Mains electricity: safety, fuses, P = IV, E = IVt, a.c./d.c. – IGCSE Physics exam questions with mark scheme

1Grade 1–33 marks

A kettle with a metal case is connected to the mains supply using a three-pin plug.

(a) Name the wire in the plug that is connected to the metal case of the kettle for safety. [1]

(b) The plug contains a fuse connected in the live wire.

Explain how the fuse protects the user if a fault causes too large a current. [2]

(a) earth (wire) allow green and yellow wire(1)
(b) the fuse (wire) melts / blows (when current is too large)(1)
(b) which breaks the circuit / disconnects the live wire / stops the current allow so the cable/appliance cannot overheat / prevents a fire; ignore 'prevents an electric shock' on its own(1)
Answer: (a) earth (b) the fuse melts and breaks the circuit, disconnecting the live supply
Examiner tip: Say what the fuse does AND what that achieves: it melts (blows) when the current is too large, which breaks the circuit in the live wire.

2Grade 4–54 marks

An electric kettle has a power rating of 2.3 kW. It is connected to the 230 V mains supply.

(a) Calculate the current in the kettle. [3]

(b) Fuses are available with ratings of 3 A, 5 A and 13 A.

State which fuse should be used for the kettle. [1]

(a) power = current × voltage or $P = IV$ (in any form)(1)
(a) $I = \dfrac{2300}{230}$ (kW converted to W)(1)
(a) 10 A(1)
(b) 13 A allow ecf: the smallest rating above their (a)(1)
Answer: (a) 10 A (b) 13 A
Examiner tip: Convert 2.3 kW to 2300 W before dividing; 2.3/230 = 0.01 A is a classic slip.

3Grade 6–76 marks

A 230 V electric heater has a metal case. The current in the heater is 4.0 A.

(a) The heater is switched on for 2.5 hours.

Calculate the energy transferred by the heater. Give your answer in joules. [3]

(b) A fault causes the live wire to touch the metal case.

Explain how the earth wire and the fuse together protect a person who touches the case. [3]

(a) $t = 2.5 \times 3600 = 9000$ s(1)
(a) substitution: $E = 4.0 \times 230 \times 9000$ allow P = 4.0 × 230 = 920 W then E = 920 × 9000; ignore the equation E = I × V × t quoted on its own (it is given); if time is left in hours (E = 4.0 × 230 × 2.5 = 2300) award this mark only(1)
(a) $8.3 \times 10^{6}$ J allow 8 280 000 J / 8.28 × 10⁶ J(1)
(b) a large current flows (from live) through the earth wire (to ground) allow earth wire has low resistance(1)
(b) the fuse melts / blows (breaking the circuit)(1)
(b) so the case is no longer connected to the live supply / is not at a high voltage, so no current passes through the person allow so the person does not get an electric shock(1)
Answer: (a) $8.3 \times 10^{6}$ J (b) large current in earth wire → fuse melts → live disconnected, case safe
Examiner tip: The earth wire alone does not make the case safe — you must say the large current melts the fuse.

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