Mesh and Nodal Analysis, and Star-Delta Conversion

By the end you should be able to: Formulate mesh current and node voltage equations in matrix form and solve them, and convert between star and delta resistive networks.

Series-parallel reduction and the division rules solve simple networks. They fail the moment a circuit has two sources, or a bridge, or any element that is neither in series nor in parallel with anything. These two methods never fail — they turn any network into simultaneous equations.

The syllabus asks for both in matrix form, so the pattern in the matrix matters as much as the answer.

Mesh current analysis

Choose loop currents, not branch currents. Assign a circulating current to each independent loop (mesh), conventionally all clockwise. A branch shared by two meshes carries the difference of the two mesh currents, and KCL is then satisfied automatically — that is the saving.

Write KVL for each mesh. The result always has the same shape:

\begin{bmatrix} R_{11} & -R_{12} \\ -R_{21} & R_{22} \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \end{bmatrix} = \begin{bmatrix} V_1 \\ V_2 \end{bmatrix}

Read the entries directly off the circuit:

$R_{11}$ — the self-resistance of mesh 1: the sum of all resistances around it. Always positive.
$R_{12} = R_{21}$ — the mutual resistance shared between meshes 1 and 2, entered as negative when both mesh currents are assigned the same rotational sense (both clockwise).
$V_1$ — the algebraic sum of source EMFs driving mesh 1, positive when the source pushes current in the assumed direction.

The matrix is symmetric for a network of passive resistors. If yours is not, you have made a sign error — a free check before solving.

Number of equations = number of meshes.

Node voltage analysis

Choose node voltages. Pick a reference node (the one with most connections, usually), set it to 0 V, and write KCL at each remaining node. Since every voltage is expressed relative to the reference, KVL is satisfied automatically.

\begin{bmatrix} G_{11} & -G_{12} \\ -G_{21} & G_{22} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \end{bmatrix} = \begin{bmatrix} I_1 \\ I_2 \end{bmatrix}

Exactly the dual of the mesh matrix, with conductances $G = 1/R$ replacing resistances and source currents replacing source EMFs:

$G_{11}$ — sum of conductances connected to node 1. Positive.
$G_{12}$ — conductance between nodes 1 and 2, entered negative.
$I_1$ — sum of current sources feeding into node 1.

Number of equations = (number of nodes − 1).

Which to choose

Count. Use mesh when meshes are fewer than nodes − 1, and nodal when the reverse. Voltage sources suit mesh analysis; current sources suit nodal.

For a circuit with 4 nodes and 3 meshes, mesh gives 3 equations against nodal's 3 — a tie. For a ladder with many series elements, mesh usually wins; for a circuit with one long bus and many branches hanging off it, nodal wins decisively.

Star-delta conversion

The syllabus notes: resistive networks only, derivation not required.

Some networks — the Wheatstone bridge above all — contain three resistors in a triangle (delta, Δ) or a Y (star, Y) that cannot be reduced by series or parallel rules. Converting one form to the other usually unlocks the rest.

Delta to star. Each star resistance is the product of the two adjacent delta resistances divided by the sum of all three:

R_1 = \frac{R_{12}R_{31}}{R_{12}+R_{23}+R_{31}}

and cyclically. The pattern: product of the two deltas touching that node, over the sum of all three.

Star to delta. Each delta resistance is the sum of the pairwise products divided by the opposite star resistance:

R_{12} = \frac{R_1R_2 + R_2R_3 + R_3R_1}{R_3}

The pattern: sum of pairwise products, over the opposite one.

A useful check on any delta-to-star conversion: the star resistances are always smaller than the delta ones. If a star value comes out larger than every delta value, the conversion has gone the wrong way.