Northern IrelandTechnology and DesignSyllabus dot point

How do sensors and potential dividers turn a physical quantity into a usable input voltage?

Input transducers (LDR, thermistor, switches) and the potential divider as a sensing subsystem.

A CCEA A-Level Technology and Design answer on input transducers such as the LDR, thermistor and switches, and how a potential divider converts a changing resistance into a changing voltage for the process subsystem to act on.

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

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

CCEA expects you to know the common input transducers (LDR, thermistor, switches and other sensors), to use the potential divider as a sensing subsystem, and to calculate its output voltage. You must explain how the divider's wiring decides whether the output rises or falls with the sensed quantity.

The answer

Input transducers

The potential divider

Which way does the output move?

Worked example: designing a dark-detecting input

Making the output rise as it gets dark

You want an input whose voltage rises in darkness, to switch a light on.

step Choose the sensor behaviour

An LDR has high resistance in the dark and low resistance in light. You want a high output in the dark.

step Position the LDR

Put the LDR at the bottom and take the output across the LDR, with a fixed resistor on top. In darkness the LDR resistance is large, so it takes the larger share of the supply and the output is high; in light its resistance is small and the output is low.

step Check with the formula

With a 10 kilohm fixed resistor, a 6 V supply, and LDR resistance 50 kilohm in the dark, the output across the LDR is $6 \times 50/(10+50) = 5.0\ \text{V}$ (high); in bright light at 1 kilohm it is $6 \times 1/(10+1) = 0.55\ \text{V}$ (low). The rising-in-dark behaviour is exactly what is needed to trigger a lamp, and confirms the design.

Examples in context

Example 1. Garden light. An LDR potential divider gives a high voltage at dusk, switching the lamp on automatically, the everyday use of a light sensor exactly as analysed above.

Example 2. Fridge over-temperature alarm. A thermistor divider arranged so the output rises when the fridge gets too warm feeds a comparator that sounds an alarm, showing the same divider principle applied to heat.

Try this

Q1. State how the resistance of an NTC thermistor changes as it gets hotter. [1 mark]

Cue. It decreases (falls).

Q2. A divider has a 2 kilohm fixed resistor in series with a 6 kilohm sensor across 8 V, output across the sensor. Find the output voltage. [2 marks]

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

Q3. How would you wire an LDR divider so the output voltage is high in bright light? [2 marks]

Cue. Put the LDR on top (in series) and take the output across the fixed resistor; in light the LDR resistance is low, so the fixed resistor takes the larger share and the output is high.

Exam-style practice questions

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

CCEA 20186 marksA potential divider uses a 10 kilohm fixed resistor in series with an LDR across a 6 V supply, with the output taken across the fixed resistor. In darkness the LDR has a resistance of 50 kilohm and in bright light 1 kilohm. Calculate the output voltage in each case and state how the circuit would be used.

Show worked answer →

The output is across the fixed 10 kilohm resistor, so

V_{\text{out}} = V_s \times \frac{R_{\text{fixed}}}{R_{\text{fixed}} + R_{\text{LDR}}}.

In darkness ( $R_{\text{LDR}} = 50\ \text{k}\Omega$ ):

V_{\text{out}} = 6 \times \frac{10}{10 + 50} = 6 \times \frac{10}{60} = 1.0\ \text{V}.

In bright light ( $R_{\text{LDR}} = 1\ \text{k}\Omega$ ):

V_{\text{out}} = 6 \times \frac{10}{10 + 1} = 6 \times \frac{10}{11} = 5.45\ \text{V}.

So the output rises as the light increases. This is a light sensor: feeding the output to a comparator or transistor lets the system switch when light crosses a threshold (for example to detect daylight). If the output were taken across the LDR instead, it would fall as light increased, suiting a dark-detecting circuit.

Markers reward the correct divider formula with the fixed resistor on top, both calculated voltages, and a correct statement that the output rises with light (light sensor).

CCEA 20204 marksExplain how swapping the positions of the fixed resistor and the thermistor in a potential divider changes whether the output rises or falls with temperature.

Show worked answer →

In a potential divider the output is the share of the supply across the lower component, and a thermistor (NTC) falls in resistance as temperature rises.

If the output is taken across the thermistor (thermistor at the bottom), then as temperature rises its resistance falls, it takes a smaller share of the supply, and the output voltage falls. This suits detecting heat by a falling voltage.
If the thermistor is at the top and the output is across the fixed resistor, then as temperature rises the thermistor's resistance falls, the fixed resistor takes a larger share, and the output voltage rises with temperature. This suits a heat sensor that triggers on a rising voltage.

So swapping the positions inverts the direction of the output change. Markers want the link between which component the output is across and whether the voltage rises or falls, using the NTC behaviour of the thermistor.

Related dot points

Sources & how we know this

CCEA GCE Technology and Design specification — CCEA (2016)