Q.State the working principle of potentiometer. Explain with the help of a circuit diagram, how the potentiometer is used to determine the internal resistance of the given primary cell. In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second 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: …