Practical Electricity is about how mains electricity is used, paid for and made safe in the home. It links what you learnt in Current of Electricity (resistance and energy conversion) to real appliances such as the kettle, oven, iron and storage water heater.
The topic is tested almost every year in Paper 2 structured questions. Typical questions ask for the operating current of an appliance, the energy used and its cost in kilowatt-hours, the correct fuse rating, and a clear explanation of why switches and fuses must be in the live wire.
What these notes cover
- •Heating effect of electricity
- •Heating elements: why nichrome?
- •Electric kettle, oven, iron and the thermostat
- •Electrical power and energy: P = VI and E = VIt
- •Deriving P = I²R and P = V² ÷ R
- •Reading an appliance rating plate
- •Paying for electricity: the kilowatt-hour
- •Electrical hazards: damaged insulation, overheating cables, damp conditions
- •Fuses and circuit breakers
- •The 3-pin plug and wire colours
- •Earthing a metal casing and double insulation
- •Common mistakes and exam technique
Key definitions
- Electrical power
- The rate at which electrical energy is converted into other forms of energy. The SI unit is the watt (W), where 1 W = 1 J/s.
- Kilowatt-hour (kWh)
- The electrical energy used by a 1 kW appliance operating for 1 hour. 1 kWh = 3.6 × 10⁶ J.
- Fuse
- A deliberately thin piece of wire connected in series with an appliance. It melts and breaks the circuit when the current exceeds its rating, which is the maximum current it can carry without melting.
- Circuit breaker
- An electromagnetic switch that automatically trips (springs open) when the current exceeds a preset value. It acts faster than a fuse and can be reset instead of replaced.
- Thermostat
- An automatic switch, commonly a bimetallic strip, that switches the current off when an appliance gets too hot and back on when it cools below the set temperature.
Key formulas
Electrical power
P = VI
P is power in watts (W), V is potential difference in volts (V) and I is current in amperes (A). It follows from P = E ÷ t, with V = E ÷ Q and I = Q ÷ t.
Electrical energy
E = Pt = VIt
E is the electrical energy converted, in joules (J). It is in joules only when t is in seconds.
Derived power forms
P = I²R = V² ÷ R
Found by substituting Ohm's law, V = IR, into P = VI. Use I²R when the current is shared (series); use V² ÷ R when the p.d. is shared (parallel, or anything connected to the 230 V mains).
Normal operating current
I = P ÷ V
Found from the rating plate. For example, 230 V, 1 800 W gives 7.8 A. This current decides the fuse rating. The operating resistance is R = V² ÷ P.
Energy in kilowatt-hours
E (kWh) = P (kW) × t (h)
Before multiplying, convert power from W to kW (÷ 1000) and time from minutes to hours (÷ 60). 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.
Cost of electricity
cost = energy used (kWh) × tariff (price per kWh)
The tariff is the price charged per kWh. Cost always uses the input energy drawn from the mains, not the useful output.
Worked example
An electric iron is rated 230 V, 1 100 W. Fuses rated 3 A, 5 A and 13 A are available. Which fuse should be fitted? Justify your choice.
- 1Find the normal operating current: I = P ÷ V = 1100 W ÷ 230 V = 4.78 A.
- 2A 3 A fuse is below the operating current, so it would melt during normal use and the iron could not work at all.
- 3A 13 A fuse is far above the operating current, so a fault current of, say, 8 A would flow without blowing the fuse. It gives no protection.
- 4A 5 A fuse is the lowest rating just above 4.78 A. It allows normal operation but melts as soon as the current rises significantly above normal.
- 5Answer: fit the 5 A fuse, because it is the smallest available rating that exceeds the normal operating current of 4.78 A.
Watch out: Never guess the fuse first. Calculate I = P ÷ V, then choose the next standard rating just above it. A rating far above the normal current cannot protect the appliance.
Common mistakes
Leaving power in watts and time in minutes
Wrong: Substituting 1500 W and 45 min straight into a kWh calculation, which gives an answer wrong by a factor of 1000 or 60.
Right: Convert first: 1500 W = 1.5 kW and 45 min = 0.75 h. One appliance used for under an hour should use a few kWh at most, never thousands.
Saying the fuse protects the user
Wrong: The fuse protects the user from electric shock.
Right: The fuse protects the wiring and appliance from overheating and fire. About 0.05 A through the body can be fatal, which is far too small to blow even a 3 A fuse. The earth wire is what protects people.
Saying the earth wire absorbs current
Wrong: The earth wire absorbs the excess current.
Right: The earth wire connects the metal casing to the ground through a very low resistance. If the live wire touches the casing, a very large current flows to earth and melts the fuse in the live wire, which disconnects the appliance from the supply.
Questions students ask
Why is nichrome used for heating elements?
Nichrome has a high resistivity, so it converts electrical energy to thermal energy rapidly and heats up quickly. Its melting point is very high (over 1 400 °C), so it does not melt when red-hot. It also does not oxidise easily, so the element lasts a long time. In an exam, pair each property with its correct consequence.
How do you calculate the cost of electricity using kWh?
Use energy (kWh) = power (kW) × time (h), then multiply by the tariff: cost = energy used × price per kWh. Convert power from watts to kilowatts and time from minutes to hours before multiplying. If the answer is in joules, use 1 kWh = 3.6 × 10⁶ J.
How do you choose the correct fuse rating for an appliance?
First calculate the normal operating current with I = P ÷ V. Then choose the next standard rating just above it (standard ratings are 1 A, 2 A, 3 A, 5 A, 10 A and 13 A). For example, a kettle drawing 10 A takes a 13 A fuse. A 10 A fuse would blow during normal use.
Why must the switch be in the live wire and not the neutral wire?
If the switch were in the neutral wire, opening it would still leave the appliance connected to the live wire at 230 V. A person touching a faulty part could then receive an electric shock even though the appliance looks 'off'. With the switch in the live wire, opening it disconnects the appliance from the 230 V supply entirely. The same rule applies to fuses and circuit breakers.
In the full PDF
- •Further worked examples: comparing heat produced in wire segments, kettle design, the thermostat, a 30-day household bill and a full-marks oven cost answer
- •The A-B-C wire-segment diagrams used throughout the heating-effect and cable-overheating sections
- •Electrical hazards in depth: damaged insulation and short circuits, overloaded sockets with multiple adaptors, long and thin cables, and the dry-skin versus wet-skin current calculation
- •The remaining misconceptions, including fuse ratings that are too low or far too high and switches placed in the neutral wire
- •Model cause → effect → consequence answers for earthing and switch placement, plus double insulation and the square-within-a-square symbol
- •A one-page revision summary with the hazards-and-protection table and the 3-pin plug wiring table