Q.(a) State the working principle of potentiometer. Explain with the help of circuit diagram how potentiometer is used to determine the internal resistance of the given primary cell.

The Intuition: A Cell Isn't a Perfect Battery

Every real cell — whether a dry cell, a lead-acid battery, or a lithium-ion cell — behaves like a perfect voltage source with a small resistor hidden inside it. That hidden resistor is the internal resistance $r$. You can't see it, but it's always there, wasting a little voltage whenever current flows.

Think of it this way: when you connect a bulb to a cell, the bulb glows. But if you measure the voltage across the cell's terminals while the bulb is on, you'll get a value slightly less than the cell's actual electromotive force (emf). That drop happens because current passing through the internal resistance $r$ causes a voltage drop $I r$ inside the cell itself. The terminal voltage $V$ is therefore:

where $\mathcal{E}$ is the cell's emf (the voltage when no current flows).

So how do we measure this hidden $r$? A voltmeter won't help directly — it draws some current and disturbs the reading. The potentiometer is the perfect tool because it draws zero current from the cell when measuring.

The Setup: Two Measurements, One Potentiometer

A potentiometer is a long uniform wire with a sliding contact. You connect the cell you're testing (with emf $\mathcal{E}$ and internal resistance $r$) across the wire through a galvanometer. By finding the balancing length $l$ where the galvanometer shows zero deflection, you measure the cell's voltage without drawing any current.

We make two measurements:

1. Without any external load — the cell is open-circuited. The potentiometer measures its full emf $\mathcal{E}$. Let the balancing length be $l_1$. Since the potentiometer wire has uniform resistance, the voltage is proportional to length:

2. With an external shunt resistor $R$ connected across the cell's terminals. Now the cell delivers current through $R$. The terminal voltage $V$ (which is less than $\mathcal{E}$) is measured by the potentiometer. Let the balancing length be $l_2$:

The Physics: Relating the Two Lengths

When the external resistor $R$ is connected, the cell supplies a current:

The terminal voltage $V$ across the cell (which is also the voltage across $R$) is:

But from the potentiometer measurements:

Substitute $V$:

The $\mathcal{E}$ cancels:

Cross-multiply: …