AS Level · Topic 10.3
A-Level 9702 / Topic 10 / AS

Tapping off a voltage.

Two resistors in series share the supply voltage in proportion to their resistances. Make one of them a sensor and the tapped-off voltage becomes a signal that responds to temperature or light.

The key idea

A potential divider uses two resistors in series to split a supply voltage. The output across R₂ is Vₒᵤₜ = Vᵢₙ × R₂ / (R₁ + R₂). A potentiometer is a continuously variable divider, used in null methods to compare potential differences. Replacing one resistor with a thermistor or LDR makes a sensor circuit whose output voltage changes with temperature or light.

R₁ R₂ Vout Vin
Fig. 1 — Potential divider: Vₒₘₜ = R₂ ÷ (R₁ + R₂) × Vₐₙ
Section 01

Slide the tap.

Move the divider tap, or swap in a thermistor or LDR, and watch the output voltage follow V × R₂ / (R₁ + R₂) as the resistances change.

Section 02

The divider rule.

The output is the supply scaled by the resistance ratio.

RelationUse
Vₒᵤₜ = V R₂/(R₁+R₂)potential divider output
potentiometera variable divider for null methods
sensor dividerthermistor or LDR sets the output
Stage 1 · Learn

Check what the sim just showed you

Four quick checks tied to this lesson. Each correct answer earns XP and lights this skill on your star map.

Quick check+10 XP

In a potential divider, the output voltage across R₂ is:

Quick check+10 XP

A 12 V supply is divided by two equal resistors. The output across one of them is:

Quick check+10 XP

A potentiometer is used to:

Quick check+10 XP

In a potential divider, a 4.0 V output is wanted from a 12 V supply across R₂ = 2.0 kΩ. The value of R₁ is:

Section 03

Building a sensor.

Sensor dividers turn a physical change into a voltage.

Examiner trap

Use the resistance of the component you take the output across in the numerator: Vₒᵤₜ = V R₂/(R₁ + R₂). When a sensor replaces one resistor, decide carefully which resistor the output is measured across, because a falling sensor resistance can make the output rise or fall depending on the arrangement. The unloaded divider equation assumes negligible current is drawn from the output.

Stage 2 · Exam

Exam-style questions

Unlocks once the checks above are done. Worth more XP, written to AS Paper 1 and 2 standard.

Finish the checks above to unlock the exam questions
Exam style+20 XP

A 9.0 V supply is connected across a 1.0 kΩ and a 2.0 kΩ resistor in series. The p.d. across the 2.0 kΩ resistor is:

Exam style+20 XP

In a thermistor potential divider, the temperature rises. The thermistor resistance falls, so the p.d. across the thermistor:

Exam style+20 XP

A potentiometer compares two potential differences by adjusting until the galvanometer reads:

Skill unlocked

Potential dividers, mastered.

This skill is now lit gold on your star map. You have finished the lessons of Topic 10; the Paper 1 set awaits.

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Go deeper · practice
Six original Cambridge-style questions
Using the potential divider equation, designing sensor circuits, and the potentiometer null method. Attempt each, then reveal the worked solution.
Stage 3 · Paper 1 readiness
D.C. circuits · Paper 1 Practice
A bank of original multiple-choice questions across the whole topic, in the style of Paper 1. You have now seen all three lessons, so this is the moment to test the unit as a whole.
Start Paper 1 Practice →