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5.3 Electrical Power and Potential Difference

GCSE Physics · AQA · Chapter 5: Electricity in the Home

A free GCSE Physics lesson for AQA (8463) from OBA School: power as energy per second, P = V I, P = I² R and choosing the right fuse. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

  • Describe power as the energy transferred per second and use P = E ÷ t
  • Calculate electrical power using P = V I, and rearrange it to find V or I
  • Calculate the power in a resistor using P = I² R, and explain why doubling the current makes the power four times bigger
  • Use the power rating and pd of an appliance to choose a suitable fuse

Key words

  • Power: The energy transferred per second. Measured in watts (W).
  • Watt (W): One joule per second. 1 kilowatt (kW) = 1000 W.
  • Power rating: The power an appliance takes from the supply when it works normally, printed on its label.
  • Fuse rating: The current (in amps) above which the fuse melts. Common plug fuses are 3 A, 5 A and 13 A.
  • Resistance heating: The heating of a wire or resistor when a current passes through it.

Power is energy per second

When an appliance is switched on, the current transfers energy to it from the supply. The power of the appliance tells you how many joules it transfers every second.

power, P (W) = energy transferred, E (J) ÷ time, t (s)

Rearranged: E = P t and t = E ÷ P. 1 W = 1 J/s 1 kW = 1000 W

Worked example: energy used by a fan

A 60 W desk fan runs for 15 minutes. Calculate the energy transferred.
t = 15 × 60 = 900 s
E = P t = 60 W × 900 s = 54 000 J

Answer: 54 000 J (54 kJ)

Calculating power from current and pd

The current is the charge flowing each second, and the pd is the energy carried by each coulomb of charge. Multiplying them gives the energy transferred each second, which is the power:

power, P (W) = potential difference, V (V) × current, I (A)

Rearranged: I = P ÷ V V = P ÷ I

To measure the power of a component, put an ammeter in series with it to measure I, and a voltmeter in parallel with it to measure V. Then multiply.

Worked example: power of a vacuum cleaner

A vacuum cleaner takes a current of 3.0 A from the 230 V mains.
P = V I = 230 V × 3.0 A = 690 W

Answer: 690 W

Worked example: current in a kettle

A kettle is rated at 2.3 kW, 230 V. Calculate the current.
P = 2.3 kW = 2300 W
I = P ÷ V = 2300 W ÷ 230 V = 10 A

Answer: 10 A

Choosing the right fuse

A fuse should melt if the current rises above the normal working current, but must not melt in normal use. So choose the fuse with the smallest rating that is bigger than the normal current.

Worked example: fuse for a toaster

A toaster is rated 920 W, 230 V. Which fuse should be fitted: 3 A, 5 A or 13 A?
I = P ÷ V = 920 ÷ 230 = 4.0 A
A 3 A fuse would melt in normal use. A 13 A fuse would let a dangerously large current flow before melting.

Answer: use the 5 A fuse

Exam tip: Always calculate the current first, then pick the next fuse rating above it. Explain both sides: why the lower fuse would blow, and why the higher fuse would not protect the cable.

Power and resistance: P = I² R

When a current passes through a resistor, electrical work is done and the resistor heats up. The energy is then dissipated to the surroundings. Since V = I R, we can replace V in P = V I:

P = V I = (I R) × I, so power, P (W) = current², I² (A²) × resistance, R (Ω)

Rearranged: R = P ÷ I² I = √(P ÷ R)

Power depends on the square of the current. If the current doubles, the power becomes 2² = 4 times bigger; if the current triples, the power is 9 times bigger.

Worked example: heating element

A current of 2.5 A passes through a 40 Ω heating element.
P = I² R = 2.5² × 40 = 6.25 × 40 = 250 W

Answer: 250 W

Worked example: (Higher demand) finding the current

A 40 Ω resistor dissipates 360 W. Calculate the current.
I² = P ÷ R = 360 ÷ 40 = 9.0
I = √9.0 = 3.0 A

Answer: 3.0 A

The same idea explains why cables waste energy. A cable of resistance 0.15 Ω carrying 20 A wastes P = 20² × 0.15 = 60 W as heat. This is why high-current appliances need thick, low-resistance cables, and why the National Grid uses small currents.

Common mistakes

  • Forgetting to change kilowatts to watts (× 1000) or minutes to seconds (× 60) before calculating.
  • Choosing a fuse rating just below the normal current. The fuse would melt every time the appliance is used.
  • Squaring the resistance instead of the current in P = I² R. Only the current is squared.
  • Thinking that doubling the current doubles the power in a resistor. With P = I² R, it is four times bigger.
  • Using P = I² R with the pd of the supply instead of the resistance.

Revision questions

Try these, then check your answers below each one.

Question 1. A vacuum cleaner takes a current of 4.0 A from the 230 V mains. Calculate its power.

Answer: 920 W

Working: P = V I = 230 V × 4.0 A = 920 W.

Question 2. A kettle is rated 2.76 kW, 230 V. Calculate the current in the kettle.

Answer: 12 A

Working: I = P ÷ V = 2760 W ÷ 230 V = 12 A.

Question 3. A food mixer is rated 460 W, 230 V. Which fuse should be fitted in its plug? (Choose from: 3 A, 5 A, 13 A.)

Answer: 3 A

Working: I = 460 ÷ 230 = 2.0 A. The 3 A fuse is the smallest rating above 2.0 A, so it will not melt in normal use but will melt in a fault.

Related lessons

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