A divider has R_1 = R_2 = 1.0\ k across 6.0\ V. Find the output across R_2. [2 marks]

WJEC Eduqas Eduqas 2019-style practice: A potential divider is made from a 4.7\ k resistor (top) in series with a 2.2\ k resistor (bottom) across a 9.0\ V supply, with the output taken across the bottom resistor. Calculate the output voltage.

Use the potential-divider equation with the output across the bottom resistor R_2: V_out = V_inR_2R_1 + R_2. Substitute: V_out = 9.0 × 2.24.7 + 2.2 = 9.0 × 2.26.9. Evaluate: 2.26.9 = 0.319, so V_out = 9.0 × 0.319 = 2.87\ V ≈ 2.9\ V. Markers reward correct substitution, the ratio, and the answer 2.9\ V. Resistor units cancel, so kilohms can be used directly. The usual error is putting the wrong resistor on top.

WJEC Eduqas Eduqas 2021-style practice: A potential divider must give an output of 3.0\ V from a 12\ V supply. The bottom resistor is fixed at 1.0\ k. Calculate the value of the top resistor required.

The output across the bottom resistor is V_out = V_inR_2R_1 + R_2. The output is a quarter of the supply ((3.0)/(12) = (1)/(4)), so the bottom resistor must be a quarter of the total resistance. Rearrange: R_2R_1 + R_2 = 3.012 = 0.25, so R_1 + R_2 = 4 R_2 = 4 × 1.0\ k = 4.0\ k. Therefore R_1 = 4.0 - 1.0 = 3.0\ k. Markers reward forming the ratio, the total resistance 4.0\ k, and R_1 = 3.0\ k (nearest preferred value 3.0\ k via 2.7 + 0.3, or simply 3.0\ k).

EnglandElectronicsSyllabus dot point

How does a potential divider produce a chosen fraction of the supply voltage?

Potential dividers: the potential-divider equation, choosing resistor values for a target output voltage, and the effect of loading the output.

An Eduqas GCSE Electronics answer on potential dividers: how two series resistors split the supply, the potential-divider equation, choosing resistor values for a target output voltage, and how connecting a load changes the output.

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

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

Eduqas wants you to use a potential divider: two resistors in series that split the supply voltage, with the potential-divider equation giving the output. You must choose resistor values for a target output voltage and explain how connecting a load changes the output. The potential divider is the input subsystem of almost every sensing circuit.

The answer

The potential divider

The potential-divider equation

Choosing resistor values

The loading effect

Designing a divider and checking loading

Design a divider to give $4.0\ \text{V}$ from a $10\ \text{V}$ supply using a $2.0\ \text{k}\Omega$ lower resistor, then find the output if a $10\ \text{k}\Omega$ load is connected.

step 1: Find the top resistor for the target output

$\dfrac{R_2}{R_1 + R_2} = \dfrac{4.0}{10} = 0.40$ , so $R_1 + R_2 = \dfrac{R_2}{0.40} = \dfrac{2.0}{0.40} = 5.0\ \text{k}\Omega$ , giving $R_1 = 5.0 - 2.0 = 3.0\ \text{k}\Omega$ .

step 2: Account for the load in parallel with R2

The $10\ \text{k}\Omega$ load is in parallel with $R_2 = 2.0\ \text{k}\Omega$ : $R_2' = \dfrac{2.0 \times 10}{2.0 + 10} = \dfrac{20}{12} = 1.67\ \text{k}\Omega$ .

step 3: Recalculate the loaded output

$V_\text{out} = 10 \times \dfrac{1.67}{3.0 + 1.67} = 10 \times \dfrac{1.67}{4.67} = 3.6\ \text{V}$ . The load has pulled the output down from $4.0\ \text{V}$ to $3.6\ \text{V}$ , the loading effect.

Examples in context

The potential divider is the input subsystem of nearly every sensing circuit. Replace one resistor with a light-dependent resistor or thermistor and the output voltage tracks light or temperature, ready for a comparator. A potentiometer is a variable divider used to set a reference voltage or a volume level. Understanding loading explains why sensor outputs feed high-input-resistance stages such as comparators and op-amps, and the non-exam assessment expects you to design dividers with real preferred values.

Try this

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

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

Q2. A divider has $R_1 = R_2 = 1.0\ \text{k}\Omega$ across $6.0\ \text{V}$ . Find the output across $R_2$ . [2 marks]

Cue. Equal resistors share equally: $V_\text{out} = 6.0 \times \frac{1}{2} = 3.0\ \text{V}$ .

Q3. State what happens to a divider's output when a low-resistance load is connected across it, and why. [2 marks]

Cue. It falls; the load in parallel with $R_2$ reduces the effective lower resistance, so $R_2$ takes a smaller share of the supply.

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 20193 marksA potential divider is made from a

4.7\ \text{k}\Omega

resistor (top) in series with a

2.2\ \text{k}\Omega

resistor (bottom) across a

9.0\ \text{V}

supply, with the output taken across the bottom resistor. Calculate the output voltage.

Show worked answer →

Use the potential-divider equation with the output across the bottom resistor $R_2$ : $V_\text{out} = V_\text{in}\dfrac{R_2}{R_1 + R_2}$ .

Substitute: $V_\text{out} = 9.0 \times \dfrac{2.2}{4.7 + 2.2} = 9.0 \times \dfrac{2.2}{6.9}$ .

Evaluate: $\dfrac{2.2}{6.9} = 0.319$ , so $V_\text{out} = 9.0 \times 0.319 = 2.87\ \text{V} \approx 2.9\ \text{V}$ .

Markers reward correct substitution, the ratio, and the answer $2.9\ \text{V}$ . Resistor units cancel, so kilohms can be used directly. The usual error is putting the wrong resistor on top.

Eduqas 20213 marksA potential divider must give an output of

3.0\ \text{V}

from a

12\ \text{V}

supply. The bottom resistor is fixed at

1.0\ \text{k}\Omega

. Calculate the value of the top resistor required.

Show worked answer →

The output across the bottom resistor is $V_\text{out} = V_\text{in}\dfrac{R_2}{R_1 + R_2}$ . The output is a quarter of the supply ( $\frac{3.0}{12} = \frac{1}{4}$ ), so the bottom resistor must be a quarter of the total resistance.

Rearrange: $\dfrac{R_2}{R_1 + R_2} = \dfrac{3.0}{12} = 0.25$ , so $R_1 + R_2 = 4 R_2 = 4 \times 1.0\ \text{k}\Omega = 4.0\ \text{k}\Omega$ .

Therefore $R_1 = 4.0 - 1.0 = 3.0\ \text{k}\Omega$ .

Markers reward forming the ratio, the total resistance $4.0\ \text{k}\Omega$ , and $R_1 = 3.0\ \text{k}\Omega$ (nearest preferred value $3.0\ \text{k}\Omega$ via $2.7 + 0.3$ , or simply $3.0\ \text{k}\Omega$ ).

Related dot points

Sources & how we know this

WJEC Eduqas GCSE (9-1) Electronics specification (C490) — WJEC Eduqas (2017)