# Electronics/Mesh Analysis

## Meshes

A 'mesh' (also called a loop) is simply a path through a circuit that starts and ends at the same place. For the purpose of mesh analysis, a mesh is a loop that does not enclose other loops.

## Mesh Analysis

Similar to nodal analysis, mesh analysis is a formalized procedure based on KVL equations. A caveat: mesh analysis can only be used on 'planar' circuits (i.e. there are no crossed, but unconnected, wires in the circuit diagram.)

Steps:

1. Draw circuit in planar form (if possible.)
2. Identify meshes and name mesh currents. Mesh currents should be in the clockwise direction. The current in a branch shared by two meshes is the difference of the two mesh currents.
3. Write a KVL equation in terms of mesh currents for each mesh.
4. Solve the resulting system of equations.

## Complication in Mesh Analysis

1. Dependent Voltage Sources

Solution: Same procedure, but write the dependency variable in terms of mesh currents.

2. Independent Current Sources

Solution: If current source is not on a shared branch, then we have been given one of the mesh currents! If it is on a shared branch, then use a 'super-mesh' that encircles the problem branch. To make up for the mesh equation you lose by doing this, use the mesh current relationship implied by the current source (i.e. ${\displaystyle I_{2}-I_{1}=4mA}$ ).

3. Dependent Current Sources

Solution: Same procedure as for an independent current source, but with an extra step to eliminate the dependency variable. Write the dependency variable in terms of mesh currents.

## Example

Given the Circuit below, find the currents ${\displaystyle I_{1}}$ , ${\displaystyle I_{2}}$ .

The circuit has 2 loops indicated on the diagram. Using KVL we get:
Loop1: ${\displaystyle 0=9-1000I_{1}-3000(I_{1}-I_{2})}$
Loop2: ${\displaystyle 0=3000(I_{1}-I_{2})-2000I_{2}-2000I_{2}}$
Simplifying we get the simultaneous equations:
${\displaystyle 0=9-4000I_{1}+3000I_{2}}$
${\displaystyle 0=0+3000I_{1}-7000I_{2}}$
solving to get:
${\displaystyle I_{1}=3.32mA}$
${\displaystyle I_{2}=1.42mA}$