How do you solve a DC circuit using mesh analysis with mesh currents?

In one line

Mesh analysis assigns a loop current to each window (mesh) of a planar circuit, writes KVL for every mesh, and solves the resulting equations for those currents.

Explained simply

Imagine each window of the circuit has its own little conveyor belt of current going round clockwise. Where two windows share a wire, both belts run through it, so the actual current there is the difference of the two. Instead of guessing many branch currents, you only find one current per window.

Technically, a mesh is a loop that contains no other loop inside it. With \( m \) meshes you get \( m \) independent KVL equations in the mesh currents \( I_1, I_2, \dots \). Any branch current is then a mesh current or the difference of two mesh currents.

How it works

1. Check the circuit is planar and count the meshes.
2. Assign clockwise mesh currents \( I_1, I_2, \dots \).
3. Apply KVL around each mesh. For a resistor shared with mesh \( k \), the drop is \( R(I_1 - I_k) \) when writing mesh 1's equation.
4. Add source EMFs with the right sign (rise in the direction of the mesh current is +).
5. Arrange in matrix form and solve (substitution or Cramer's rule).
6. Find branch currents and voltages from the mesh currents.

Shortcut for circuits with only independent voltage sources: diagonal entries = total resistance of the mesh, off-diagonal = minus the shared resistance.

Key formula

For two meshes: \( \begin{bmatrix} R_{11} & -R_{12} \\ -R_{21} & R_{22} \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \end{bmatrix} = \begin{bmatrix} E_1 \\ E_2 \end{bmatrix} \)

\( R_{11}, R_{22} \) = total resistance in each mesh (\( \Omega \)), \( R_{12} = R_{21} \) = shared resistance (\( \Omega \)), \( E_1, E_2 \) = net EMF in each mesh in the direction of its current (V), \( I_1, I_2 \) = mesh currents (A).

Quick example

Mesh 1: 10 V source, 2 \( \Omega \), and a 4 \( \Omega \) resistor shared with mesh 2. Mesh 2: the 4 \( \Omega \) and a 2 \( \Omega \) resistor, no source.

Mesh 1: \( 6I_1 - 4I_2 = 10 \). Mesh 2: \( -4I_1 + 6I_2 = 0 \Rightarrow I_1 = 1.5I_2 \). Substituting: \( 9I_2 - 4I_2 = 10 \Rightarrow I_2 = 2 \) A, \( I_1 = 3 \) A. Current in the 4 \( \Omega \) resistor \( = I_1 - I_2 = 1 \) A.

Common mistakes

• Writing the shared resistor drop as \( R I_1 \) only, forgetting the other mesh current.
• Mixing clockwise and anticlockwise mesh currents, which flips signs of the off-diagonal terms.
• Reporting a mesh current as the current of a shared branch.

How MU asks it

A standard 7–10 mark numerical: "Find the current through the 6 \( \Omega \) resistor using mesh analysis." (MU syllabus excludes supermesh, so only independent voltage sources or current sources on outer branches appear.) For full marks: redraw the circuit with mesh currents marked, write each KVL equation fully, show the matrix, solve neatly, and state the final branch current with direction and unit.