A divider has a 2.0\ k top resistor and a 3.0\ k bottom resistor across 10\ V. Find the output across the bottom resistor. [2 marks]

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How does a potential divider set a reference voltage and turn a sensor into a usable signal?

Potential dividers: the divider equation, loading effects, and sensing circuits using thermistors, light-dependent resistors and strain gauges to convert a physical quantity into a voltage.

An Eduqas A-Level Electronics answer on potential dividers and sensing: the potential-divider equation, how loading a divider changes its output, and how thermistors, light-dependent resistors and strain gauges form the input subsystem of a sensing circuit that converts temperature, light or strain into a voltage.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

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

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Quick answer

A potential divider is two resistors in series across a supply; the output taken across $R_2$ is the supply voltage shared in the ratio of the resistances:

V_\text{out} = V_\text{in}\frac{R_2}{R_1 + R_2}.

Replace one resistor with a transducer whose resistance depends on a physical quantity, a thermistor (resistance falls as temperature rises, for the common NTC type), a light-dependent resistor (resistance falls as light increases), or a strain gauge (resistance rises as it is stretched), and the output becomes a voltage that tracks that quantity. This is the sensing front end of a system. Loading the output with a low-resistance next stage pulls the output below its unloaded value; use a high-impedance load or a voltage-follower buffer to avoid this.

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What this dot point is asking
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What this dot point is asking

Eduqas wants you to use the potential-divider equation, explain and account for loading effects, and design sensing circuits that turn a physical quantity (temperature, light, strain) into a voltage using a thermistor, a light-dependent resistor or a strain gauge. The potential divider is the input subsystem of countless electronic systems.

The answer

The potential-divider equation

Sensing with a thermistor and a light-dependent resistor

Sensing with a strain gauge

Loading effects

Designing a divider to give a chosen output

Design a potential divider from a $9.0\ \text{V}$ supply that produces $3.0\ \text{V}$ at the output, using a $4.7\ \text{k}\Omega$ resistor for the lower arm.

step 1: Write the divider equation for the unknown

$V_\text{out} = V_\text{in}\dfrac{R_2}{R_1 + R_2}$ , so $3.0 = 9.0 \times \dfrac{4700}{R_1 + 4700}$ .

step 2: Rearrange for $R_1$

$\dfrac{4700}{R_1 + 4700} = \dfrac{1}{3}$ , so $R_1 + 4700 = 3 \times 4700 = 14100$ , giving $R_1 = 9400\ \Omega$ .

step 3: Choose a preferred value

The nearest E24 preferred value is $9.1\ \text{k}\Omega$ , which gives $V_\text{out} = 9.0 \times \frac{4700}{13800} = 3.07\ \text{V}$ , close enough for most designs.

Examples in context

The potential divider is everywhere: a volume control is a variable divider, a thermostat compares a thermistor divider against a reference, a camera light meter uses an LDR divider, and electronic scales use a strain-gauge bridge. In a microcontroller system, a divider scales a sensor's range to the input range of the analogue-to-digital converter. The loading problem is exactly why op-amp voltage followers are so common between a sensor and the next stage.

Try this

Q1. A divider has a $2.0\ \text{k}\Omega$ top resistor and a $3.0\ \text{k}\Omega$ bottom resistor across $10\ \text{V}$ . Find the output across the bottom resistor. [2 marks]

Cue. $V_\text{out} = 10 \times \frac{3}{5} = 6.0\ \text{V}$ .

Q2. State what happens to the resistance of an LDR as the light level increases. [1 mark]

Cue. It decreases.

Q3. State one reason to buffer a potential-divider output with a voltage follower. [1 mark]

Cue. The follower's very high input resistance stops the next stage loading the divider and pulling the output down.

Exam-style practice questions

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

Eduqas 20205 marksA potential divider is made from a

10\ \text{k}\Omega

fixed resistor in series with a thermistor, connected across a

5.0\ \text{V}

supply, with the output taken across the thermistor. At

25\ ^\circ\text{C}

the thermistor resistance is

15\ \text{k}\Omega

. Calculate the output voltage, and explain how it changes as the temperature rises.

Show worked answer →

Output across the thermistor (up to 3 marks): $V_\text{out} = V_\text{in}\dfrac{R_\text{th}}{R_\text{th} + R_\text{fixed}} = 5.0 \times \dfrac{15}{15 + 10} = 5.0 \times \dfrac{15}{25} = 3.0\ \text{V}$ .

Effect of temperature (up to 2 marks): a thermistor (NTC) has a resistance that falls as temperature rises. The thermistor therefore takes a smaller share of the supply voltage, so the output voltage across it decreases as the temperature rises.

Markers reward the divider ratio, the output $3.0\ \text{V}$ , and the correct direction (output falls as temperature rises because the NTC resistance drops).

Eduqas 20224 marksExplain what is meant by the loading effect on a potential divider, and state one way to reduce it when the output drives a later stage.

Show worked answer →

Loading effect (up to 3 marks): connecting a load (the next stage) across the output resistor places its input resistance in parallel with that resistor. This lowers the effective resistance of the lower arm, so the output voltage falls below the unloaded divider value. The smaller the load resistance compared with the divider resistors, the larger the drop.

Reducing it (up to 1 mark): use a load whose input resistance is much larger than the divider resistors, or buffer the output with a voltage follower (an op-amp with near-infinite input resistance), so the divider sees effectively no load.

Markers reward the parallel-loading explanation, the resulting drop in output, and a valid fix (high-impedance load or a voltage-follower buffer).

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Sources & how we know this

Eduqas GCE AS/A Level Electronics specification (A410QS) — WJEC Eduqas (2017)