Simple Start with Millman's

It has always been my contention that in order for a circuit analysis tool to be of any value to me, it must be scalable. This means it needs to solve the simpliest problem as well as the most complex problem by using the same repeatable algorithm.

My experience has been that Millman's Theorem meets this criteria fully.

For example, take a simple problem such as a resistive divider (see figure insert above). Those familar with the simple resistive divider circuit will recognize that the expression below is used to calculate the attenuation of input voltage \(\normalsize V_I\) expressed as \(\normalsize V_O\).

\(\frac{V_{O}}{V_{I}}\ =\ \Large\frac{R2}{R1+R2}\)

To show the Millman's at work you simply apply Millman's to the two resistor divider circuit.

\(\normalsize V_{O}\ =\ \Large\frac{\frac{V_I}{R1}}{\frac{1}{R1} + \frac{1}{R2}}\)

Next, multiply the numerator and the denominator of the expression on the right of the equal sign by the least common denominator factor R1*R2 .

\(\normalsize V_{O}\ =\ \Large\frac{\frac{V_I}{R1}}{\frac{1}{R1}+ \frac{1}{R2}}\ * \ \Large\frac{R1*R2}{R1*R2}\)

\(V_{O}\ =\ \Large\frac{\frac{V_I*R1*R2}{R1}}{\frac{R1*R2}{R1}+\frac{R1*R2}{R2}}\)

\(\normalsize V_{O}\ =\ \Large\frac{V_I*R2}{R2+R1}\)

Divide both sides of the expression by \(\ \normalsize V_{I}\ \) and rearrange the denominator and you end up with the familiar expression that is identical to the one that is associated with a standard two resistor divider.

\(\frac{V_{O}}{V_I}\ =\ \Large\frac{R2}{R1+R2}\)

This example illustrates that Millman's Theorem can be used to solve uncomplicated circuits.

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