Electronics/Mesh Analysis

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Electronics | Foreword | Basic Electronics | Complex Electronics | Electricity | Machines | History of Electronics | Appendix | edit

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Contents 1 Meshes 2 Mesh Analysis 3 Complication in Mesh Analysis 4 Example

[edit] 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.

[edit] 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.

[edit] 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. I₂ − I₁ = 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.

[edit] Example

Given the Circuit below, find the currents I₁, I₂.

The circuit has 2 loops indicated on the diagram. Using KVL we get:
Loop1: 0 = 9 - 1000I₁ - 3000(I₁ - I₂)
Loop2: 0 = 3000(I₁ - I₂) - 2000I₂ - 2000I₂
Simplifying we get the simultaneous equations:
0 = 9 - 4000I₁ + 3000I₂
0 = 0 + 3000I₁ - 7000I₂
solving to get:
I₁ = 3.32mA
I₂ = 1.42mA