Millman's Theorem

The fastest way to solve parallel source branches. When several branches — each a source with a resistance — share the same two terminals, Millman's theorem gives the terminal voltage in one line: V = ΣVG / ΣG. Learn the formula, its derivation, a step-by-step method, a fully solved example and how it compares with Thevenin, Norton and superposition.

Complete Learning Path — Millman's Theorem

From the statement and the V = ΣVG/ΣG formula, to conductance, a KCL derivation, the step-by-step method, a solved example, current sources, sign rules and comparisons

What is Millman's Theorem?

Millman's theorem (named after Jacob Millman) is a fast circuit-analysis shortcut. It says that when several branches — each containing a source in series with a resistance — are connected in parallel across the same two terminals, you can find the voltage across those terminals with a single formula, without writing full mesh or nodal equations.

It is really nodal analysis at one node, packaged as a ready-made formula — perfect for circuits with many parallel batteries or supplies feeding a common load.

Millman's theorem concept: several parallel source branches reduced to one equivalent voltage source across the terminals using V equals sum of V G over sum of G
The big idea: reduce many parallel source branches to a single equivalent source across the terminals — V = ΣVG / ΣG.
∥
Branches in parallel
V = ΣVG/ΣG
One formula
G = 1/R
Conductance weights
1 node
Single-node method

Consider three branches with sources V1, V2, V3 and resistances R1, R2, R3, all tied between terminals A and B. Millman's theorem finds the voltage VAB in one step.

Millman's theorem circuit with three parallel branches each having a voltage source in series with a resistor across common terminals A and B
The Millman configuration: parallel source branches (V1/R1, V2/R2, V3/R3) sharing terminals A and B.
When Millman shines

Whenever you see many parallel branches each with its own source and resistance across a common pair of nodes — parallel batteries, multiple supplies, or a mix — Millman gives the answer far faster than mesh or superposition.

The Formula & Conductance

The heart of the theorem is one compact equation:

V = (V1G1 + V2G2 + V3G3 + …) / (G1 + G2 + G3 + …)
The terminal voltage is the conductance-weighted average of the source voltages. Each source counts in proportion to its branch conductance.
Millman's theorem formula V equals sum of V times G divided by sum of G, with the equivalent form using V over R
The Millman formula and its equivalent form V = (Σ Vk/Rk) / (Σ 1/Rk).

Everything hinges on conductance, the reciprocal of resistance:

G = 1 / R   (unit: siemens, S)
A low-resistance branch has a high conductance, so it has more “pull” on the final terminal voltage.
Conductance G equals one over R measured in siemens, with example values for 2, 5 and 10 ohm branches
Conductance G = 1/R is the weight each branch carries in the Millman average.

Derivation from Kirchhoff's Current Law

Millman's theorem is not magic — it drops straight out of Kirchhoff's current law (KCL) applied at the single common node.

Derivation of Millman's theorem: applying Kirchhoff current law at the common node where the branch currents sum to zero
Each branch current is Ik = (Vk − V)/Rk; KCL says they sum to zero at the node.

Let the common node sit at voltage V. The current flowing into the node from branch k is (Vk − V)/Rk. KCL requires the branch currents to sum to zero:

Σ (Vk − V)/Rk = 0

Expanding and grouping the V terms gives ΣVk/Rk = V · Σ(1/Rk), so:

V = (Σ Vk/Rk) / (Σ 1/Rk) = (Σ VkGk) / (Σ Gk)
The Millman formula — derived in three lines from KCL.

The Step-by-Step Method

Applying Millman's theorem is a short, repeatable recipe.

Five steps of Millman's theorem: check the layout, find each conductance, weight each source, apply the formula, and find the branch currents
The five-step Millman procedure.

The procedure in words

  1. Check the layout — all branches must be in parallel across the same two terminals.
  2. Find each conductance, Gk = 1/Rk (in siemens).
  3. Weight each source by its conductance: form each VkGk (with the correct sign).
  4. Apply the formula, V = (ΣVkGk) / (ΣGk), to get the terminal voltage.
  5. Find the currents with Ik = (Vk − V)/Rk and Ohm's law for any load.

Solved Example

Find the terminal voltage for three parallel branches: V1 = 12 V, R1 = 4 Ω; V2 = 6 V, R2 = 2 Ω; V3 = 4 V, R3 = 4 Ω.

Solved Millman's theorem example with three branches 12V 4 ohm, 6V 2 ohm and 4V 4 ohm giving a terminal voltage of 7 volts
Plugging the numbers into V = ΣVG/ΣG gives V = 7 V.

Conductances

G1 = 1/4 = 0.25 S,   G2 = 1/2 = 0.50 S,   G3 = 1/4 = 0.25 S.

Weighted sources (VkGk)

12×0.25 = 3,   6×0.50 = 3,   4×0.25 = 1  →  ΣVG = 7;   ΣG = 0.25+0.50+0.25 = 1.0 S.

V = 7 / 1.0 = 7 V
The common terminal voltage VAB.
Branch currents (a quick check)

I1 = (12−7)/4 = +1.25 A,   I2 = (6−7)/2 = −0.5 A,   I3 = (4−7)/4 = −0.75 A. With no external load they sum to 0 A — exactly what KCL demands.

The Millman Equivalent Source & a Load

Millman's result is really a single equivalent source: an EMF Veq = ΣVG/ΣG in series with a resistance Req = 1/ΣG. That equivalent then drives whatever load you connect.

Millman equivalent source: an equivalent EMF Veq in series with equivalent resistance Req driving a load resistor RL
The network reduces to Veq in series with Req = 1/ΣG, feeding the load.

From the solved example, Veq = 7 V and Req = 1/1.0 = 1 Ω. Connect a load RL = 6 Ω across the terminals and the load current and voltage follow at once:

IL = Veq / (Req + RL) = 7 / (1 + 6) = 1 A  →  VL = 1 × 6 = 6 V
The 6 Ω load pulls the terminal voltage down from 7 V (open circuit) to 6 V.

Millman with Current Sources

Millman's theorem is not limited to voltage sources. Parallel current sources with conductances fit the same pattern.

Millman's theorem with parallel current sources and conductances, giving V equal to sum of currents over sum of conductances
With parallel current sources, V = ΣIk / ΣGk.

A voltage source V in series with R is equivalent to a current source I = V/R in parallel with R (source transformation). So any Millman branch can be written either way, and the numerator becomes the sum of the branch short-circuit currents.

V = (I1 + I2 + I3 + …) / (G1 + G2 + G3 + …)
Same denominator (ΣG); the numerator is now the sum of the source currents.

Sign Convention — Watch the Polarity

The single most common Millman mistake is a wrong sign. Each source's contribution VkGk carries a sign set by its polarity relative to the chosen terminal.

Millman sign convention: a source aiding the assumed terminal polarity is positive while an opposing source is negative
A source whose + terminal faces the node adds (+V·G); one facing the other way subtracts (−V·G).
Rule of thumb

Pick a reference terminal (say A). For every branch, if its source drives current toward A, take VkGk as positive; if it opposes, take it negative. A branch that is just a resistor (no source) contributes only to ΣG, not to the numerator.

Millman vs Thevenin, Norton & Superposition

Millman is one of several network theorems. Knowing when to reach for each saves time.

Comparison of Millman's theorem with Thevenin, Norton and superposition theorems
Each theorem has a sweet spot — Millman is fastest for parallel source branches.
TheoremBest forWhat it gives
MillmanParallel source branches on two common terminalsThe node voltage in one formula
TheveninAny two-terminal linear networkVeq in series with Req
NortonAny two-terminal linear networkIeq in parallel with Req
SuperpositionCircuits with several sourcesSum of each source's contribution

Advantages & Limitations

Advantages

  • Very fast — one formula, no simultaneous equations.
  • Ideal for many parallel sources on a common load.
  • Gives a clean equivalent source (Veq, Req).
  • Handles voltage and current sources via source transformation.

Limitations

  • Only for branches all in parallel across the same two nodes.
  • Applies to linear circuits.
  • Not a fit for general multi-node networks.
  • Dependent sources need extra care.

Applications

Millman's theorem is a working shortcut wherever parallel sources meet a common node.

Parallel batteries

Combine cells or supplies of different EMF and internal resistance.

Multi-source DC

Solve DC networks with several parallel sources fast.

Op-amp inputs

Find the node voltage of a resistive summing/averaging network.

Voltage averaging

Weighted-average circuits and sensor summing nodes.

Power distribution

Feeders from several sources onto a common bus.

Exams & homework

A huge time-saver on parallel-source problems.

Key Terms — Glossary

TermMeaning
Millman's theoremV = (ΣVG)/(ΣG) for parallel source branches on two common terminals.
Conductance (G)Reciprocal of resistance, G = 1/R, in siemens (S).
Siemens (S)The SI unit of conductance (1 S = 1 A/V = 1/Ω).
Common node / terminalsThe two points all branches share; Millman finds the voltage between them.
BranchOne parallel path: a source in series with (or a current source in parallel with) a resistance.
Equivalent EMF (Veq)The Millman result ΣVG/ΣG, the open-terminal voltage.
Equivalent resistance (Req)1/ΣG — the resistance seen by a load.
Source transformationConverting a voltage source + series R into a current source I = V/R + parallel R.
KCLKirchhoff's current law — the sum of currents at a node is zero; the basis of the proof.

Frequently Asked Questions

Quick, expert answers to the questions people ask most about Millman's theorem.

What is Millman's theorem in simple words?

It is a shortcut for circuits where several branches, each with a voltage source and a resistance, are connected in parallel across the same two points. It gives the voltage across those points directly as V = (ΣVG)/(ΣG).

What is the formula for Millman's theorem?

V = (V₁G₁ + V₂G₂ + V₃G₃ + …) / (G₁ + G₂ + G₃ + …), where each G = 1/R is the conductance of that branch. Equivalently, V = (ΣVₖ/Rₖ)/(Σ1/Rₖ).

What is conductance in Millman's theorem?

Conductance G is the reciprocal of resistance, G = 1/R, measured in siemens (S). Each source is weighted by its branch conductance, so lower-resistance branches influence the result more.

When can Millman's theorem be used?

For linear circuits where the branches are all in parallel between the same two terminals, each branch containing a source with a resistance. It is ideal for many parallel voltage sources feeding a common load.

How is Millman's theorem derived?

From KCL at the single common node: write each branch current as (Vₖ − V)/Rₖ, set their sum to zero, and solve for V. This gives V = (ΣVₖ/Rₖ)/(Σ1/Rₖ).

Can Millman's theorem be used with current sources?

Yes. Convert a voltage source with a series resistance to a current source I = V/R in parallel with that resistance. With parallel current sources the formula becomes V = (ΣIₖ)/(ΣGₖ).

What are the limitations of Millman's theorem?

It works only when the branches are all in parallel across the same two terminals, applies to linear circuits, and does not directly handle general multi-node networks or dependent sources without extra work.

How is Millman's theorem different from Thevenin's theorem?

Both reduce a network to an equivalent source, but Millman is a fast single-formula method for parallel source branches sharing two terminals, while Thevenin applies to any two-terminal linear network and gives a voltage source in series with a resistance.

Conclusion & Key Takeaways

Millman's theorem turns a cluster of parallel source branches into a single equivalent source with one elegant formula — the quickest route to the terminal voltage.

Parallel branches

Same two terminals.

V = ΣVG/ΣG

Conductance-weighted average.

G = 1/R

Weights in siemens.

From KCL

One-node derivation.

Veq, Req

A clean equivalent source.

Mind signs

Polarity sets each term's sign.

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