Mesh-Current Method

One KVL equation per mesh gives you every loop current in the circuit.

Circuit AnalysisElectrical Engineering Year 1
⏱️ About 18 min

A circuit has two loops sharing a resistor, and you need every branch current — without writing six separate KCL equations.

💡
The big idea: Assign one current per mesh, let shared resistors carry the difference, and write one KVL equation per mesh.
🎯 By the end, you'll be able to
  • Identify meshes in a planar circuit and assign clockwise mesh currents
  • Write KVL equations for each mesh in terms of mesh currents
  • Solve the resulting system for all mesh currents
  • Extract any branch current from the mesh current solutions
📎 Helpful to know first

From Node Voltages to Loop Currents

The node-voltage method uses KCL to solve for voltages. The mesh-current method is its dual: it uses KVL to solve for loop currents. Instead of one equation per node, you write one equation per mesh — an innermost loop of the circuit. For circuits with many nodes but few loops, mesh analysis can be significantly faster.

🔑 Mesh vs. loop

A mesh is a loop that contains no other loops inside it — an innermost window of the circuit. Mesh analysis applies only to planar circuits (circuits that can be drawn with no crossing wires).

The Procedure

Step 1 — Identify meshes. Find every innermost loop in the planar circuit.

Step 2 — Assign mesh currents. Label each mesh with a clockwise current (I1, I2, …).

Step 3 — Write KVL per mesh. Walk clockwise around each mesh. For a resistor that belongs to only one mesh, the voltage drop is R·I. For a shared resistor between two meshes, the net current is the difference of the two mesh currents — if both flow clockwise, the shared resistor carries (I1 − I2) in the direction of I1.

Step 4 — Solve the resulting linear system for all mesh currents, then extract any branch current as an algebraic sum of the mesh currents through that branch.

\[ V_{\text{drop}} = R\,(I_1 - I_2) \]
Voltage drop across a resistor shared by two clockwise meshes. The net current through it is the difference of the two mesh currents.
📝 Worked example: A planar circuit has two adjacent meshes. The left mesh contains a 10 V source (positive terminal up) and R1 = 2 Ω along the top. The middle branch (shared by both meshes) has R2 = 2 Ω. The right mesh has R3 = 4 Ω along the top. Assign clockwise mesh currents I1 (left) and I2 (right). Find I1, I2, and the current through the shared resistor R2.
  1. Step 1 — Two meshes identified. Assign clockwise I1 (left) and I2 (right).
  2. Step 2 — KVL mesh 1 (clockwise): 10 − 2·I1 − 2·(I1 − I2) = 0
  3. Step 3 — Simplify: 10 − 4·I1 + 2·I2 = 0 → 2·I1 − I2 = 5 …(1)
  4. Step 4 — KVL mesh 2 (clockwise): −2·(I2 − I1) − 4·I2 = 0 → 2·I1 − 6·I2 = 0 → I1 = 3·I2 …(2)
  5. Step 5 — Substitute (2) into (1): 2(3·I2) − I2 = 5 → 5·I2 = 5 → I2 = 1 A
  6. Step 6 — Back-substitute: I1 = 3(1) = 3 A
  7. Step 7 — Current through R2 (downward): I1 − I2 = 3 − 1 = 2 A
✓ I1 = 3 A, I2 = 1 A, current through R2 = 2 A (downward)

Extracting Branch Currents

Mesh currents are mathematical constructs — they don't always correspond to a single physical current. To find the actual current in any branch, look at which mesh currents flow through it and add them algebraically. A resistor on the outer edge of the circuit carries exactly one mesh current. A shared resistor carries the difference of two. The sign depends on the reference direction you choose for the branch current.

✏️ Practice: A two-mesh planar circuit has a 16 V source in the left mesh, R1 = 4 Ω along the top of the left mesh, R2 = 4 Ω in the shared middle branch, and R3 = 2 Ω along the top of the right mesh. Assign clockwise mesh currents I1 (left) and I2 (right). Find I2.
A
Solution
  1. KVL mesh 1: 16 − 4·I1 − 4·(I1 − I2) = 0 → 8·I1 − 4·I2 = 16 → 2·I1 − I2 = 4
  2. KVL mesh 2: −4·(I2 − I1) − 2·I2 = 0 → 4·I1 − 6·I2 = 0 → I1 = 3·I2/2
  3. Substitute: 2(3·I2/2) − I2 = 4 → 2·I2 = 4 → I2 = 2 A

Check your understanding

1. In mesh analysis, what defines a 'mesh'?
A mesh is an innermost loop — one that does not enclose any other loops. Mesh analysis applies only to planar circuits.
2. Two clockwise mesh currents I1 = 5 A and I2 = 2 A share a 4 Ω resistor. What is the current through the shared resistor in the direction of I1?
When two clockwise mesh currents pass through a shared resistor in opposite directions, the net current is their difference: I1 − I2 = 5 − 2 = 3 A.
🔎 Related lessons
Supernodes & Supermeshes · soon
✅ Key takeaways
  • A mesh is an innermost loop in a planar circuit — assign one clockwise current per mesh.
  • For a shared resistor, the net current is the difference of the two mesh currents.
  • Each mesh gets one KVL equation; the result is a linear system.
  • Branch currents are algebraic sums of the mesh currents flowing through that branch.
➡️ Up next: supernodes and supermeshes — techniques for handling voltage sources between non-reference nodes and current sources on shared branches.
Want to test yourself on this? Try the Electrical Aptitude test →