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Electricity and magnetism · Subtopic 5 of 8

Electrical Power, Energy and Safety

Electrical power is $P=IV$ , and the energy transferred is $E=IVt$ . Electricity companies charge per kilowatt-hour: the energy used by a $1$ kW appliance in $1$ hour. Fuses and circuit breakers cut off the current if it gets too big; earthing a metal case makes a fault blow the fuse instead of electrocuting you.

Cambridge 5054 Paper 1Cambridge 5054 Paper 2Cambridge 0625 Papers 2, 4Cambridge 0625 Papers 1, 3Edexcel 4PH1 Paper 1PEdexcel 4PH1 Paper 2P≈ 30 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use $P=IV$ and $E=IVt$ .
Calculate energy in kWh and its cost.
Choose a suitable fuse rating.
Explain fuses, circuit breakers, earthing and double insulation, and identify electrical hazards.

Pause and think

A washing machine has a metal case. Inside, a loose live wire touches the case. Why might nothing happen until someone touches the machine, and what stops that being deadly?

Section 2 of 14

Core idea

Start hereThe electricity bill

Your bill doesn't count joules; it counts kilowatt-hours. A 2 kW kettle running for an hour uses 2 kWh. Knowing power and time lets you predict cost, and knowing current lets you choose a fuse that keeps you safe.

The big idea

$P=VI=I^2R$ ; $E=Pt$ ; 1 kWh $=3.6\times10^6$ J. Fuse rating just above normal current; earthing protects against shocks.

1. Power and cost

P=VI,\qquad E=Pt,\qquad1\text{ kWh}=1000\times3600\text{ J}

$P$: power (W).
$E$: energy (J or kWh).

Worked exampleKettle

2 kW, 15 min a day, 20 cents per kWh.

Goal: Daily cost; energy in joules.
Strategy: kW × hours.

$2\times0.25=0.5\text{ kWh}$
15 min = 0.25 h.
$0.5\times20=10\text{ cents}$
Cost.
$2000\times900=1.8\times10^6\text{ J}$
In SI units.

Answer

10

cents a day;

1.8

MJ.

✓ Check

0.5\times3.6\times10^6=1.8\times10^6

✓.

2. Safety

Inside a three-pin plug

The fuse is in the live wire: if too much current flows, it melts and cuts off the supply. The earth wire carries current safely away if a metal case becomes live.

Worked exampleChoosing a fuse

2300 W appliance on 230 V. Fuses: 3, 5, 13 A.

Goal: Fuse.
Strategy: Normal current, then next size up.

$I=\frac{2300}{230}=10\text{ A}$
$I=P/V$ .
$13\text{ A}$
Smallest rating above 10 A.

Answer

13

✓ CheckA 5 A fuse would blow in normal use ✓.

Live (brown), neutral (blue), earth (green/yellow).
Fuse and switch in the live wire.
Earth carries fault current away, blowing the fuse.
Double insulation: no earth needed.

Pause and think

Why not use a 30 A fuse everywhere?

Summary

$P=VI$ ; $E=Pt$ .
Fuse just above normal current.

Kilowatt-hour: Energy used by 1 kW for 1 hour.

HistoryEdison's fuse. Thomas Edison patented an early fuse in

1890

: a thin wire that melted to protect his new electrical lighting systems from overheating. The basic idea is still used in plugs and cars today.

Fun factA lightning bolt for a day. A lightning bolt transfers around

1

billion joules, about

280

kWh. That sounds huge, but at typical prices it's worth only about

\$50

of electricity, and it's over in a fraction of a second.

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