Two equal 10 k resistors form a divider across 5.0 V. Calculate the output across the lower resistor. [2 marks]

A divider has R_1 = 3.0 k (upper) and R_2 = 1.0 k across 12 V. Find the output across R_2. [2 marks]

ScotlandEngineering ScienceSyllabus dot point

How do we turn a physical condition like light or heat into a changing voltage?

The voltage divider and input transducers: calculating the output voltage of a divider and using the LDR and thermistor to make light- and temperature-sensing circuits.

An SQA National 5 Engineering Science answer on the voltage divider and input transducers, covering the divider equation, how the output voltage splits in proportion to resistance, and how an LDR or thermistor makes a light- or temperature-sensing circuit that produces a changing voltage.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-15

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 key area is asking
The voltage divider
Input transducers: the LDR and the thermistor
Making a sensing circuit
Why this matters
Try this

What this key area is asking

The SQA wants you to calculate the output voltage of a voltage divider and to explain how an LDR or thermistor in a divider turns a change in light or temperature into a changing voltage that can control a circuit.

The voltage divider

The supply voltage is shared between the two resistors in proportion to their resistances. The larger a resistor's share of the total resistance, the larger its share of the voltage. If the two resistors are equal, the output is exactly half the supply.

Input transducers: the LDR and the thermistor

The two you must know are:

Light-dependent resistor (LDR): its resistance is high in the dark and falls as the light gets brighter.
Thermistor: its resistance is high when cold and falls as it gets hotter (the common type used at National 5).

Making a sensing circuit

Replace one resistor of a voltage divider with an LDR or thermistor and the output voltage now changes with the condition. The position matters:

If the sensor is the upper resistor, its share of the voltage rises as its resistance rises, so the output across the lower resistor falls when its resistance rises.
If the sensor is the lower resistor, the output is taken across it, so the output rises as the sensor's resistance rises.

This changing voltage is then fed to a process sub-system - a transistor or a comparator - to switch an output device automatically.

Designing a dark-sensing output

You need an output voltage that rises as it gets darker, to trigger a light. Choose the LDR's position.

Step 1: decide what "gets darker" does to the LDR

As it gets darker the LDR's resistance increases.

Step 2: choose the position so the output rises

You want $V_{out}$ to rise when the LDR resistance rises. The output rises with a resistor's resistance only when the output is taken across that resistor, so make the LDR the lower resistor and take the output across it.

Step 3: check the behaviour

In the dark the LDR resistance is high, so it takes a large share of the supply and $V_{out}$ is high - which is what is needed to switch the light on. In bright light the LDR resistance is low and $V_{out}$ falls, switching the light off.

Why this matters

The voltage divider with a sensor is the heart of nearly every automatic control circuit at National 5: streetlights, frost alarms, fans and thermostats all start here. Master the divider equation and the LDR/thermistor behaviour and you can design the input stage of almost any control system.

Try this

Q1. Two equal $10 \text{ k}\Omega$ resistors form a divider across $5.0 \text{ V}$ . Calculate the output across the lower resistor. [2 marks]

Cue. $V_{out} = \frac{10}{10+10}\times 5.0 = 2.5 \text{ V}$ (half the supply).

Q2. State what happens to a thermistor's resistance as it gets hotter. [1 mark]

Cue. Its resistance decreases.

Q3. A divider has $R_1 = 3.0 \text{ k}\Omega$ (upper) and $R_2 = 1.0 \text{ k}\Omega$ across $12 \text{ V}$ . Find the output across $R_2$ . [2 marks]

Cue. $V_{out} = \frac{1.0}{3.0+1.0}\times 12 = 3.0 \text{ V}$ .

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 N5 style3 marksA voltage divider is made from a 4.0 kohm resistor (R1) and a 2.0 kohm resistor (R2) across a 9.0 V supply. Calculate the output voltage across R2.

Show worked answer →

Use the voltage divider equation for the lower resistor.

Relationship: $V_{out} = \dfrac{R_2}{R_1 + R_2} \times V_s$ .

Substitution: $V_{out} = \dfrac{2.0}{4.0 + 2.0} \times 9.0 = \dfrac{2.0}{6.0} \times 9.0 = 3.0 \text{ V}$ .

Markers reward selecting the divider equation, substituting correctly (the kilo-ohms cancel, so there is no need to convert), and a final answer in volts. The output is one third of the supply because R2 is one third of the total resistance.

SQA N5 style2 marksA voltage divider uses a light-dependent resistor (LDR) as the upper resistor and a fixed resistor across the output. Describe what happens to the output voltage as the surroundings get darker.

Show worked answer →

Reason from how the LDR's resistance changes.

As it gets darker, the LDR's resistance increases.

The LDR is the upper resistor, so it takes a larger share of the supply voltage, leaving a smaller share across the fixed output resistor. Therefore the output voltage decreases as it gets darker.

Markers reward stating the LDR resistance rises in the dark and correctly reasoning that the output across the lower fixed resistor falls. (If the LDR were the lower resistor the output would rise instead.)

Related dot points

Sources & how we know this

SQA National 5 Engineering Science Course Specification — SQA (2017)
SQA Engineering Science Data Booklet National 4/5 — SQA (2017)