A 6 V supply pushes 2 A through a resistor. Calculate its resistance. [2 marks]

AQA AQA 2018-style practice: A 12 V supply drives a current of 0.5 A through a heater. Calculate the resistance of the heater and the power it dissipates.

A good answer applies Ohm's law and the power formula with units. From Ohm's law, R = VI = 120.5 = 24. Power is P = V × I = 12 × 0.5 = 6 W (you could also use P = I^2 R = 0.5^2 × 24 = 6 W). So the heater has a resistance of 24 and dissipates 6 W. Markers reward the correct rearrangement of Ohm's law, the correct power, and the right units ( and W).

EnglandEngineeringSyllabus dot point

How do we calculate voltage, current, resistance and power in a circuit?

Ohm's law, electrical power, series and parallel resistance, and using the correct units in electrical calculations.

A focused answer to AQA GCSE Engineering on Ohm's law, electrical power, resistors in series and parallel, and using the correct units for voltage, current, resistance and power.

Generated by Claude Opus 4.89 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
Ohm's law
Electrical power
Series and parallel resistance
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What this dot point is asking

AQA wants you to use Ohm's law and the power formula, find the total resistance of resistors in series and in parallel, rearrange to find any unknown, and always use the correct electrical units. Multi-step circuit questions reward a clear order of working: find total resistance, then current, then power.

Ohm's law

The three forms are the same relationship rearranged, so the skill is choosing the right form for the unknown you want. A formula triangle with $V$ on top and $I$ and $R$ below helps: cover the unknown to read off the calculation.

Electrical power

The three power forms let you pick whichever two quantities you already know. For example, if you know the current and resistance but not the voltage, $P = I^2 R$ gives the power in one step without finding $V$ first. Power matters in design because it sets how much heat a component must get rid of: a resistor rated at $0.25 \text{ W}$ will overheat and burn out if asked to dissipate $1 \text{ W}$ , so an engineer calculates the power first and then chooses a component with a high enough power rating. Energy use over time follows from power: a $6 \text{ W}$ device left on for $1 \text{ hour}$ ( $3600 \text{ s}$ ) uses $E = P \times t = 6 \times 3600 = 21600 \text{ J}$ .

Series and parallel resistance

For example, two $10 \text{ }\Omega$ resistors give $20 \text{ }\Omega$ in series but $5 \text{ }\Omega$ in parallel. The parallel result being smaller than either resistor is a useful check: if your parallel answer is larger than the smallest resistor, you have made an error.

Worked example: a series circuit, finding current and power

A $9 \text{ V}$ battery is connected to two resistors in series, $200 \text{ }\Omega$ and $400 \text{ }\Omega$ . Find the total resistance, the current, and the power dissipated by the $400 \text{ }\Omega$ resistor.

step Find the total resistance

Series resistances add: $R_{total} = 200 + 400 = 600 \text{ }\Omega$ .

step Find the current

The same current flows through a series circuit. Using Ohm's law, $I = \dfrac{V}{R_{total}} = \dfrac{9}{600} = 0.015 \text{ A}$ (which is $15 \text{ mA}$ ).

step Find the voltage across the 400 ohm resistor

The current through it is $0.015 \text{ A}$ , so its voltage is $V = I \times R = 0.015 \times 400 = 6 \text{ V}$ .

step Find the power in the 400 ohm resistor

$P = V \times I = 6 \times 0.015 = 0.09 \text{ W}$ (which is $90 \text{ mW}$ ). You could also use $P = I^2 R = 0.015^2 \times 400 = 0.09 \text{ W}$ , which confirms the answer. Working in order (total resistance, then current, then power) keeps a multi-step circuit question under control.

Try this

Q1. A $6 \text{ V}$ supply pushes $2 \text{ A}$ through a resistor. Calculate its resistance. [2 marks]

Cue. $R = V/I = 6/2 = 3 \text{ }\Omega$ .

Q2. Two $4 \text{ }\Omega$ resistors are connected in series. State the total resistance. [1 mark]

Cue. $4 + 4 = 8 \text{ }\Omega$ .

Exam-style practice questions

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

AQA 20184 marksA

12 \text{ V}

supply drives a current of

0.5 \text{ A}

through a heater. Calculate the resistance of the heater and the power it dissipates.

Show worked answer →

A good answer applies Ohm's law and the power formula with units.

From Ohm's law, $R = \dfrac{V}{I} = \dfrac{12}{0.5} = 24 \text{ }\Omega$ .

Power is $P = V \times I = 12 \times 0.5 = 6 \text{ W}$ (you could also use $P = I^2 R = 0.5^2 \times 24 = 6 \text{ W}$ ).

So the heater has a resistance of $24 \text{ }\Omega$ and dissipates $6 \text{ W}$ . Markers reward the correct rearrangement of Ohm's law, the correct power, and the right units ( $\Omega$ and W).

AQA 20226 marksTwo resistors,

30 \text{ }\Omega

and

60 \text{ }\Omega

, are connected in parallel across a

9 \text{ V}

supply. Calculate the total resistance, the total current drawn from the supply, and the total power dissipated.

Show worked answer →

A good answer works in order: total resistance, then current, then power, with units.

Parallel resistance: $\dfrac{1}{R_{total}} = \dfrac{1}{30} + \dfrac{1}{60} = \dfrac{2}{60} + \dfrac{1}{60} = \dfrac{3}{60} = \dfrac{1}{20}$ , so $R_{total} = 20 \text{ }\Omega$ (smaller than the smallest resistor, as expected).

Total current: $I = \dfrac{V}{R_{total}} = \dfrac{9}{20} = 0.45 \text{ A}$ .

Total power: $P = V \times I = 9 \times 0.45 = 4.05 \text{ W}$ .

Markers reward the parallel-resistance method giving $20 \text{ }\Omega$ , the current of $0.45 \text{ A}$ , and the power of about $4.05 \text{ W}$ , each with units.

Related dot points

Sources & how we know this

AQA GCSE Engineering (8852) specification — AQA (2017)