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Millman's theorem gives a quick shortcut for finding the voltage at a node where several voltage-source branches (each a source in series with a resistor) meet in parallel. Instead of writing full nodal equations, you sum each branch's short-circuit current contribution (Vi/Ri) and divide by the sum of branch conductances (1/Ri). The result, VM, together with the combined Millman resistance RM (all branch resistances in parallel), forms a single Thevenin-equivalent source for everything connected beyond that node.
| Quantity | Formula |
|---|---|
| Millman node voltage | VM = (V1/R1+V2/R2+V3/R3) / (1/R1+1/R2+1/R3) |
| Millman equivalent resistance | RM = 1 / (1/R1+1/R2+1/R3) = R1∥R2∥R3 |
| Load current (add RL at the node) | IL = VM/(RM+RL) |
Millman's theorem is essentially a compact, single-step nodal analysis for the special case of multiple voltage sources feeding one common node — the same answer superposition or nodal analysis would give, computed in one pass.
It quickly finds the voltage at a single node where several parallel voltage-source branches meet, without writing full nodal or mesh equations.
VM = Σ(Vi/Ri) / Σ(1/Ri) — the sum of each branch's short-circuit current divided by the sum of branch conductances.
RM = 1/Σ(1/Ri), which is simply all the branch resistances combined in parallel — this becomes the Thevenin resistance of the combined source.
Yes conceptually — it is nodal analysis specialized to the case where every branch is a voltage source with a series resistor meeting at one common node, expressed as a single convenient formula.
Yes, the formula extends to any number of branches: just keep summing Vi/Ri in the numerator and 1/Ri in the denominator for every branch present.
Treat it as a branch with Vi=0; its term contributes 0 to the numerator but still adds 1/Ri to the denominator, pulling the node voltage down toward ground.
Treat VM and RM as a Thevenin source and use IL=VM/(RM+RL), exactly like any Thevenin-equivalent load calculation.
It requires the branches to be independent voltage sources with series resistance (not current sources) all meeting at the same two nodes; other topologies need general nodal analysis instead.
Its 1/Ri term is large in both sums, so it pulls VM strongly toward its own voltage — a "stiffer" (lower-impedance) source has more influence on the shared node.
Yes, using complex impedances Zi in place of Ri and phasor voltages Vi, following the same formula structure.
VM simply equals that common voltage (the weighted average degenerates to a plain average), and RM equals that resistance divided by the number of branches.
It gives a good first-order estimate using each battery's open-circuit voltage and internal resistance, useful for predicting shared terminal voltage and any circulating current between mismatched cells.
Superposition finds each source's individual contribution one at a time with others deactivated; Millman's theorem computes the combined node voltage directly in a single formula for the specific parallel-branch topology.
No — a zero-resistance branch would force the node voltage to exactly that branch's source voltage (an ideal voltage source dominates completely), and the formula's division by Ri becomes undefined; use a very small but nonzero resistance in that case.
Superposition Theorem • Thevenin Equivalent Circuit • Source Transformation • All Calculators