 0

Or i_{1 }= 2i_{2 }– 1 ——————(b)

From equation (a) and (b) using substitution method,

6i_{2} – 3 – 2i_{2} = 1

Or i_{2} = 1A

Putting the value of i_{2} in equation (b),

I_{1} = i_{1 }= 1A

Thus I_{2} = i_{2} = 1A