Mesh Analysis in AC Circuits

Mesh analysis is a circuit analysis technique that turns the currents in a circuit into a series of simultaneous equations using Kirchhoff's Voltage Law. Mesh analysis allows you to find the current or voltage anywhere in the circuit by solving for the current at every point in the circuit.
Mesh analysis only works on circuits called "planar" circuits which means they can be drawn without any wires crossing over another (unless they join at the crossing point).
The steps to performing mesh analysis are:
  1. Find the impedance for all resistors, capacitors and inductors
  2. Replace all elements with their phasor representations
  3. Assign each loop in the circuit a current
  4. Write Kirchhoff's voltage law for each loop
  5. Solve the equations simultaneously to find all the currents

Use mesh analysis to analyse the following circuit and find the current through the capacitor:

First we find the impedances of each of the elements. Since \(\omega = 2\pi\): \(Z_C = -j\frac{1}{\omega C} = -j159\Omega\) And the impedances of the resistors are of course, unchanged. Now we replace all elements in the circuit with their phasor representations. The voltage sources are replaced with DC symbols just to avoid any confusion with changing voltage polarities during the analysis. Now we assign each of the loops in the circuit (that don't contain another loop) a "loop current". Next we write Kirchhoff's voltage law around our first loop with current \(I_1\). \(3\angle 0^\circ - 1000\angle 0^\circ I_1 - 159\angle -90^\circ (I_1-I_2) = 0\) The impedance of the capacitor was multiplied by \(I_1-I_2\) because the current through it is given by \(I_c = I_1-I_2\). Now the second equation for the second loop: \(-1\angle 0^\circ - 159\angle -90^\circ (I_2-I_1) - 200\angle 0^\circ I_2 = 0\) If we solve these two equations simultaneously we'll find that: \(I_1 = 0.0025\angle -15.7^\circ\) \(I_2 = 0.0038\angle 119^\circ\) I won't solve them here (or anywhere else on this page) to avoid adding confusing detail. If you're not sure how to solve simultaneous equations go now and learn, you're going to need it. Finally the current running through the capacitor is: \(I_c = I_1 - I_2 = 0.0058\angle -44^\circ\) \(I_c = 0.0058\cos(2\pi t - 44^\circ)\)
It's a lot of work to do mesh analysis on a circuit. However without mesh analysis your job would be longer and the steps less clear. There are a few special cases we need to add. We need to know what to do in the presence of current sources, especially when they're in between two loops.
If there is a current source on a branch that is part of only one loop, the current in that loop is given by the current source.

Find the loop current \(I_1\) in the following circuit:

The current source \(I = 5\)A only touches the loop with current \(I_1\) which means that \(I_1 = 5\)A.
If there is a current source on a branch shared by two different loops we:
  1. Replace it with an open circuit and write Kirchhoff's voltage law around the two loops sharing the current source.
  2. Write an equation linking the currents of two loops that share the source

Find the loop currents in the following circuit:

Since the current source \(I = 3\angle 0^\circ\) is shared by both \(I_1\) and \(I_2\) we replace the current source with an open circuit and write KVL around both \(I_1\) and \(I_2\). \(-159\angle -90^\circ I_1 - 1k\angle 0^\circ I_1 - 200\angle 0^\circ I_2 - 2\angle 0^\circ = 0\) Then write \(I_2 - I_1 = 2\angle 0^\circ\) as our second equation. Solving both equations simultaneously gives us: \(I_1 = 0.329\angle 172.5^\circ\) \(I_2 = 1.67\angle 1.5^\circ\)
practice problems