Animated Solution for Physics - Current Electricity: In the given circuit, an ideal voltmeter connected across the $10\Omega$ resistance reads 2 V. The internal resistance $r$, of each cell is

The journey to solving this circuit problem begins with a careful analysis of its topology. At first glance, the circuit might seem a bit tangled, but breaking it down into smaller, manageable chunks reveals a beautiful simplicity.

Imagine you are a tiny electron navigating this circuit. You start at the positive terminal of the battery and flow through the main wire. Soon, you reach a junction where the path splits into two. One path has a $15\Omega$ resistor, and the other has a $10\Omega$ resistor. Because the current divides and then recombines, these two resistors are in a parallel combination.

After the paths recombine, the total current flows through a single $2\Omega$ resistor. This means the $2\Omega$ resistor is in series with the parallel combination. Finally, the current returns to the power source, which consists of two identical cells connected in series. Each cell has an electromotive force (EMF) of $1.5\text{ V}$ and an unknown internal resistance $r$.

To make sense of the circuit, our first goal is to simplify the parallel part. Let's calculate the equivalent resistance of the $15\Omega$ and $10\Omega$ resistors, which we will call $R_p$.

Using the standard formula for two parallel resistors:

Substituting our values into the equation:

Now, the problem gives us a crucial piece of information: an ideal voltmeter connected across the $10\Omega$ resistor reads $2\text{ V}$. The word "ideal" is a massive hint. It means the voltmeter has infinite resistance and draws absolutely zero current, so it doesn't alter the circuit's behavior.

Because the $10\Omega$ and $15\Omega$ resistors are in parallel, they share the same nodes. Therefore, the voltage drop across the entire parallel combination is exactly the same as the voltage across the $10\Omega$ resistor.

With the voltage across the parallel section and its equivalent resistance known, we can use Ohm's law to find the total current $I$ flowing through the main circuit:

Now that we know the main current, we can look at the circuit as a whole. The two cells are connected in series, so their EMFs simply add up to give the total driving voltage of the circuit:

The total equivalent resistance of the entire circuit, $R_{eq}$, is the sum of all the series components: the parallel equivalent $R_p$, the $2\Omega$ resistor, and the internal resistances of the two cells ($r+r=2r$).

Applying Ohm's law to the complete circuit, we relate the total current, total EMF, and total resistance:

Substituting the values we've found:

This is a simple linear equation. Cross-multiplying gives us:

Subtracting 8 from both sides:

And there we have it! The internal resistance of each cell is $0.5\Omega$. By systematically breaking down the circuit and applying Ohm's law at both the component level and the global level, the solution naturally unfolds.