One Socket, Two Appliances and a Melting Fuse
A workshop runs a 2400 W heater and a 1200 W kettle from a single 240 V, 13 A wall socket through a 4-way extension lead. The lead has 8 m of cable with a total resistance of 0.16 ohm and a 13 A fuse in its plug, and it is left coiled on a reel. The wall socket itself is protected by a 32 A miniature circuit breaker on the ring main. An apprentice plugs in both appliances and switches them on together; after a few minutes the lead feels warm and the fuse in the lead's plug melts. He replaces it with another 13 A fuse, and later asks why the 32 A breaker did not trip first, and whether it would be safe to use a piece of 5 A fuse wire instead. The heater has a metal case connected to the earth pin of its plug.
1. Calculate the current drawn by each appliance and the total, and explain the fuse failure and why the 32 A breaker did not trip.
Read the worked answer
Heater: I = P/V = 2400 / 240 = 10 A Kettle: I = 1200 / 240 = 5 A Total through the lead = 15 A
The 13 A fuse carries the whole 15 A, so it is overloaded by about 15%. A fuse is a thermal device: it does not melt instantly at 13.1 A but after a time that depends on the overload, which is exactly what 'after a few minutes' describes.
The 32 A breaker protects the fixed ring-main cable, which is designed to carry 32 A. It sees 15 A, well below its rating, so it has no reason to trip. Protection is graded: the 13 A fuse protects the flexible lead, the 32 A breaker protects the wiring in the wall.
Correct action: split the appliances between two sockets, or use a lead whose cable and fuse are rated for 15 A. Replacing the fuse with a larger one would defeat the protection of the lead.
2. Explain the purpose of the earth wire, and say whether replacing the fuse with 5 A fuse wire would be safe.
Read the worked answer
The earth wire connects the metal case to earth. If a live wire works loose and touches the case, a large current flows to earth, the fuse melts quickly, and the case is held at about 0 V. Without the earth, the case becomes live, and a person touching it while earthed provides a path through the body - possibly fatal.
A 5 A fuse would be unuseful, and in a modified form unsafe:
- At 15 A it would melt immediately, so the equipment would not run at all.
- The temptation is then to fit a thicker fuse wire, foil or a nail. With no effective fuse, the 0.16 ohm lead carries 15 A continuously; the 32 A breaker will not intervene because 15 A is a normal current for it. The lead overheats, and because it is coiled the heat cannot escape, so the insulation can soften, short and ignite.
The correct response is to reduce the load, not to change the fuse.
3. Calculate the power wasted in the lead and the voltage reaching the appliances, and comment on the significance.
Read the worked answer
Power wasted in the cable: P = I^2 R = 15^2 x 0.16 = 36 W Voltage drop: V = IR = 15 x 0.16 = 2.4 V, so the appliances receive about 240 - 2.4 = 237.6 V
Comment:
- The heater then dissipates about 2400 x (237.6/240)^2 = 2352 W, slightly less than its rating, so it heats a little more slowly.
- The 36 W is only about 1% of the 3.6 kW total, so losses are small; the 1% voltage drop is acceptable.
- The 36 W is what makes the lead feel warm. The real danger is not the size of the loss but that 15 A exceeds the rating of the 13 A fuse and cable, and a coiled lead cannot get rid of even 36 W easily.
- Plug and fuse contacts add further resistance and can become the hottest points.