A component has 12\ V across it and draws 0.50\ A. Find the power. [2 marks]

WJEC Eduqas Eduqas 2019-style practice: A resistor carries a current of 50\ mA and has a resistance of 120\. Calculate the power it dissipates and state a suitable standard power rating for it from 0.25\ W, 0.5\ W and 1\ W.

Use P = I^2 R because the current and resistance are known. Convert: I = 50\ mA = 0.050\ A. Substitute: P = (0.050)^2 × 120 = 0.0025 × 120 = 0.30\ W. A resistor must be rated above the power it dissipates, so the 0.25\ W rating is too low; choose the 0.5\ W rating. Markers reward the correct equation and substitution, the answer 0.30\ W, and the 0.5\ W rating with the reason that the rating must exceed the dissipated power.

WJEC Eduqas Eduqas 2022-style practice: A 5.0\ V supply drives a heating element of resistance 10\ for 2.0 minutes. Calculate the energy transferred.

First find the power. With voltage and resistance known, use P = V^2R = (5.0)^210 = 2510 = 2.5\ W. Convert the time to seconds: 2.0\ minutes = 120\ s. Energy transferred: E = Pt = 2.5 × 120 = 300\ J. Markers reward finding the power 2.5\ W, converting the time to seconds, and the energy 300\ J. Forgetting to convert minutes to seconds is the usual error.

EnglandElectronicsSyllabus dot point

How do you calculate the electrical power dissipated by a component and the energy it transfers?

Electrical power and energy: power as the rate of energy transfer, the equations relating power to voltage, current and resistance, and resistor power ratings.

An Eduqas GCSE Electronics answer on electrical power and energy: power as the rate of energy transfer, the three power equations, calculating energy transferred over time, and why resistor power ratings matter.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

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What this dot point is asking

Eduqas wants you to define electrical power as the rate of energy transfer, use the three power equations to calculate the power dissipated by a component, work out the energy transferred over a time, and explain why a resistor's power rating must exceed the power it dissipates. Power calculations decide component ratings, heat-sinking and battery life.

The answer

Power as a rate of energy transfer

The three power equations

Energy transferred over time

Power ratings

Choosing a resistor's power rating

A $100\ \Omega$ resistor will have $6.0\ \text{V}$ across it in a circuit. Find the power and choose a rating from $0.25\ \text{W}$ , $0.5\ \text{W}$ , $1\ \text{W}$ .

step 1: Pick the right power equation

You know the voltage and resistance, so use $P = \dfrac{V^2}{R}$ .

step 2: Calculate the power

$P = \dfrac{(6.0)^2}{100} = \dfrac{36}{100} = 0.36\ \text{W}$ .

step 3: Choose a rating above the dissipated power

$0.36\ \text{W}$ is more than the $0.25\ \text{W}$ rating, so a quarter-watt resistor would overheat. The next standard rating up is $0.5\ \text{W}$ , which is safely above $0.36\ \text{W}$ , so choose 0.5 W.

Examples in context

Power calculations appear throughout both papers. The current-limiting resistor for an LED dissipates power you must check against its rating. A transistor switching a motor dissipates power and may need a heat sink. The mains-driven heating element in a controller transfers energy you calculate with $E = Pt$ . Estimating how long a battery lasts in a portable design starts from the power drawn by each subsystem. Fluent power calculation protects your circuits and earns method marks.

Try this

Q1. State the unit of power and what one of that unit equals. [1 mark]

Cue. The watt (W); one watt is one joule per second.

Q2. A component has $12\ \text{V}$ across it and draws $0.50\ \text{A}$ . Find the power. [2 marks]

Cue. $P = VI = 12 \times 0.50 = 6.0\ \text{W}$ .

Q3. A $2.0\ \text{W}$ lamp runs for $30$ seconds. Find the energy transferred. [2 marks]

Cue. $E = Pt = 2.0 \times 30 = 60\ \text{J}$ .

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 resistor carries a current of

50\ \text{mA}

and has a resistance of

120\ \Omega

. Calculate the power it dissipates and state a suitable standard power rating for it from

0.25\ \text{W}

0.5\ \text{W}

and

1\ \text{W}

Show worked answer →

Use $P = I^2 R$ because the current and resistance are known.

Convert: $I = 50\ \text{mA} = 0.050\ \text{A}$ .

Substitute: $P = (0.050)^2 \times 120 = 0.0025 \times 120 = 0.30\ \text{W}$ .

A resistor must be rated above the power it dissipates, so the $0.25\ \text{W}$ rating is too low; choose the $0.5\ \text{W}$ rating.

Markers reward the correct equation and substitution, the answer $0.30\ \text{W}$ , and the $0.5\ \text{W}$ rating with the reason that the rating must exceed the dissipated power.

Eduqas 20223 marksA

5.0\ \text{V}

supply drives a heating element of resistance

10\ \Omega

for

2.0

minutes. Calculate the energy transferred.

Show worked answer →

First find the power. With voltage and resistance known, use $P = \dfrac{V^2}{R} = \dfrac{(5.0)^2}{10} = \dfrac{25}{10} = 2.5\ \text{W}$ .

Convert the time to seconds: $2.0\ \text{minutes} = 120\ \text{s}$ .

Energy transferred: $E = Pt = 2.5 \times 120 = 300\ \text{J}$ .

Markers reward finding the power $2.5\ \text{W}$ , converting the time to seconds, and the energy $300\ \text{J}$ . Forgetting to convert minutes to seconds is the usual error.

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Sources & how we know this

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