# Potentiometer

> Potentiometer for NEET Physics: potential gradient, comparison of emfs, measuring internal resistance by the null method, sensitivity physics.

- Canonical URL: https://prepelephant.com/topics/neet-ug/physics/potentiometer-neet
- Exam / course: NEET-UG · Subject: Physics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Potentiometer", PrepElephant, https://prepelephant.com/topics/neet-ug/physics/potentiometer-neet

## Direct answer

Comparing two emfs without drawing current from either is the potentiometer's whole purpose. The instrument is a length of uniform wire carrying a steady current, with potential gradient k = V/L volts per centimetre; when the jockey finds the null point, the unknown emf equals k times the balance length. Because no current is drawn from the cell under test at null, the potentiometer reads the true emf — the property that elevates it over every voltmeter. Its two standard jobs are comparing emfs, ε1/ε2 = l1/l2, and measuring internal resistance, r = R(l1 - l2)/l2 from the null shift when a known resistance R discharges the cell; the driver cell must exceed the tested emf, or no null exists.

## What you must remember

- **Potential gradient:** k = V/L (volts per unit length) set by the driver cell and the wire; the sensitivity improves as k falls — a longer wire or a smaller driver voltage stretches the null point out.
- **Null principle:** at balance the unknown emf equals the drop across the balance length, and the test cell delivers no current — hence true emf, undisturbed circuits.
- **Comparison of emfs:** ε1/ε2 = l1/l2 with the same driver setting; the driver cell's emf itself never enters the ratio.
- **Internal resistance:** r = R(l1 - l2)/l2, where l1 is the null without the resistance R across the cell and l2 with it.
- **Driver condition:** ε_driver must exceed the largest emf to be tested, otherwise the wire's full drop cannot balance the cell and no null point appears.
- **Volts-to-length identity:** with k = 10 mV/cm, a null at 60 cm reads 0.6 V — the arithmetic that runs almost every numerical.

## Measuring a cell's internals, one null at a time

A cell balances at l1 = 75 cm on a potentiometer wire. Now connect a resistance R = 9.5 Ω across the cell's terminals and the null shifts to l2 = 60 cm. With R discharging the cell, the terminal voltage is what the wire now balances: V = ε l2/l1. Terminal voltage is also V = ε R/(R + r). Equating the two: R/(R + r) = l2/l1, which rearranges to r = R(l1 - l2)/l2 = 9.5 × (75 - 60)/60 = 9.5 × 0.25 = 2.375 Ω. The elegance is that ε itself was never needed — the ratio of two lengths did all the work, which is exactly why the potentiometer survives as the exam's emblem of the null method.

## How the exam asks about the null

The recurring conceptual question is the comparison with the voltmeter: a voltmeter across a cell reads V = ε - Ir for its own small draw of current, so it always under-reads; the potentiometer at null draws nothing, and this difference — not any claim of better manufacturing — is the accepted justification. The second favourite is the missing-null scenario: a candidate searches the whole wire for a balance and finds none; the diagnosis is a driver emf smaller than the tested emf (or a dead driver cell), and the remedy is to raise the driver voltage or use a longer wire — the options test whether you think in terms of kL versus ε. Third, sensitivity: decreasing the potential gradient spreads the null point along more centimetres, so the same 1 mm of judgement error costs less voltage — the reason longer boards and smaller driver currents are praised. Finally, watch the internal-resistance formula's parts: l1 is always the open-circuit (larger) length, and interchanging l1 and l2 produces the reciprocal answer sitting attractively among the options.

## Frequently asked questions

### Why is a potentiometer preferred over a voltmeter for measuring emf?

At the null point it draws no current from the cell under test, so it reads the true emf; a voltmeter always draws some current and shows only the reduced terminal voltage.

### What is the potential gradient of a potentiometer wire?

The drop per unit length, k = V/L, fixed by the driver cell and the wire's resistance; multiplying it by the balance length gives the balanced voltage.

### How are two emfs compared with a potentiometer?

By taking two balance lengths with the same gradient: ε1/ε2 = l1/l2, eliminating the gradient and the driver cell entirely.

### How is the internal resistance of a cell measured?

From the shift of the null when a known R discharges the cell: r = R(l1 - l2)/l2, where l1 and l2 are the balance lengths without and with R connected.

### Why might no null point be found anywhere on the wire?

The driver cell's emf (setting the full wire drop kL) is less than the emf being tested, so no length can balance it — increase the driver voltage or lengthen the wire.