EnglandPhysicsSyllabus dot point

How can we use resistors to supply a chosen fraction of a source voltage?

The potential divider as a way of producing a required potential difference, the divider equation, the use of variable resistors and potentiometers, and divider circuits using thermistors and LDRs as sensors.

A focused answer to AQA A-Level Physics 3.5.1.5, covering the potential divider as a way of supplying a chosen potential difference, the divider equation, the use of variable resistors and potentiometers, and sensing circuits built with thermistors and light-dependent resistors.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The potential divider
Variable resistors and potentiometers
Sensor circuits
Try this

What this dot point is asking

AQA specification point 3.5.1.5 wants you to explain how a potential divider supplies a chosen potential difference, use the divider equation, describe the role of variable resistors and potentiometers, and analyse divider circuits that use thermistors or LDRs as sensors.

The potential divider

The supply voltage is shared between the resistors in proportion to their resistances, so the larger resistor takes the larger share. The equation follows directly from the series rules: the same current $I = \dfrac{V_{\text{in}}}{R_1 + R_2}$ flows through both, and the output is $V_{\text{out}} = IR_2$ . Substituting gives the divider equation. Note that the result depends only on the ratio of resistances, so any consistent units (ohms or kilohms) can be used.

A crucial practical point is loading. When a real load (which has its own resistance) is connected across the output, it sits in parallel with $R_2$ , lowering the effective resistance and reducing $V_{\text{out}}$ . The divider works best when the load resistance is much larger than $R_2$ , so the unloaded equation remains a good approximation.

Variable resistors and potentiometers

A simple variable resistor (rheostat) instead uses two terminals to control the current in series with a load. This gives a more limited control of output voltage than a potentiometer, because a rheostat alone cannot reduce the output all the way to zero and its output depends on the load.

Sensor circuits

If one resistor is replaced by a thermistor, the output voltage changes as the temperature changes; with an LDR, it changes with light intensity. As the sensor's resistance falls, its share of the supply voltage falls and the other resistor's share rises. These circuits drive alarms, thermostats, automatic street lighting and frost detectors, often by feeding the output into a comparator or transistor that switches at a chosen threshold.

Designing a temperature-warning sensor

A circuit must give a rising output voltage as a room warms, to switch on a cooling fan above a set temperature.

step 1: Choose the arrangement

Put an NTC thermistor in series with a fixed resistor across the supply, and take the output across the fixed resistor (which acts as $R_2$ ).

step 2: Reason about the thermistor change

When the temperature rises, the thermistor's resistance ( $R_1$ ) falls, so it takes a smaller share of the supply voltage.

step 3: Apply the divider equation

With a smaller $R_1$ , the ratio $\dfrac{R_2}{R_1 + R_2}$ increases, so the output across the fixed resistor rises.

step 4: Use the output

Feed the rising voltage to a comparator or transistor base so the fan switches on once the output passes a preset threshold.

Try this

Q1. Write the potential divider equation for the output across $R_2$ . [1 mark]

Cue. $V_{\text{out}} = V_{\text{in}} \dfrac{R_2}{R_1 + R_2}$ .

Q2. In a divider with an LDR, state what happens to the LDR's resistance as the light gets brighter. [1 mark]

Cue. Its resistance decreases.

Q3. State one everyday device that uses a potentiometer. [1 mark]

Cue. A volume control or a dimmer switch.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20193 marksA potential divider connects a

4.0 \text{ k}\Omega

resistor in series with an

8.0 \text{ k}\Omega

resistor across a

12 \text{ V}

supply. Calculate the potential difference across the

8.0 \text{ k}\Omega

resistor.

Show worked answer →

Apply the potential divider equation, taking the output across the $8.0 \text{ k}\Omega$ resistor.

$V_{\text{out}} = V_{\text{in}} \times \dfrac{R_2}{R_1 + R_2} = 12 \times \dfrac{8.0}{4.0 + 8.0}$ .

$V_{\text{out}} = 12 \times \dfrac{8.0}{12.0} = 8.0 \text{ V}$ .

Markers reward correct identification of which resistor the output is taken across (the larger resistor takes the larger share) and correct substitution. Because the resistances are in the same units (kilohms) the ratio is unaffected.

AQA 20214 marksDescribe how a potential divider containing a thermistor can be used to produce an output voltage that increases as the temperature rises. Explain your reasoning.

Show worked answer →

Connect a fixed resistor in series with an NTC thermistor across the supply, and take the output across the fixed resistor.

As the temperature rises, the thermistor's resistance falls, so its share of the supply voltage falls. By conservation of energy around the loop, the remaining share across the fixed resistor must rise, so the output voltage increases.

Using the divider equation with the fixed resistor as $R_2$ , a smaller thermistor resistance $R_1$ increases the ratio $\dfrac{R_2}{R_1 + R_2}$ and hence $V_{\text{out}}$ .

Markers reward the correct circuit (output across the fixed resistor), the falling thermistor resistance, and linking the changed ratio to a rising output.

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Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)