ScotlandEngineering ScienceSyllabus dot point

How does a potential divider turn a sensor's changing resistance into a changing voltage for the rest of a circuit?

The potential divider relationship and its use with input transducers such as the thermistor and the light-dependent resistor to produce a sensing voltage.

An SQA Higher Engineering Science answer on the potential divider relationship and how it is used with input transducers such as the thermistor and light-dependent resistor to produce a voltage that responds to temperature or light.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

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Quick answer

A potential divider is two resistors in series across a supply; the supply voltage divides between them in proportion to their resistances. The voltage across one resistor is $V_{\text{out}} = V_S \times \dfrac{R}{R_1 + R_2}$ , where $R$ is the resistor you take the output across. An input transducer is a sensor whose resistance changes with a physical quantity: a thermistor's resistance falls as temperature rises, and an LDR's resistance falls as light increases (high in the dark). Putting a transducer and a fixed resistor in a divider gives an output voltage that responds to temperature or light, ready to drive the next stage. Swapping which resistor the output is across reverses the direction of the change, and changing the fixed resistor shifts the trigger point.

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What this key area is asking
The potential divider
Input transducers
Working out the sensing voltage
Setting and changing the trigger point
Examples in context
Try this

What this key area is asking

The SQA wants you to use the potential divider relationship and to build an input sub-system from a fixed resistor and an input transducer (a sensor whose resistance changes), such as a thermistor or a light-dependent resistor (LDR). The potential divider is the workhorse input stage of nearly every control circuit, turning a changing resistance into a changing voltage that a comparator, amplifier or transistor can use.

The potential divider

The rule follows directly from Ohm's law in a series circuit: the current is the same through both resistors, so each resistor's voltage is its share of the total resistance times the supply. The resistor whose voltage you want goes on top of the fraction.

flowchart TB Vs["+ supply Vs"] --> R1["R1 (top)"] --> M(("Vout node")) --> R2["R2 (bottom)"] --> G["0 V"] M --> O["Vout taken here<br/>(across R2)"]

Input transducers

An input transducer converts a physical quantity into an electrical one. The two the course relies on change their resistance:

Thermistor. A temperature sensor whose resistance falls as temperature rises (the common negative-temperature-coefficient type). Cold means high resistance; hot means low resistance.
Light-dependent resistor (LDR). A light sensor whose resistance falls as light increases. Dark means high resistance (often megohms); bright means low resistance.

On their own these give a changing resistance, which a circuit cannot act on directly. Placed in a potential divider with a fixed resistor, they produce a changing voltage, which is exactly what the processing stage needs.

Working out the sensing voltage

Thermistor divider output

A thermistor is in series with a fixed 1 kilohm resistor across a 6 V supply, with the output taken across the fixed resistor. The thermistor reads 3 kilohms when cold and 0.5 kilohms when hot. Find the output voltage in each case.

step 1 Cold case (thermistor 3 kilohm)

Output is across the fixed resistor, so it goes on top: $V_{\text{out}} = 6 \times \dfrac{1}{1 + 3} = 6 \times \dfrac{1}{4} = 1.5\ \text{V}$ .

step 2 Hot case (thermistor 0.5 kilohm)

$V_{\text{out}} = 6 \times \dfrac{1}{1 + 0.5} = 6 \times \dfrac{1}{1.5} = 4.0\ \text{V}$ .

step 3 Interpret

As the thermistor warms, its resistance falls, so the fixed resistor's share (the output) rises from 1.5 V to 4.0 V. This rising voltage can switch a cooling fan on once it passes a set threshold.

Setting and changing the trigger point

The whole input sub-system feeds a stage (a comparator or a transistor) that switches when the output voltage crosses a threshold. The light or temperature at which switching happens is set by the fixed resistor: making it larger or smaller shifts the divider output for a given transducer resistance, so the same circuit triggers at a different level. Using a variable resistor in place of the fixed one lets a user adjust the trigger point, which is how the sensitivity dial on a dusk-to-dawn light or a thermostat works.

Examples in context

A dusk-to-dawn light uses an LDR divider: as it darkens the LDR resistance rises, the output (taken to switch the lamp on) crosses the threshold, and the lamp comes on. Turning the sensitivity dial changes the fixed (variable) resistor and so the darkness level at which it triggers. A fridge thermostat uses a thermistor divider: as the inside warms the thermistor resistance falls, the output rises, and once it passes the threshold the compressor switches on to cool again. In both, the potential divider is the bridge from a sensor's changing resistance to a usable control voltage.

Try this

Q1. Two equal resistors form a divider across 9 V. State the output voltage across one of them. [1 mark]

Cue. Equal resistors share equally: $V_{\text{out}} = 9 \times \frac{R}{2R} = 4.5\ \text{V}$ .

Q2. State how a thermistor's resistance changes as it gets hotter. [1 mark]

Cue. It decreases (for the common negative-temperature-coefficient thermistor).

Q3. A 2 kilohm fixed resistor is in series with a sensor of 6 kilohms across 8 V, output across the sensor. Find the output. [2 marks]

Cue. $V_{\text{out}} = 8 \times \frac{6}{6 + 2} = 8 \times \frac{6}{8} = 6\ \text{V}$ .

Exam-style practice questions

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

SQA Higher (specimen)3 marksA potential divider is made of a fixed 4.7 kilohm resistor in series with a sensor across a 5 V supply. When the sensor resistance is 2.3 kilohms, calculate the voltage across the sensor.

Show worked answer →

Use the potential divider relationship for the voltage across the sensor.

Relationship: $V_{\text{sensor}} = V_S \times \dfrac{R_{\text{sensor}}}{R_{\text{sensor}} + R_{\text{fixed}}}$ .

Substitution: $V_{\text{sensor}} = 5 \times \dfrac{2.3}{2.3 + 4.7}$ (the kilohm units cancel in the ratio).

Working: $V_{\text{sensor}} = 5 \times \dfrac{2.3}{7.0} = 5 \times 0.329 = 1.64\ \text{V}$ .

Markers reward the correct form of the divider rule (the resistor whose voltage you want on top), a clean substitution with consistent units, and the final answer. Putting the wrong resistor on top is the usual error.

SQA Higher (specimen)4 marksA light-dependent resistor is connected in series with a fixed resistor across a supply, with the output taken across the fixed resistor. Explain how the output voltage changes as it gets darker, and state how to make the circuit trigger at a different light level.

Show worked answer →

As it gets darker the LDR's resistance increases (an LDR has high resistance in the dark, low in bright light).

With the output taken across the fixed resistor, the LDR's growing share of the supply voltage means the fixed resistor's share, the output, falls. So as it gets darker the output voltage decreases. (If the output were taken across the LDR instead, it would rise as it darkened.)

To trigger at a different light level, change the value of the fixed resistor (or use a variable resistor): this shifts the light level at which the output reaches the switching threshold.

Markers reward the LDR resistance rising in the dark, a correct link to the output voltage based on where the output is taken, and adjusting the fixed (or variable) resistor to change the trigger point.

Related dot points

Sources & how we know this

SQA Higher Engineering Science Course Specification — SQA (2019)
Higher Engineering Science Course Specification (PDF) — SQA (2019)