Animated Solution for Physics - Current Electricity: Two batteries with emf $12\text{ V}$ and $13\text{ V}$ are connected in parallel across a load resistor of $10\Omega$. The internal resistances of the two batteries are $1\Omega$ and $2\Omega$, respectively. The voltage across the load lies between

Imagine you are an electrical engineer tasked with powering a delicate instrument. You have two batteries at your disposal: one provides $12\text{ V}$ with an internal resistance of $1\Omega$, and the other provides $13\text{ V}$ with an internal resistance of $2\Omega$. You connect them in parallel to supply a $10\Omega$ load resistor. The burning question is: what exact voltage will your instrument receive?

At first glance, this looks like a classic Kirchhoff's laws problem. You could set up two loops, define your currents, write down the equations, and solve the simultaneous linear equations. While that method is foolproof, it can be time-consuming and prone to algebraic silly mistakes.

Is there a more elegant way? Absolutely. Enter Millman's Theorem.

Millman's Theorem

Millman's Theorem is a spectacular shortcut for circuits with parallel branches. It allows us to compress any number of parallel voltage sources into a single equivalent battery with one equivalent EMF ($E_{eq}$) and one equivalent internal resistance ($r_{eq}$).

The formula for the equivalent EMF is a weighted average of the individual EMFs, where the weights are the conductances (reciprocal of resistance) of each branch:

Let's plug in our values. For the first branch, $E_1=12\text{ V}$ and $r_1=1\Omega$. For the second branch, $E_2=13\text{ V}$ and $r_2=2\Omega$.

Simplifying the numerator and denominator:

To make the math cleaner, let's multiply the top and bottom by 2:

Now, what about the equivalent internal resistance? Since the internal resistors are effectively in parallel when we look back into the source, we calculate $r_{eq}$ using the standard parallel resistance formula:

Substituting our values:

The Single Loop

By using Millman's Theorem, we have transformed a multi-loop nightmare into a serene, single-loop circuit. We now have a single battery of $\frac{37}{3}\text{ V}$ with an internal resistance of $\frac{2}{3}\Omega$ connected in series with our $10\Omega$ load resistor.

To find the voltage across the load, we first need the total current $I$ flowing through the circuit. According to Ohm's law for a complete loop:

Let's substitute our equivalent values:

Finding a common denominator for the bottom term:

The $3$s cancel out beautifully, leaving us with:

Finally, the voltage $V$ across the load resistor is simply the current multiplied by the load resistance:

Let's do the final division:

Looking at our options, $11.5625\text{ V}$ perfectly lies between $11.5\text{ V}$ and $11.6\text{ V}$.

This problem beautifully demonstrates how knowing the right theorem can turn a tedious calculation into a swift and satisfying victory!