12 min readConcept sheetSection B long answer (5 marks), often with a choice: principle, circuit and use to compare EMFs or find internal resistance; Section A objective parts on potential gradient; also a NEET favourite

Potentiometer: comparing EMFs and finding internal resistance

Along a uniform wire carrying a steady current, V = kl. At the null point no current is drawn, so E₁/E₂ = l₁/l₂ and r = R(l₁/l₂ − 1).

PrincipleFor a steady current in a uniform wire, the potential drop is proportional to the length: V = kl

Potential gradientk = V/l, the potential drop per unit length of the wire, in V m⁻¹

Comparing EMFsE₁/E₂ = l₁/l₂, where l₁ and l₂ are the balance lengths

Internal resistancer = R(l₁/l₂ − 1); l₁ with the cell on open circuit, l₂ with R joined across it

Why a potentiometer?

A voltmeter must draw a little current to work. So across a cell it reads the terminal voltage, a little less than the EMF. A potentiometer measures a voltage without drawing any current, by a null method. It has a long uniform wire of manganin or nichrome, usually ten 1 m lengths (four in some laboratory models) joined by thick copper strips, beside a metre scale. A jockey makes contact at any point of the wire. A steady driver cell (an accumulator) and a rheostat send a steady current through the wire. Then the potential falls evenly along it: V = kl, where k is the potential gradient. The EMF of the driver cell must be more than any voltage to be measured, or no balance point is found.

How to draw the potentiometer circuits

1
Main circuit: draw a long line AB for the wire. Join the + terminal of the driver cell to A, and its − terminal through a key K and a rheostat Rh to B.
2
Comparing EMFs: join the + terminals of cells E₁ and E₂ to A. Join their − terminals to the two sides of a two-way key, and the middle of the key through a galvanometer G to the jockey J.
3
Internal resistance: join the + terminal of cell E₁ to A, and its − terminal through G to the jockey. Put a resistance box R and a key K₁ directly across the cell.
4
Mark the balance lengths l₁ and l₂ measured from A. Show the + terminals of all the cells joined to the same end, A.

Activity: comparing the EMFs of two cells

Aim: to compare the EMFs of a Leclanche cell (E₁) and a Daniell cell (E₂) with a potentiometer.

1
Set up the main circuit and the two-way key circuit as drawn. Clean all wire ends with sandpaper.
2
Close K and connect E₁. Touch the jockey at A and then at B: the galvanometer must deflect opposite ways. If not, raise the current with the rheostat.
3
Slide the jockey to find the point of no deflection for E₁. Record the balance length l₁ from A.
4
Without touching the rheostat, connect E₂ instead and find its balance length l₂.
5
Change the rheostat setting and repeat twice more, recording l₁ and l₂ each time.
6★
Result: E₁ = kl₁ and E₂ = kl₂, so E₁/E₂ = l₁/l₂. Take the mean of the ratios.

★ key step

Activity: internal resistance of a cell

Aim: to find the internal resistance r of a primary cell, such as a Leclanche cell, of EMF E₁.

1
Set up the main circuit, and the cell circuit with a resistance box R and a key K₁ across the cell.
2
With K₁ open, no current flows through the cell. Find the balance length l₁. Then E₁ = kl₁.
3
Take out a small resistance R, say 5 Ω, and close K₁. The cell now drives a current through R, so its terminal voltage V falls.
4
Quickly find the new balance length l₂, then open K₁. Then V = kl₂.
5
Since E₁/V = (R + r)/R = l₁/l₂, the internal resistance is r = R(l₁/l₂ − 1).
6★
Result: repeat for R = 6, 7, 8, 9 and 10 Ω. Observe: l₂ rises as R rises. Take the mean value of r.

★ key step

Worked example: three potentiometer numericals

Problem

(a) A cell of EMF 1.10 V balances at 330 cm. Another cell balances at 450 cm on the same wire. Find its EMF. (b) A cell balances at 300 cm on open circuit and at 250 cm when a 10 Ω resistor is joined across it. Find its internal resistance. (c) A 4 m wire carries a steady current with 2.0 V across it. Find the potential gradient.

Working

(a) E₂/E₁ = l₂/l₁, so E₂ = 1.10 V × 450 ÷ 330 = 1.10 V × 1.364 = 1.50 V.
(b) r = R(l₁/l₂ − 1) = 10 Ω × (300/250 − 1) = 10 Ω × 0.20 = 2.0 Ω.
(c) k = V/l = 2.0 V ÷ 4 m = 0.50 V m⁻¹.

Answer

(a) 1.50 V; (b) 2.0 Ω; (c) 0.50 V m⁻¹

Potentiometer vs voltmeter

PotentiometerDraws no current at balance, so it measures the true EMF. It uses a null method, so the galvanometer need not be calibrated. A longer wire makes it more accurate.

VoltmeterMust draw a small current, so across a cell it reads the terminal voltage, a little less than the EMF. It is quick and handy, but less accurate.

A longer wire gives a smaller potential gradient, so each centimetre stands for a smaller voltage. That makes the potentiometer more sensitive.

In real life

The volume knob of a radio is a small potentiometer: a sliding contact on a resistance picks off part of the voltage. Laboratories use wire potentiometers to check the EMF of cells accurately, without drawing current from them.

Exam angle

1 marks: “What is the potential gradient of a potentiometer wire?” The potential drop per unit length of the wire, k = V/l, in V m⁻¹.

2 marks: “Why is a potentiometer preferred to a voltmeter for measuring the EMF of a cell?” At the balance point the potentiometer draws no current from the cell, so there is no Ir drop and it reads the true EMF. A voltmeter draws some current and reads a little less.

5 marks: “Explain the principle of a potentiometer. With a circuit diagram, describe how you would find the internal resistance of a cell with it.” Principle: V = kl for a steady current in a uniform wire. Balance at l₁ with the cell open (E = kl₁). Join R across the cell and balance at l₂ (V = kl₂). E/V = (R + r)/R = l₁/l₂, so r = R(l₁/l₂ − 1). Draw the main circuit and the cell circuit.

Remember it

No current, no lie: at the null point the cell gives no current, so the potentiometer reads its true EMF. Compare lengths, not meter readings.