What is not rounding to a preferred value?

A designed resistor must be rounded to the nearest real (E24) value, then the output re-checked.

What is algebra slips when rearranging?

Cross-multiply carefully when solving for an unknown resistor.

A divider has two equal 10\,k resistors across a 6.0\,V supply. State the output voltage. [1 mark]

A divider with R_1 = 4.0\,k and R_2 = 2.0\,k is across a 9.0\,V supply, output across R_2. Calculate the output. [2 marks]

WJEC Eduqas style-style practice: A potential divider has a 2.0\,k resistor and a 4.0\,k resistor in series across a 9.0\,V supply. The output is taken across the 4.0\,k resistor. Calculate the output voltage.

A Component 1 Calculate question on the potential divider equation. Use V_out = V_in × R_2R_1 + R_2 with R_2 = 4.0\,k (the resistor the output is taken across) and R_1 = 2.0\,k (1 mark for the equation with the right resistor on top). Substitute: V_out = 9.0 × 4.02.0 + 4.0 = 9.0 × 4.06.0 = 6.0\,V (2 marks for the substitution and the answer). Markers reward the correct resistor in the numerator and the value. A common error is to put the wrong resistor on top.

WalesElectronicsSyllabus dot point

How does a potential divider produce a chosen output voltage, and how do we design one?

The potential divider: how two resistors share a supply voltage, the potential divider equation, and designing and analysing a divider to produce a required output voltage.

A focused answer to WJEC Eduqas GCSE Electronics on potential dividers, covering how two resistors share a supply voltage, the potential divider equation, and designing and analysing a divider for a required output voltage.

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

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

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What this topic is asking
How a potential divider works
The potential divider equation
Designing a divider
Try this

What this topic is asking

WJEC Eduqas wants you to understand the potential divider (voltage divider): two resistors in series that share a supply voltage so you can tap off a smaller, chosen voltage. You must use the potential divider equation to find the output voltage, and rearrange it to design a divider that gives a required output. This is one of the most important circuits in the whole course.

How a potential divider works

The divider works because the resistors are in series, so they obey the series voltage rule: the two voltages add up to the supply. The voltage across each resistor is its share of the total, set by its fraction of the total resistance. By choosing the two resistor values, you choose exactly how the supply voltage splits, and so the output voltage at the junction.

The potential divider equation

The key is which resistor goes on top: $R_2$ is the one the output is measured across. If the output is across the lower resistor, that resistor is $R_2$ . The fraction $\dfrac{R_2}{R_1 + R_2}$ is the share of the supply voltage that appears at the output. If the two resistors are equal, the output is half the supply; if $R_2$ is much larger than $R_1$ , the output is close to the full supply.

Designing a divider

Designing means working backwards: you know $V_{\text{in}}$ and the wanted $V_{\text{out}}$ , and you fix one resistor, then solve for the other. After finding the ideal value, you round to a real preferred value, which slightly changes the output, so you re-check it is close enough. The resistor values also set the current through the divider: very large resistors draw little current (good for battery life) but make the output more easily disturbed by whatever it drives; smaller resistors give a "stiffer" output but waste more power.

Finding and then designing a divider output

Part A: a divider has $R_1 = 3.0\,\text{k}\Omega$ (top) and $R_2 = 1.0\,\text{k}\Omega$ (bottom) across a $12\,\text{V}$ supply, output across $R_2$ . Part B: change $R_1$ so the output becomes $4.0\,\text{V}$ .

step 1: Part A - apply the equation

$V_{\text{out}} = V_{\text{in}} \times \dfrac{R_2}{R_1 + R_2} = 12 \times \dfrac{1.0}{3.0 + 1.0} = 12 \times \dfrac{1.0}{4.0} = 3.0\,\text{V}$ .

step 2: Part B - set up the design equation

We want $V_{\text{out}} = 4.0\,\text{V}$ with $R_2 = 1.0\,\text{k}\Omega$ : $4.0 = 12 \times \dfrac{1.0}{R_1 + 1.0}$ .

step 3: Rearrange for the unknown

$\dfrac{4.0}{12} = \dfrac{1.0}{R_1 + 1.0}$ , so $R_1 + 1.0 = \dfrac{1.0 \times 12}{4.0} = 3.0$ , giving $R_1 = 2.0\,\text{k}\Omega$ .

step 4: State the answers

Part A gives $3.0\,\text{V}$ ; Part B needs $R_1 = 2.0\,\text{k}\Omega$ , which is an E24 preferred value, so no rounding is needed. Reducing the top resistor increased the output, because $R_2$ now takes a larger share of the supply.

Try this

Q1. A divider has two equal $10\,\text{k}\Omega$ resistors across a $6.0\,\text{V}$ supply. State the output voltage. [1 mark]

Cue. Equal resistors share equally: $V_{\text{out}} = 3.0\,\text{V}$ .

Q2. A divider with $R_1 = 4.0\,\text{k}\Omega$ and $R_2 = 2.0\,\text{k}\Omega$ is across a $9.0\,\text{V}$ supply, output across $R_2$ . Calculate the output. [2 marks]

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

Exam-style practice questions

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

Eduqas style3 marksA potential divider has a

2.0\,\text{k}\Omega

resistor and a

4.0\,\text{k}\Omega

resistor in series across a

9.0\,\text{V}

supply. The output is taken across the

4.0\,\text{k}\Omega

resistor. Calculate the output voltage.

Show worked answer →

A Component 1 Calculate question on the potential divider equation. Use $V_{\text{out}} = V_{\text{in}} \times \dfrac{R_2}{R_1 + R_2}$ with $R_2 = 4.0\,\text{k}\Omega$ (the resistor the output is taken across) and $R_1 = 2.0\,\text{k}\Omega$ (1 mark for the equation with the right resistor on top). Substitute: $V_{\text{out}} = 9.0 \times \dfrac{4.0}{2.0 + 4.0} = 9.0 \times \dfrac{4.0}{6.0} = 6.0\,\text{V}$ (2 marks for the substitution and the answer). Markers reward the correct resistor in the numerator and the value. A common error is to put the wrong resistor on top.

Eduqas style4 marksA divider must give an output of

3.0\,\text{V}

from a

12\,\text{V}

supply. The lower resistor is

1.0\,\text{k}\Omega

. Calculate the value of the upper resistor.

Show worked answer →

A Component 1 design question. The output is across the lower resistor $R_2 = 1.0\,\text{k}\Omega$ , so $V_{\text{out}} = V_{\text{in}} \times \dfrac{R_2}{R_1 + R_2}$ gives $3.0 = 12 \times \dfrac{1.0}{R_1 + 1.0}$ (1 mark). Rearrange: $\dfrac{3.0}{12} = \dfrac{1.0}{R_1 + 1.0}$ , so $R_1 + 1.0 = \dfrac{1.0 \times 12}{3.0} = 4.0$ , giving $R_1 = 3.0\,\text{k}\Omega$ (2 marks for the rearrangement and the value). The nearest E24 preferred value is $3.0\,\text{k}\Omega$ , which is available (1 mark for a sensible real value). Markers reward the rearrangement, the value and a preferred value. A common error is an algebra slip in the rearrangement.

Related dot points

Sources & how we know this

WJEC Eduqas GCSE Electronics specification (from 2017) — WJEC Eduqas (2017)