Learn9702Paper 2 · AS Level Structured QuestionsD.C. circuitsPotential dividers

D.C. circuits · Subtopic 3 of 3

Potential Dividers

Two resistors in series share the supply p.d. in proportion to their resistances: $V_{\text{out}}=\dfrac{R_2}{R_1+R_2}V_{\text{in}}$ . Replacing one resistor with a thermistor or LDR makes the output depend on temperature or light. A potentiometer is a uniform resistance wire whose p.d. per unit length lets you compare e.m.f.s by finding the balance length, where a galvanometer reads zero.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH12AQA 7408 Paper 1≈ 35 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use the potential divider equation and the ratio $\dfrac{V_1}{V_2}=\dfrac{R_1}{R_2}$ .
Explain how a thermistor or LDR in a divider gives a sensing output.
Describe the potentiometer and the null method using a galvanometer.
Compare e.m.f.s from balance lengths.

Pause and think

A night-light switches on when it gets dark. The sensing circuit is just a fixed resistor and an LDR in series. Which component should the output be taken across?

Section 2 of 14

Core idea

Start hereStreet lights that switch on at dusk

A street light turns itself on when it gets dark. Inside is a light sensor in a potential divider: as light fades, the sensor's resistance changes, shifting how the voltage is shared, and the output crosses a threshold.

The big idea

$V_{\text{out}}=\frac{R_2}{R_1+R_2}V_{\text{in}}$ . Sensors change the share. Potentiometer: p.d. ∝ length.

1. The formula

A potential divider

V_{\text{out}}=\dfrac{2}{6}\times12=4

V across

R_2

Potential divider equation

$I=\frac{V_{\text{in}}}{R_1+R_2}$
Same current through both.
$V_{\text{out}}=IR_2=\frac{R_2}{R_1+R_2}V_{\text{in}}$
p.d. across $R_2$ .

Worked example12 V across 4 kΩ and 2 kΩ

Output across 2 kΩ; then loaded by a 2 kΩ voltmeter.

Goal: Two outputs.
Strategy: Loading puts 2 kΩ in parallel.

$\frac26\times12=4\text{ V}$
Unloaded.
$2\parallel2=1\text{ k}\Omega$
Voltmeter in parallel.
$\frac15\times12=2.4\text{ V}$
Loaded.

Answer

4

2.4

✓ CheckLoading always lowers the output ✓.

2. Sensors

Output across a fixed 5 kΩ resistor with an NTC thermistor above it

As the thermistor's resistance falls, it takes a smaller share, so

V_{\text{out}}

rises: a temperature sensor.

LDR at the top of a divider

Drawing diagram…

Swap the positions to make the output rise in the dark instead.

3. Potentiometers

Worked exampleComparing e.m.f.s

Balance lengths 60.0 cm (standard 1.50 V cell) and 72.0 cm (unknown).

Goal: Unknown e.m.f.
Strategy: $\frac{E_1}{E_2}=\frac{l_1}{l_2}$ (no current at balance).

$E=1.50\times\frac{72.0}{60.0}$
Ratio.
$=1.80\text{ V}$
Evaluate.

Answer

1.80

✓ CheckLonger balance length, larger e.m.f. ✓.

Common misconceptionWrong resistor on top

V_{\text{out}}

uses the resistor it's measured across.

Summary

$V_{\text{out}}=\frac{R_2}{R_1+R_2}V_{\text{in}}$ .
Sensors change the share.

Potentiometer: Variable potential divider.

HistoryPoggendorff's null method. Johann Poggendorff devised the potentiometer principle in

1841

. Because it measures at zero current, it reads a true e.m.f. Before digital voltmeters with huge input resistance existed, potentiometers were the most accurate way to measure voltage, and standards labs relied on them for a century.

Fun factVolume knobs. The volume knob on older radios and guitars is literally a potentiometer used as a potential divider: turning it slides a contact along a carbon track, choosing what fraction of the signal voltage goes to the amplifier. Musicians still call them "pots".

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