# Millmans Theorem

### What is Millman's Theorem?

Millman's theorem is a method used to reduce any number of parallel voltage sources to one.
This permits finding the current through or voltage across $R_L$ without having to apply a method such as mesh analysis, nodal analysis, superposition, and so on. The theorem can best be described by applying it to the network in Fig. 1, where three voltage sources can be reduced to one.
Fig.1: Demonstrating the effect of applying Millman's theorem.

### How did Millman's Theorem Work?

Basically, three steps are included in its application.
Step 1: Convert all voltage sources to current sources as shown in Fig.2.
Fig.2: Converting all the sources in Fig. 1 to current sources.
Step 2: Combine parallel current sources. The resulting network is shown in Fig. 3, where $$I_T = I_1 + I_2 + I_3$$ and $$G_T = G_1 + G_2 + G_3$$
Step 3: Convert the resulting current source to a voltage source, and the desired single-source network is obtained, as shown in Fig. 4.
Fig.3: Reducing all the current sources in Fig. 3 to a single current source.
Fig.4: Converting the current source in Fig. 3 to a voltage source.

### Millman's Theorem Equation

In general, Millman's theorem states that for any number of parallel voltage sources, $$E_{eq} = \frac{I_T}{G_T} = \frac{\pm I_1 \pm I_2 \pm I_3 \pm . . . \pm I_N}{G_1 + G_2 + G_3 + ...+G_N}$$ or $$\bbox[10px,border:1px solid grey]{E_{eq} = \frac{\pm E_1G_1 \pm E_2G_2 \pm E_3G_3 \pm ... \pm E_NG_N}{G_1 + G_2 + G_3 + ... +G_N}} \tag{1}$$ The plus-and-minus signs appear in Eq. (1) to include those cases where the sources may not be supplying energy in the same direction.
The equivalent resistance is $$\bbox[10px,border:1px solid grey]{E_{eq} = \frac{1}{G_T} = \frac{1}{G_1 + G_2 + G_3 + ... +G_N}} \tag{2}$$ In terms of the resistance values, $$\bbox[10px,border:1px solid grey]{E_{eq} = \frac{\pm \frac{E_1}{R_1} \pm \frac{E_2}{R_2} \pm \frac{E_3}{R_3} \pm ... \pm \frac{E_N}{R_N}}{G_1 + G_2 + G_3 + ... +G_N}} \tag{3}$$ $$\bbox[10px,border:1px solid grey]{R_{eq}=\frac{1}{\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_3}+ . . . +\frac{1}{R_N}} }\tag{4}$$

### Who developed Millman's theorem?

Millman's theorem was developed by Jacob Millman, who was an expert on radar, electronic circuits and pulse-circuit techniques, was born in Russia and came to the United States in 1913. He was also a professor of Electrical Engineering at Columbia University. His most notable achievement was the formulation of Millman's Theorem (otherwise known as the Parallel generator theorem), which is named after him.
Jacob Millman
Example 1: Using Millman's theorem, find the current through and voltage across the resistor $R_L$ in Fig. 5.
Fig.5: For example 1.
Solution: By Eq. (3), $$E_{eq} = \frac{\frac{+ E_1}{R_1} - \frac{E_2}{R_2}+ \frac{E_3}{R_3}}{ \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$$ The minus sign is used for $E_2/R_2$ because that supply has the opposite polarity of the other two. The chosen reference direction is therefore that of $E_1$ and $E_3$. The total conductance is unaffected by the direction, and $$\begin{split} E_{eq} &= \frac{\frac{+ 10}{5} - \frac{16}{4}+ \frac{8}{2}}{ \frac{1}{5}+\frac{1}{4}+\frac{1}{2}}\\ &=\frac{2A - 4A + 4A}{0.2S+0.25S+0.5S}\\ &=\frac{2A}{0.95 S} = 2.11 V \end{split}$$ with $$\begin{split} R_{eq} &= \frac{1}{\frac{1}{5 Ω} +\frac{1}{4 Ω} + \frac{1}{2 Ω}}\\ &= \frac{1}{ 0.95 S} = 1.05 Ω \end{split}$$ The resultant source is shown in Fig. 6, and $$\begin{split} I_L &= \frac{2.11 V}{1.05 Ω + 3 Ω} = \frac{2.11 V}{4.05Ω}\\ &= 0.52 A \end{split}$$ with $$V_L = I_LR_L = (0.52 A)(3 Ω) = 1.56 V$$
Fig.6: The result of applying Millman's theorem to the network in Fig. 5.