PhysicsISC Class 12 · Current Electricity

The Potentiometer: Principle, Potential Gradient and Applications

Updated 15 October 2026 · Fact-checked

A potentiometer works on the principle that, for a steady current through a uniform wire, the potential drop across any length is proportional to that length: $V = kl$, where k is the potential gradient. At balance no current flows from the cell under test, so it measures emf exactly. It compares emfs ($\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}$) and finds internal resistance.

At balance the galvanometer reads zero, so no current is drawn from the test cell and its emf equals the potential drop kl across length AJ.

Illustration drawn for QuestionBuddy.

Understand Potentiometer: Principle and Applications

Set-up. A long uniform wire AB (often 4 to 10 m, folded on a board) is connected in series with a driver cell (or battery), a key and often a rheostat or series resistance. This primary circuit drives a steady current I through the wire. A secondary circuit connects the cell under test from A, through a galvanometer, to a jockey that slides along the wire. The positive terminals of the driver cell and the test cell must both be connected to A.

Principle. For a uniform wire (constant cross-section and material) carrying a steady current, the resistance of a length l is $ρ \frac{l}{A}$ , so the p.d. across it is
$V = I ρ \frac{l}{A} = k l, k = \frac{V _{A B}}{L} = \frac{I R _{w i r e}}{L} .$
k is the potential gradient (V m⁻¹).

Balance. Slide the jockey until the galvanometer shows no deflection. Then the p.d. across AJ exactly equals the emf of the test cell, and no current is drawn from it:
$ε = k l .$
This is why a potentiometer measures the true emf, while a voltmeter (which draws some current) reads the slightly smaller terminal p.d. It behaves as an ideal voltmeter of infinite resistance.

Finding the potential gradient. With driver emf $ε_{D}$ , internal resistance $r_{D}$ , series resistance $R_{s}$ and wire resistance $R_{w}$ :
$I = \frac{ε _{D}}{R _{w} + R _{s} + r _{D}}, k = \frac{I R _{w}}{L} .$
The ISC 2024 examiners noted candidates who left out the wire's own resistance from this sum.

Comparing two emfs. Balance each cell in turn with the same current in the wire: $ε_{1} = k l_{1}$ , $ε_{2} = k l_{2}$ , so
$\frac{ε _{1}}{ε _{2}} = \frac{l _{1}}{l _{2}} .$
(Sum and difference method: cells aiding give $l_{1}$ , opposing give $l_{2}$ ; $\frac{ε _{1}}{ε _{2}} = \frac{l _{1} + l _{2}}{l _{1} - l _{2}}$ .)

Internal resistance of a cell. Connect a resistance box R across the test cell with a key.
1. Key open: the cell is on open circuit; balance length $l_{1}$ gives $ε = k l_{1}$ .
2. Key closed: current flows through R; balance length $l_{2}$ gives the terminal p.d. $V = k l_{2}$ .
3. Since $\frac{ε}{V} = \frac{R + r}{R}$ ,
$r = R (\frac{l _{1}}{l _{2}} - 1) = R \frac{l _{1} - l _{2}}{l _{2}} .$

Conditions for a balance point. The p.d. across the whole wire must be greater than the emf being measured (so the driver emf must exceed it), and the positive terminals must go to the same end. If no balance is found, the galvanometer deflects the same way all along the wire.

Sensitivity. A potentiometer is more sensitive when it gives a larger change in balancing length for a small change in emf, that is, when k is smaller. Use a longer wire, or reduce the current with a larger series resistance (provided a balance is still possible).

Advantages over a voltmeter. Measures emf (no current drawn), can be made very precise by using a long wire, and the reading does not depend on the resistance of the instrument. Disadvantage: slower, needs a steady driver current.

Potentiometer: Principle and Applications: formulas and key facts

Principle: $V = k l$; Steady current, uniform wire.
Potential gradient: $k = \frac{I R _{w}}{L}, I = \frac{ε _{D}}{R _{w} + R _{s} + r _{D}}$; Include the wire resistance.
Comparing emfs: $\frac{ε _{1}}{ε _{2}} = \frac{l _{1}}{l _{2}}$; Same k for both.
Sum and difference: $\frac{ε _{1}}{ε _{2}} = \frac{l _{1} + l _{2}}{l _{1} - l _{2}}$; $l_{1}$ aiding, $l_{2}$ opposing.
Internal resistance: $r = R (\frac{l _{1}}{l _{2}} - 1)$; $l_{1}$ open circuit, $l_{2}$ with R.
Balance possible: $k L > ε$; Driver p.d. across the wire must exceed the emf.

How to solve Potentiometer: Principle and Applications questions

Almost every potentiometer numerical is "find k, then multiply by a length".

1Find the current in the potentiometer wire from the primary circuit, including wire, series and internal resistances.
2Find the p.d. across the whole wire, $I R_{w}$ , and divide by its length to get k (keep units: V m⁻¹ or V cm⁻¹).
3For an unknown emf, $ε = k l$ ; for a comparison, use the ratio of lengths.
4For internal resistance, use $r = R \frac{l _{1} - l _{2}}{l _{2}}$ with $l_{1}$ the open-circuit length.
5Check that $k L$ exceeds the emf, otherwise no balance is possible.
6State answers with units; balancing lengths in cm and m must not be mixed.

Potentiometer: Principle and Applications in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Comparing a cell with a reference

Suppose a laboratory checks a new cell against a reference cell of emf 1.018 V. On the same potentiometer the reference balances at 254.5 cm and the new cell at 375.0 cm.

Find the emf of the new cell.

Try it first, then show the solutionSolution

$\frac{ε}{1.018} = \frac{375.0}{254.5}$ .
$ε = 1.018 \times \frac{375.0}{254.5} = 1.500$ V.

Answer: 1.500 V.

Memory hook: Emfs are in the ratio of balancing lengths.

Why a voltmeter reads low

Imagine a cell of emf 1.500 V and internal resistance 2 Ω measured with a voltmeter of resistance 1000 Ω and with a potentiometer.

What does each instrument read?

Try it first, then show the solutionSolution

Voltmeter: current $\frac{1.5}{1002}$ A, reading $1.5 \times \frac{1000}{1002} \approx 1.497$ V.
Potentiometer: no current at balance, so it reads the full 1.500 V.

Answer: Voltmeter ≈ 1.497 V; potentiometer 1.500 V.

Memory hook: No current, no lost volts.

Tricks to remember Potentiometer: Principle and Applications

Short memory aids. Say each one aloud once, then use it on a question the same day.

k times l
Every emf or p.d. on a potentiometer is "gradient × balancing length".
Use it when: All numericals.
Positive to positive
Driver + and test + both at end A, or there is no balance.
Use it when: Circuit diagrams and "no balance" questions.
Long wire, small k, high sensitivity
Smaller gradient spreads the scale.
Use it when: Sensitivity questions.
Open over closed minus one
$r = R (\frac{l _{o p e n}}{l _{c l ose d}} - 1)$ .
Use it when: Internal resistance.

Potentiometer: Principle and Applications: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Improving sensitivity · 1 mark · MCQBoard level

The sensitivity of a potentiometer can be increased by:
(a) increasing the current in the wire
(b) using a shorter wire
(c) using a longer wire with the same driver cell
(d) using a driver cell of larger emf

Try it first, then show the solutionSolution

A longer wire with the same p.d. has a smaller potential gradient.

Answer: (c).

Example 2: Comparing emfs · 2 marks · Short answerBoard level

A standard cell of emf 1.08 V balances at 270 cm on a potentiometer wire. A second cell balances at 375 cm with the same driver current. Find its emf.

Try it first, then show the solutionSolution

$\frac{ε _{2}}{1.08} = \frac{375}{270}$ . (1)
$ε_{2} = 1.5$ V. (1)

Answer: 1.5 V.

Example 3: Potential gradient from the circuit · 3 marks · Short answerHarder board level

A potentiometer wire 5 m long has a resistance of 15 Ω. It is connected in series with a 4.5 Ω resistor and a driver cell of emf 4 V and internal resistance 0.5 Ω. Find the current in the wire, the potential gradient, and the emf of a cell that balances at 2.5 m.

Try it first, then show the solutionSolution

$I = \frac{4}{15 + 4.5 + 0.5} = 0.2$ A. (1)
$V_{w i r e} = 0.2 \times 15 = 3$ V; $k = \frac{3}{5} = 0.6$ V m⁻¹. (1)
$ε = 0.6 \times 2.5 = 1.5$ V. (1)

Answer: 0.2 A; 0.6 V m⁻¹; 1.5 V.

Example 4: Derivation: internal resistance · 3 marks · Long answerHarder board level

Draw a circuit and derive an expression for the internal resistance of a cell using a potentiometer.

Try it first, then show the solutionSolution

Circuit: driver cell, key and rheostat across wire AB; the test cell (positive to A) with a galvanometer and jockey; a resistance box R with key K₂ across the test cell. (1)
K₂ open: no current from the cell, balance at $l_{1}$ : $ε = k l_{1}$ . K₂ closed: current flows through R, balance at $l_{2}$ : $V = k l_{2}$ . (1)
$\frac{ε}{V} = \frac{I ( R + r )}{I R} = \frac{l _{1}}{l _{2}}$ , so $r = R (\frac{l _{1}}{l _{2}} - 1)$ . (1)

Answer: $r = R \frac{l _{1} - l _{2}}{l _{2}}$ .

Example 5: Internal resistance from readings · 2 marks · Short answerHarder board level

With the cell on open circuit, the balancing length is 320 cm. When a 10 Ω resistor is connected across the cell, it becomes 280 cm. Find the internal resistance.

Try it first, then show the solutionSolution

$r = 10 \times \frac{320 - 280}{280}$ . (1)
$r = \frac{400}{280} \approx 1.43$ Ω. (1)

Answer: About 1.43 Ω.

Example 6: Designing a potentiometer for small emfs · 4 marks · Case-basedHOTS / competency

Suppose a student wants to measure small emfs of a few tens of millivolts using a 10 m potentiometer wire of resistance 20 Ω and a 2 V driver cell of negligible internal resistance. She wants a potential gradient of 0.01 V m⁻¹.
(i) What current must flow in the wire? (1 mark)
(ii) What series resistance is needed? (1 mark)
(iii) A source balances at 6.2 m. Find its emf. (1 mark)
(iv) Why is a potentiometer better than a voltmeter here? (1 mark)

Try it first, then show the solutionSolution

(i) $V_{w i r e} = 0.01 \times 10 = 0.1$ V; $I = \frac{0.1}{20} = 5$ mA.
(ii) Total resistance $\frac{2}{0.005} = 400$ Ω, so $R_{s} = 400 - 20 = 380$ Ω.
(iii) $ε = 0.01 \times 6.2 = 0.062$ V = 62 mV.
(iv) At balance it draws no current, so it reads the true emf; a small gradient also spreads small voltages over a long length, giving precise readings.

Answer: (i) 5 mA; (ii) 380 Ω; (iii) 62 mV; (iv) no current drawn and high sensitivity.

Where students lose marks in Potentiometer: Principle and Applications

The slips examiners see most often, and what to write instead.

Mistake: Leaving out the wire's resistance
In ISC 2024 candidates found the wire current using an incomplete total resistance, missing the 28 Ω of the wire itself.
Wrong working
$I = \frac{6}{20 + 2}$
Right working
$I = \frac{6}{28 + 20 + 2} = 0.12$ A
Do this: Add every resistance in the primary circuit: wire, series resistor and internal resistance.
Usually costs: 1 to 2 marks.
Mistake: Not finding the p.d. across the wire
Also in 2024: without $V = I R_{w}$ candidates could not get k and the unknown emf.
Do this: Write k = $\frac{I R _{w}}{L}$ as a separate step before using any balancing length.
Usually costs: 1 mark.
Mistake: Misreading the balancing length
The 2024 and 2025 examiners noted confusion over the balancing length and loop formation.
Do this: Balancing length is measured from the end A where both positive terminals are joined.
Usually costs: 1 mark.
Mistake: Inverting the internal-resistance formula
Swapping $l_{1}$ and $l_{2}$ gives a negative r.
Do this: $l_{1}$ (open circuit) is always the longer length.
Usually costs: 1 mark.

Exam tips

Draw both the primary and secondary circuits with polarities marked.
State the principle in words: potential drop ∝ length for a uniform wire with steady current.
Keep balancing lengths in the same unit (cm or m) throughout a ratio.
If asked why no balance is found, mention both causes: driver p.d. too small, or polarities reversed.

Practice questions on Potentiometer: Principle and Applications

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
At the balance point of a potentiometer, the current through the cell being tested is:
(a) maximum
(b) zero
(c) equal to the wire current
(d) half the wire current
Hint
The galvanometer shows no deflection.
Answer
(b).
Q2 · 2 marks · Short answerBoard level
A cell of emf 1.5 V balances at the 40 cm mark of a potentiometer wire. Another cell balances at the 72 cm mark. Find its emf. (A variation on ISC 2025.)
Hint
$\frac{E _{1}}{E _{2}} = \frac{l _{1}}{l _{2}}$ .
Answer
$1.5 \times \frac{72}{40} = 2.7$ V.
Q3 · 2 marks · Short answerHarder board level
What happens to the balancing length for a given cell if (a) the series resistance in the primary circuit is increased, (b) the driver cell is replaced by one of larger emf?
Hint
Think about k.
Answer
(a) k decreases, so the balancing length increases. (b) k increases, so the balancing length decreases.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A potentiometer measures the emf of a cell more accurately than a voltmeter.
Reason (R): At balance, the potentiometer draws no current from the cell.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Lost volts.
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
Two cells connected in series so that their emfs add balance at 600 cm; connected so that they oppose, they balance at 200 cm. Find the ratio of their emfs.
Hint
$\frac{l _{1} + l _{2}}{l _{1} - l _{2}}$ .
Answer
$\frac{800}{400} = 2 : 1$ .
Q6 · 3 marks · Short answerHOTS / competency
A driver cell of emf 6 V and internal resistance 1 Ω is connected to a 10 m potentiometer wire of resistance 24 Ω through a 35 Ω resistor. Find the potential gradient, the emf of a cell balancing at 6.25 m, and the largest emf this set-up can measure.
Hint
Total resistance 60 Ω.
Answer
I = 0.1 A; $V_{w i r e} = 2.4$ V; k = 0.24 V m⁻¹; ε = 1.5 V; largest 2.4 V.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a potentiometer has a gradient of 0.5 V m⁻¹. A cell balances at 3.0 m on open circuit and at 2.5 m when a 5 Ω resistor is connected across it.
(i) Find the emf of the cell. (1 mark)
(ii) Find its internal resistance. (1 mark)
(iii) Find the current through the 5 Ω resistor. (1 mark)
(iv) Check your answers using $I = \frac{ε}{R + r}$ . (1 mark)
Hint
Terminal p.d. = 0.5 × 2.5.
Answer
(i) 1.5 V. (ii) $r = 5 \times \frac{0.5}{2.5} = 1$ Ω. (iii) $\frac{1.25}{5} = 0.25$ A. (iv) $\frac{1.5}{5 + 1} = 0.25$ A, agreeing.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the principle of a potentiometer.Show answer
For a steady current through a uniform wire, the potential drop across any length is proportional to that length.
2.Define potential gradient.Show answer
The fall of potential per unit length of the wire (V m⁻¹).
3.k = 0.4 V m⁻¹, balance at 3.5 m: emf?Show answer
1.4 V.
4.Formula for internal resistance using a potentiometer?Show answer
$r = R (\frac{l _{1}}{l _{2}} - 1)$ .
5.How can sensitivity be increased?Show answer
Use a longer wire or reduce the current (smaller potential gradient).

Potentiometer: Principle and Applications: frequently asked questions

Why is a potentiometer called an ideal voltmeter?

At balance it draws no current from the source being measured, as if it had infinite resistance, so it reads the true emf.

Why must the wire be uniform?

The principle V ∝ l holds only if every centimetre of wire has the same resistance, which needs the same material and cross-section throughout.

Why might the galvanometer deflect the same way everywhere?

Either the p.d. across the whole wire is less than the emf under test, or the positive terminals of the driver and test cells are not connected to the same end.

Which uses of the potentiometer are in the ISC 2027 syllabus?

Measuring potential difference, comparing the emfs of two cells and determining the internal resistance of a cell, along with potential gradient and sensitivity.

How was this tested recently?

In 2024 a 3-mark numerical asked for the wire current and an unknown emf; in 2025 a 2-mark question compared two emfs from balancing lengths.