Animated Solution for Physics - Current Electricity: In a potentiometer experiment, the balancing with a cell is at length $240\text{ cm}$. On shunting the cell with a resistance of $2\Omega$, the balancing length becomes $120\text{ cm}$. The internal resistance of the cell is

Welcome to this classic potentiometer problem! A potentiometer is a beautiful and highly precise device used to measure the electromotive force (EMF) and internal resistance of a cell. Unlike a standard voltmeter, which draws a small current and alters the very voltage it tries to measure, a potentiometer operates on a null-deflection principle. At the balance point, it draws absolutely zero current from the test cell, giving us a perfectly accurate measurement.

Here, we have a primary circuit that sets up a uniform potential gradient across the main wire $AB$, and a secondary circuit containing our test cell, its internal resistance $r$, and a shunt resistance $R$ connected via a switch.

Let's look at the first case. When the switch $K$ is open, the shunt resistance is disconnected. The cell is in an open circuit, meaning no current is drawn from it. In this state, the balancing length we find on the wire corresponds directly to the true EMF of the cell, $E$.

The problem states this initial balancing length, $l_1$, is $240\text{ cm}$. Mathematically, we can write:

Now, what happens when we close the switch? The cell is now shunted by the $2\Omega$ resistance. Current starts flowing in this local secondary loop. Because of the internal resistance $r$, some voltage is 'lost' inside the cell (the $ir$ drop). The voltage available across the external terminals drops from its EMF to a lower terminal voltage, $V$.

The new balancing length, $l_2$, which is $120\text{ cm}$, now corresponds to this terminal voltage:

Let's connect the physics to the math. The EMF $E$ is the total voltage driving the current $I$ through both the external resistance $R$ and internal resistance $r$. So, we have:

The terminal voltage $V$ is just the voltage across the external resistance, so:

Dividing these two equations, we get a beautiful ratio:

Since the potentiometer principle tells us that voltages are proportional to their balancing lengths, the ratio $\frac{E}{V}$ is exactly equal to the ratio $\frac{l_1}{l_2}$. Substituting this into our previous relation, we can easily rearrange it to find the master formula for internal resistance:

This is a very important formula to remember for your exams!

We have all the pieces of the puzzle now. Let's plug them in. The initial balancing length $l_1$ is $240\text{ cm}$, the shunted balancing length $l_2$ is $120\text{ cm}$, and the shunt resistance $R$ is $2\Omega$. Substituting these into our formula, we get:

The math here is beautifully simple. $240$ divided by $120$ is exactly $2$. Then, $2$ minus $1$ is $1$. Finally, multiplying by the external resistance $2$, we find that the internal resistance of the cell is exactly:

And that matches option (d) perfectly! Before we wrap up, think about this: what would happen if we used a much larger shunt resistance? As $R$ increases, the terminal voltage $V$ gets closer to the EMF $E$, which means the new balancing length $l_2$ would increase and approach $l_1$. Understanding these limits helps you master the true physical behavior of the circuit. Keep exploring!