Short notes

Mesh and Nodal Analysis Short Notes

Given a circuit with two voltage sources and three resistors forming two loops, mesh analysis assigns clockwise loop currents I₁ and I₂, then writes two KVL equations — the system matrix has diagonal entries equal to total resistance in each mesh and off-diagonal entries equal to the negative of shared resistances. Solving that 2×2 system by Cramer's rule or substitution gives branch currents without ever computing node voltages. That systematic matrix formulation is exactly what examiners want to see.

EEE, ECE, EI

How it works

Mesh analysis steps: (1) identify and label meshes with assumed clockwise currents; (2) write KVL for each mesh: for mesh k, (sum of resistors in mesh k)×I_k − (shared resistors)×adjacent mesh currents = net voltage rise. In matrix form: [R][I] = [V]. For a current source shared by two meshes (supermesh): write KVL around the periphery of both meshes combined, then add the constraint that the current source value equals I₁ − I₂. Nodal analysis: (1) select reference node; (2) assign node voltages V₁, V₂...; (3) write KCL at each non-reference node expressing currents as (V_node − V_adjacent)/R.

Key points to remember

Nodal analysis uses N−1 equations for N nodes; mesh analysis uses B − N + 1 equations. Choose nodal analysis when the circuit has more loops than nodes; choose mesh when it has more nodes than loops. For a circuit with a voltage source between two non-reference nodes, form a supernode: combine KCL for both nodes and add the voltage source constraint equation V₁ − V₂ = V_s. The conductance matrix [G] in nodal analysis has diagonal entries equal to sum of conductances at each node and off-diagonals equal to negative of shared conductance — this mirrors the resistance matrix in mesh analysis. Dependent sources add a controlled term to the equations without changing the method.

Exam tip

The examiner always asks you to solve a two-mesh or two-node circuit by the matrix method and verify with power balance — calculate total power delivered by sources and total power absorbed by resistors separately, because they must be equal for the solution to be correct.

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