Formula & Calculator

Millman's Theorem

Gives the voltage at a common node of several parallel branches, each with its own source and conductance.

Circuit AnalysisNetwork Theorem

Millman's Theorem Calculator V = (ΣIₖ) / (ΣGₖ)

V = (Σ Ik) / (Σ Gk)
V = voltage at common node  ·  Ik = current source of branch k  ·  Gk = conductance of branch k (1/R)
⟹ Solve V, Vk, Rk
Vk (V)
Rk (Ω)
Ik = V/R (A)
1
V
Ω
A
2
V
Ω
A
3
V
Ω
A
4
V
Ω
A
ΣIₖ = 0.000 A
ΣGₖ = 0.000 S
V = V
Please fix the errors above.
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Output Voltage (V)
ΣIₖ: ΣGₖ: V:
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V = (ΣVₖ/Rₖ) / (Σ1/Rₖ)  ·  Millman's theorem combines parallel branches with voltage sources and series resistances.

Interpretation

Millman's theorem: multiple parallel voltage sources (with series resistances) can be combined into a single equivalent voltage source.
The equivalent voltage is the sum of each branch's current (V/R) divided by the sum of conductances (1/R).
Example: Branches with currents 2A,3A,1A and conductances 0.1S,0.2S,0.05S → V = (2+3+1)/(0.1+0.2+0.05) ≈ 17.14V.

V = (ΣI_k) / (ΣG_k)
Millman's Theorem

Variables

SymbolQuantityUnit
VVoltage at the common nodeV
I_kBranch current sourceA
G_kBranch conductanceS (siemens)

What it means

Millman’s theorem provides a method to find the voltage at a common node in a network of parallel voltage sources, each with a series resistance. It states that the equivalent voltage V is the sum of each branch current (V_i / R_i) divided by the sum of branch conductances (1/R_i). The formula is V = (Σ (V_i/R_i)) / (Σ (1/R_i)). This is useful for simplifying circuits with multiple sources feeding a common node. It is particularly convenient for circuits with many parallel branches, reducing them to a single equivalent source. Millman’s theorem is also known as the parallel‑source theorem and is often applied in electronics for summing voltages in op‑amp circuits. It can be derived from both KCL and Ohm’s law. Example: Branches with voltages 2V,3V,1V and resistances 10Ω,5Ω,20Ω give currents 0.2A,0.6A,0.05A and conductances 0.1,0.2,0.05 S. Sum currents = 0.85A, sum conductances = 0.35S, so V = 0.85/0.35 = 2.43V.

Worked example

Millman's Theorem – Practical Example

Real‑World
Scenario: Three voltage sources (5 V, 10 V, 15 V) with series resistances (1 Ω, 2 Ω, 3 Ω) are connected in parallel. Find the common output voltage using Millman's theorem.
ParameterValue
V₁, R₁5 V, 1 Ω
V₂, R₂10 V, 2 Ω
V₃, R₃15 V, 3 Ω
FormulaV = (Σ Vk/Rk) / (Σ 1/Rk)
1Compute each V/R: 5/1=5, 10/2=5, 15/3=5 → sum = 15
2Compute sum of 1/R: 1+0.5+0.333 = 1.833
3V = 15 / 1.833 =V ≈ 8.18 V
Final Design V ≈ 8.18 V ✓ Common output voltage
Why: Millman's theorem simplifies finding the voltage at a node where multiple branches meet – it's a weighted average of the branch voltages.

Common mistakes

  • I_k: The current contribution of each branch – V_k/R_k, where V_k is the branch source voltage.
  • G_k: Conductance of each branch (1/R_k) – not resistance.
  • Summation: The numerator is the sum of currents, denominator is the sum of conductances.
  • Sign: Use the correct sign for current directions (sources aiding the output are positive).
  • Applicability: Only for circuits with multiple parallel voltage sources and series resistances.

Applications

Millman's theorem provides a method for combining multiple parallel voltage sources (each with its series resistance) into a single equivalent voltage source. It is particularly useful in circuits with several branches, such as in power distribution networks or multi‑source analog circuits. Engineers use it to simplify analysis, to determine the voltage at a common node, and to design circuits with redundant supplies. By applying Millman's theorem, they can quickly calculate the equivalent voltage and resistance without solving simultaneous equations. This theorem is a practical tool in circuit analysis, especially when dealing with non‑ideal voltage sources. Understanding Millman's theorem enhances analytical capabilities in complex circuits.

  • Analysis of multi‑source circuits and power supplies
  • Redundant supply systems and load sharing
    • Analog circuit design with multiple reference voltages
    • Simplification of network for simulation
    • Educational insight into source transformations

    Frequently Asked Questions

    Q01What is Millman's theorem and what does it calculate?
    A01

    Millman's theorem gives the voltage at a common node of several parallel branches, each with its own source and conductance: V = (Σ I_k) / (Σ G_k), where I_k are equivalent current sources and G_k are conductances.

    Q02What do I_k and G_k represent?
    A02

    I_k = equivalent current source in each branch (voltage source in series with resistance converted to current source), G_k = conductance of each branch (1/R_k).

    Q03How do you convert a voltage source to a current source for Millman?
    A03

    A voltage source V in series with resistance R becomes a current source I = V/R in parallel with conductance G = 1/R.

    Q04What is the advantage of Millman's theorem?
    A04

    It simplifies the analysis of parallel branches to find the voltage at a common node without solving multiple equations.

    Q05Can Millman's theorem be applied to AC circuits?
    A05

    Yes, using phasor notation and admittances (complex conductances).

    Q06What is the result if the denominator is zero?
    A06

    If ΣG_k = 0 (no conductance), the voltage is infinite, which is not physically possible; it indicates an open circuit condition.

    Q07How is Millman's theorem related to the node voltage method?
    A07

    It is essentially a shortcut for the node voltage method when multiple branches connect to a single node with no series resistance between sources.

    Q08What are the limitations of Millman's theorem?
    A08

    It applies only to circuits with independent sources and linear elements, and only for the voltage at a common node.

    Q09What are some practical applications of Millman's theorem?
    A09

    Analysis of power distribution networks, resistor ladder networks, and circuits with multiple parallel branches.

    Q10What are the common mistakes when using Millman's theorem?
    A10

    Common errors include: 1) forgetting to convert voltage sources to current sources, 2) using the wrong sign for sources, 3) not summing conductances correctly, 4) applying the theorem to nodes with series resistance, and 5) using it for non-linear circuits.