Choosing an Extension Lead for a 13 A Heater
A 15 m extension lead is used continuously at 13 A to supply a 3 kW fan heater from a 230 V socket. Two cables are available: 1.0 mm² copper, resistance 18.1 mΩ per metre of conductor, rated at 10 A; and 2.5 mm² copper, resistance 7.4 mΩ per metre, rated at 20 A. Both live and neutral conductors carry the current. The maximum acceptable voltage drop along the lead is 5% of 230 V. The lead is usually left tightly coiled on its reel while the heater runs. The socket circuit is protected by a 13 A fuse, and the heater is switched on for hours at a time.
1. Calculate the voltage drop and the power dissipated in each cable at 13 A. State which cable should be used and justify the choice.
Read the worked answer
The current passes through live and neutral, so the conductor length is 2 × 15 = 30 m.
1.0 mm²: R = 18.1 × 10⁻³ × 30 = 0.543 Ω.
V = IR = 13 × 0.543 = 7.1 V (3.1% of 230 V).
P = I²R = 13² × 0.543 = 92 W.
2.5 mm²: R = 7.4 × 10⁻³ × 30 = 0.222 Ω.
V = 13 × 0.222 = 2.9 V (1.3%).
P = 13² × 0.222 = 37.5 W.
On voltage drop alone both pass the 5% (11.5 V) limit. But the 1.0 mm² cable is rated at 10 A and the load is 13 A, so it must be rejected on rating grounds. Use the 2.5 mm² cable. Note that 3000 W / 230 V ≈ 13 A, so the current is genuine and continuous, not a short surge.
2. Explain why the coiled reel matters, and how it changes the decision even for the thicker cable.
Read the worked answer
The 1.0 mm² cable dissipates 92 W over 30 m, about 3 W per metre; the 2.5 mm² dissipates about 1.25 W per metre. In free air that is modest, but a tight reel leaves the inner turns surrounded by warm cable, so heat leaves mainly by conduction along the cable rather than by convection. The centre of the reel can therefore reach a temperature well above the insulation's rating, softening the plastic and eventually allowing live and neutral to contact.
Heating is set by I²R per unit length and by how easily heat escapes, not by the percentage voltage lost. So a cable can satisfy the 5% rule and still overheat. Practical conclusion: uncoil the lead fully whenever it carries a heavy, long-duration load; if it must stay reeled, derate it well below 13 A.
3. A colleague argues: 'The drop is only 3%, well under 5%, so the thin cable is fine.' Assess this reasoning.
Read the worked answer
The colleague has checked the less important criterion and conflated two different questions.
The 5% figure is a performance guide: it limits how much the supply voltage falls at the appliance, so the heater runs at close to its rated power. It is not a safety limit for the cable. Cable safety is governed by the current rating, which reflects how much heat the insulation can survive, and by how the heat is removed. A cable can carry a small voltage drop and still overheat if it is thin, coiled, or buried in insulation.
Here the 1.0 mm² cable is rated 10 A and the load is 13 A, so it fails the safety criterion outright. The correct procedure is to satisfy both tests: current rating first, then voltage drop. The colleague's reasoning would also fail to notice the extra hazard from coiling.