AQA AQA 2019-style practice: An electric motor is connected to a 230\,V supply and draws a current of 3.0\,A. Calculate the power of the motor, and calculate the energy it transfers when it runs for 4.0 minutes.

Use P = VI = 230 × 3.0 = 690\,W (2 marks: one for the equation, one for the value with the unit watt). Then use E = Pt. Convert the time to seconds: 4.0 minutes is 240\,s (1 mark for the conversion, a common slip). So E = 690 × 240 = 165,600\,J, which is about 1.66 × 10^5\,J or 166\,kJ (2 marks). Markers reward selecting P = VI, converting the time to seconds, and quoting energy in joules. This is a routine multi-step calculation on AQA Physics Paper 1.

EnglandPhysicsSyllabus dot point

How do we calculate electrical power and energy, and why does the grid use high voltages?

Electrical power and the national grid: the power and energy equations, charge and energy transfer, and why step-up and step-down transformers make transmission efficient.

A focused answer to AQA GCSE Physics 4.2.3 and 4.2.5, covering the electrical power equations, the energy transferred by an appliance, and why the national grid uses step-up and step-down transformers to transmit power efficiently.

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
Electrical power
Energy transferred
The national grid
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What this dot point is asking

AQA wants you to use the electrical power and energy equations, calculate the charge and energy transferred by a component, and explain why the national grid uses transformers to transmit electricity efficiently. This covers topics 4.2.3 and 4.2.5 of the AQA GCSE Physics (8463) specification.

Electrical power

Power is the rate at which energy is transferred. In an electrical component, the energy carried by the charge is transferred to other stores (thermal, kinetic, light and so on) as the charge passes through. The two power equations come from combining the definitions of current and potential difference, and you choose between them based on the quantities given in the question.

The second form, $P = I^2R$ , follows from substituting $V = IR$ into $P = VI$ . It is the key to understanding transmission losses: because the wasted power depends on the current squared, halving the current cuts the power loss to a quarter.

Energy transferred

The form $E = QV$ shows directly what potential difference means: it is the energy transferred per coulomb of charge that passes. So a $230\,V$ supply gives $230\,J$ of energy to each coulomb of charge flowing through an appliance. The two energy equations are linked because power is energy per second and charge is current times time, so substituting gives consistent results either way. In all of these calculations the time must be in seconds, so a duration given in minutes or hours has to be converted first, which is one of the most common places marks are lost.

Worked example: energy used by a heater

A $2000\,W$ heater runs for $5$ minutes. Calculate the energy it transfers.

Step 1: Convert the time to seconds

The energy equation $E = Pt$ requires time in seconds. Minutes must be converted before substituting, as this is one of the most common places marks are lost:

t = 5 \times 60 = 300\,\text{s}

Step 2: Identify the equation and substitute

Power is the rate of energy transfer, so the total energy transferred over a time $t$ is:

E = Pt = 2000 \times 300

Step 3: Calculate and state the answer with units

E = 600\,000\,\text{J} = 600\,\text{kJ}

The watt is joules per second, so multiplying watts by seconds always gives joules.

Final answer: $E = 600\,\text{kJ}$ .

The national grid

The national grid is the system of cables and transformers that carries electricity from power stations to consumers.

This combination of high transmission voltage and low current makes the grid an efficient way to move large amounts of power over long distances. Transformers only work with alternating current, which is one of the reasons the mains supply is a.c. rather than d.c. A transformer changes the potential difference but, in an ideal transformer, conserves power: the power going in equals the power coming out, so raising the voltage by a factor of (say) ten lowers the current by the same factor. The grid therefore moves the same total power but with a far smaller current in the long-distance cables, where the resistance and hence the $I^2R$ heating would otherwise be large. Pylons carry the cables high above the ground partly because $400\,kV$ would be lethal at ground level, underlining why the step-down transformer to about $230\,V$ is essential before the supply reaches homes.

Try this

Q1. Write down the two equations for electrical power. [2 marks]

Cue. $P = VI$ and $P = I^2R$ .

Q2. Explain why the national grid transmits electricity at high voltage and low current. [2 marks]

Cue. Power lost in the cables is $P = I^2R$ , so a low current reduces the energy wasted as heat.

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 20195 marksAn electric motor is connected to a

230\,\text{V}

supply and draws a current of

3.0\,\text{A}

. Calculate the power of the motor, and calculate the energy it transfers when it runs for

4.0

minutes.

Show worked answer →

Use $P = VI = 230 \times 3.0 = 690\,\text{W}$ (2 marks: one for the equation, one for the value with the unit watt). Then use $E = Pt$ . Convert the time to seconds: $4.0$ minutes is $240\,\text{s}$ (1 mark for the conversion, a common slip). So $E = 690 \times 240 = 165{,}600\,\text{J}$ , which is about $1.66 \times 10^{5}\,\text{J}$ or $166\,\text{kJ}$ (2 marks). Markers reward selecting $P = VI$ , converting the time to seconds, and quoting energy in joules. This is a routine multi-step calculation on AQA Physics Paper 1.

AQA 20224 marksExplain why the national grid transmits electrical power at very high potential difference and low current, referring to the equation for power loss in the transmission cables.

Show worked answer →

The power wasted in the transmission cables is given by $P = I^2 R$ , where $R$ is the resistance of the cables (1 mark). Because the wasted power depends on the square of the current, reducing the current greatly reduces the energy lost as heat in the cables (1 mark). A step-up transformer raises the potential difference to around $400\,\text{kV}$ , and because the power transmitted is fixed at $P = VI$ , raising the voltage lowers the current for the same power (1 mark). The much lower current means far less power is wasted, making transmission efficient; a step-down transformer then lowers the voltage to a safe level before it reaches homes (1 mark). Markers reward citing $P = I^2 R$ and the link between high voltage, low current, and reduced loss.

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Sources & how we know this

AQA GCSE Physics (8463) specification — AQA (2016)