The mesh current method simplifies and shortens the solution of problems by allowing you to write only Kirchhoff's Voltage Law. Kirchhoff's Current Law is implicit. Consider the circuit below.

Using branch currents, we can write KVL:

-V_{S} + i_{1}R_{1} + i_{3}R_{3}
= 0

-i_{3}R_{3} + i_{2}R_{2} = 0

We can also write KCL at the top node, which gives:

-i_{1} + i_{2} + i_{3} = 0

Which gives:

i_{3} = i_{1} - i_{2}

Substituting into KVL gives:

-V_{S} + i_{1}R_{1} + (i_{1} - i_{2})R_{3}
= 0

(i_{2} - i_{1})R_{3} + i_{2}R_{2}
= 0

With the mesh current method, we skip several steps, and just
write the two equations above. i_{1} and i_{2}
are then considered to be mesh currents rather than branch
currents. Nevertheless, if an element is not shared by two
meshes, such as R_{1}, the mesh current is the same as
the branch current.

Now consider a more complicated circuit.

We want to write mesh current equations for this circuit, so
we define currents in the three meshes. The top and right meshes
are labeled I_{1} and I_{2}, respectively. The
left mesh can be labeled 6 A, since we already know the current.
We also define a voltage across the 4 A source. Next, we write KVL
equations for the two unknown mesh currents. The convention for
doing this is to put sources on the left, and loads on the right.

5 - V_{I} = 2(I_{1} - 6)

V_{I} = 3(I_{2} - 6) + 7I_{2}

We need a third equation, one relating 4 A to the mesh
currents. Observe that the branch current in the 4 A source is
made up of two mesh currents, I_{2} in the direction of
the 4 A current, and I_{1} opposite to the 4 A current.
Therefore:

4 = I_{2} - I_{1}

With three equations and three unknowns, we can solve for all the unknowns.

Another solution technique uses the concept of a "supermesh", and saves the writing of one equation. The circuit below is the same as before, but it is labeled differently.

The 4 A source is unlabeled (it still has a voltage, but we don't have to know what it is). The supermesh is shown by the red line. We write KVL around this supermesh as follows:

5 = 7I_{2} + 3(I_{2} - 6) + 2(I_{1} -
6)

We still need the equation for the current in the 4 A source:

4 = I_{2} - I_{1}

Now we have two equations and two unknowns. Use either solution method you wish.

Before going on to the homework, you should complete
Tutorial 5
on **the mesh current method**.

- Use the mesh current method to write enough equations to
solve for the mesh currents. Do not solve.

- E1 = 195 V, E2 = 20 V, R1 = 5 Ω, R2 = 40 Ω,
R3 = 20 Ω, R4 = 50 Ω,
R5 = 80 Ω. Use the mesh current method to find I. The answer is in
integer.

- Ix = 5 A, Iy = 7 A, E = 76 V, R1 = 12 Ω, R2
= 20 Ω, R3 = 8 Ω.
Use the mesh current method to find V. The answer is in
integer.

- Is = 1.5I, E1 = 88 V, E2 = 52 V, R1 = 7 Ω, R2
= 3 Ω, R3 = 10 Ω, R4
= 30 Ω. Use the mesh current method to find I. The answer is in
integer.