Kirchhoff's Law – Mesh Current Method
As defined earlier, a mesh is the simplest possible closed path in a circuit. If we revisit our now very familiar circuit in figure 10 below, the two inside loops (ABEF and BCDE) are also meshes. The outside loop (ABCDEF) is not a mesh because it contains a branch (segment BE).
A mesh current is a hypothetical current flowing around each mesh. In figure 10, mesh current IA flows around Mesh A starting at V1 to R3 to R1 before returning to V1 without dividing. Mesh current IB flows from V2 to R2 to R3 then back to V2, circumnavigating Mesh B. The fact that mesh currents don't divide at branch points differentiates them from real currents. They are still useful because solving the mesh currents allows you to solve the actual branch currents and the voltage drops.
Mesh Current Equations
Most electrical engineering textbooks use Conventional Current when solving a circuit by the Mesh Current Method. This tutorial will continue to use Electron Flow. We will use Kirchhoff's Voltage Law to define the equations we need. Since there is one mesh current for each mesh in a circuit, we need one mesh current equation per mesh.
The direction chosen for the mesh currents is completely arbitrary. You can have them all going clockwise or all counterclockwise or even some clockwise while others go counterclockwise. I believe you avoid more mistakes by setting all the mesh currents in the same direction, so that is what we will do here. We will assume a counterclockwise path for each mesh current.
As you move around a mesh in the direction of the mesh current, you add the voltage drops caused by the mesh's own current. We will use Mesh A in figure 10 as an example. Begin at any point on Mesh A (we will start at point F). Move around the mesh in the direction of the mesh current flow (counterclockwise). The voltage drops in Mesh A caused by current IA are 10IA for R3 and 20IA for R1. Combined, these two drops equal 30IA.
Mesh A shares a resistor, R3, with Mesh B. Since the mesh current IB flows through R3 in the opposite direction as IA, the 10IB voltage drop across R3 caused by IB must be subtracted in the Mesh A equation. (If IB flowed through R3 in the same direction as IA, you would add the voltage drop caused by IB.)
Finally, we must deal with any voltage sources in a mesh. We don't care about the polarity of voltage drops, but the polarity of each voltage source is important. Since we are using Electron Flow, if the mesh current leaves the negative terminal of a voltage source and returns to the positive terminal, that voltage source is considered positive. Similarly, if the current leaves the positive terminal and returns to the negative terminal, that source is considered negative.
Returning to our Mesh A example from figure 10, since the mesh current IA flows into the negative terminal of V1, that voltage source is considered negative in the mesh equation for Mesh A. We also know that the sum of all the voltage drops in Mesh A must equal the sum of all the voltage sources in that mesh. We get the following mesh current equation for Mesh A:
R1IA + R3IA − R3IB = −V1
(R1 + R3)IA − R3IB = −V1
(20 + 10)IA − 10IB = −130
30IA − 10IB =−130
3IA − IB =−13
Continuing with the figure 10 example, we next calculate the mesh current equation for Mesh B. Here, mesh current IB causes two voltage drops: 5IB across R2 and 10IB across R3. Combined, these voltage drops equal 15IB. Since Mesh A shares resistor R3 with Mesh B, the voltage drop across R3 caused by IA (flowing in the opposite direction) must be subtracted in the Mesh B equation. Finally, all these drops must equal the voltage source, V2 located in Mesh B. Note that this voltage source is considered positive since the IB current flows from the negative to the positive terminals of V2. The mesh current equation for Mesh B is:
R2IB + R3IB − R3IA = V2
(R2 + R3)IB − R3IA = V2
(5 + 10)IB − 10IA = 20
15IB − 10IA = 20
3IB − 2IA = 4
Let's do one last thing. We will rearrange the terms in the Mesh B equation to be in the same order as the Mesh A equation.
-2IA + 3IB = 4
Now we have two equations with two variables and can solve for the mesh currents.
3IA − IB = −13
-2IA + 3IB = 4
Note that each equation contains one term for each mesh in the circuit. Solving the two equations, we get:
IA = −5 amps
IB = −2 amps
The negative signs for IA and IB indicate that we got the direction of both mesh currents wrong. They actually flow clockwise around each mesh! Figure 11 below shows the correct direction for both mesh currents.
We now know the mesh currents for the circuit, but where do we go from here? Well, we know that the entire real branch current I1 flows through R1 and that the entire mesh current IA does the same. Therefore, I1 and IA must have the same value. Similarly, the entire I2 branch current flows through R2 as does IB, so they also have the same value. We also know that branch current I3 flowing through R3 is a combination of IA and IB which flow in the opposite directions to one another. Therefore,
I3 = IA − IB = 5 − 2 = 3 amps in the direction from node B to node E. Now that we've solved all the branch currents, just use Ohm's Law to solve the voltage drops and we're done!
A Three Mesh Example
For this last example, we will tackle a slightly more difficult circuit shown in figure 12, below. It is made up of three meshes and the center mesh, Mesh B, has no voltage source. What do we do about that? If there is no voltage source in a mesh, then the value of the voltage source for that mesh is zero. With that in mind, the problem is only slightly more difficult. Since it has three meshes, it will require three mesh current equations (each with three terms) to solve.
We formulate the mesh current equations for each mesh as we've done before, but for clarity, we will put each term inside a set of brackets .
[(R6 + R7 + R1)IA] − [R7IB] +  = V1
[−R7IA] + [(R2 + R7 + R5 + R8)IB] − [R8IC] = 0
 − [R8IB] + [(R3 + R8 + R4)IC] = −V2
You should notice that the third term in the Mesh A equation is zero. This is because Mesh A does not share any voltage drops with Mesh C and that is noted by placing a zero as a placeholder in the equation. The same thing is true for the first term in the Mesh C equation (Mesh C shares nothing with Mesh A). Also note that the Mesh B equation equals zero because it contains no voltage source. Substituting known values and simplifying, we get:
15IA − 5IB + 0 = 100
−5IA + 20IB − 5IC = 0
0 − 5IB + 15IC = −50
We now have three equations with three unknowns and can solve for the mesh currents:
IA = 7
IB = 1
IC = −3
From here, we just solve for the branch currents and voltage drops as before.