Mesh Analysis (AC Analysis)

MESH ANALYSIS:

Another method of analyzing circuits and the counterpart of nodal analysis is the mesh analysis. Also known as loop analysis or the mesh-current method. A mesh is a loop which does not contain any other loops within it. And we can recall that a loop in an electronic circuit is a closed path with no node passed more than once. The current through a mesh is known as mesh current.

Mesh analysis makes use of Kirchhoff’s Voltage Law (KVL) in forming equations to solve for unknown currents in a given circuit. If it’s Vx for nodal analysis, it’s Ix for mesh analysis. Don’t you find it ironic that nodal analysis applies KCL to find unknown voltages, while mesh analysis applies KVL to find unknown current? Though mesh analysis isn’t quite as general as nodal analysis because it is only applicable to planar circuits**.

**one that can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar.

Here are the steps in determining mesh currents:

Assign the meshes found in the circuit as mesh currents i1, i2, … , in.
Apply KVL to each of the assigned meshes to form equations. Use Ohm’s law (V=IR) to express the voltages in terms of the mesh currents.
Solve the formed equations from step 2 to get the value of the unknown mesh currents.
Let’s take this very familiar circuit for this tutorial:

Assuming that all the values for the elements are already given. And that we are supposed to determine the value of the voltage across the capacitor Z3 (Vz3).

Again, as mentioned above, a mesh is a loop which does not contain any other loops within it. By looking at this simple circuit, we can already say that there are two meshes present in the circuit. The bigger loop outside the circuit cannot be considered as a mesh, because it contains loops within it. The direction of the mesh current can either be clockwise or counterclockwise and it will not affect the validity of the solution, as long as you follow the rules in mesh analysis. And we will encounter those rules as we proceed in solving this circuit.

I have assumed that both meshes i1 and i2 loop at a clockwise direction. You can have meshes at different directions (one goes clockwise, another goes counter) but for ethicality purposes, it’s better to have them at the same direction.

To determine Vz3, we have to first get the current that passes through the capacitor Z3. Then we apply Ohm’s law to solve for Vz3. If you look at Z3, you can observe that there are two currents that are passing through it but not in the same direction as each other: i1 and i2. We can also see that i2 passes through a current source Is but in a different position. This means we can conclude that i2 = -Is.

In nodal analysis, the voltage of a non-reference node directly connected to a voltage source is already equal to that source. Unless the node is connected to another voltage source, then that’d be a supernode. In mesh analysis, the current of a mesh that passes through a current source is already equal to that source (positive if they go the same direction, otherwise negative). Unless the mesh shares the current source with another mesh, then that’d be a supermesh.

So, now that we know mesh i2 = -Is, we’ll only need the current of mesh i1. And we can form its equation using KVL this way:

@mesh i1:

+Vs = 0

In nodal analysis, a current source is considered a constant* and its sign depends on its direction. In mesh analysis, the voltage source is the one considered as a constant* and its sign depends on the direction of the mesh that enters it.

*unless it’s a dependent source.

In this case, Vs is positive since mesh i1 came out of the positive terminal of the source (this is one of the rules I was talking about earlier). Moving on as mesh i1 passes through the resistor Z1..

+Vs – (i1 · Z1) = 0

You might think that the i1 · Z1 is quite familiar. That is actually just Ohm’s law, where V = I · R. So, it’s Vz1 = i1 · Z1. See the resemblance? Now as mesh i1 passes through the capacitor Z3..

+Vs – (i1 · Z1) – Z3(i1 – i2) = 0

As you can see from the last expression of the equation, mesh i1 is subtracted by mesh i2 and is multiplied to Z3. This is because both meshes go through the capacitor but with a difference in direction. Maybe you’d add the meshes if they go to the same direction, but i’ve never tried it before since i’m ethical in solving my circuits.

..and since i2 = -Is..

+Vs – (i1 · Z1) – Z3(i1 – (-Is)) = 0

..or..

+Vs – (i1 · Z1) – Z3(i1 + Is) = 0

..simplifying further..

Vs – (i1 · Z1 ) – (i1 · Z3) – (Is · Z3) = 0

Vs – (Is · Z3) = (i1 · Z1) + (i1 · Z3)

Vs – (Is · Z3) = i1(Z1 + Z3)

[Vs – (Is · Z3)] / (Z1 + Z3) = i1

i1 = [Vs – (Is · Z3)] / (Z1 + Z3)

And now that we have solved i1, we substitute it to:

Vz3 = Z3 · (i1 – i2)

..or..

Vz3 = Z3 · (i1 – Is)

Tadaa! And that is how you do mesh analysis, without the presence of a supermesh. I hope you were able to comprehend all that. If you have some questions, leave them below!