One Socket, Three Appliances: Judging an Overloaded Extension Cord
A classroom in Nagano has a single 100 V wall socket. A 5 m extension cord rated at 15 A is used to run three appliances at the same time: a fan heater marked 1200 W, an electric kettle marked 700 W, and a desk lamp marked 40 W. The two wires inside the cord have a combined resistance of 0.10 Ω. After ten minutes the teacher notices that the cord is warm and that the plug of the heater is slightly discoloured, although the heater itself works normally. The heater contains a fan driven by a small electric motor. One student proposes buying a 20 m cord of exactly the same type so that the appliances can be spread further apart; another proposes unplugging the 40 W lamp, because 'the lamp is the extra one'. The teacher wants a decision that can be defended with numbers rather than a guess.
1. Calculate the total current drawn from the socket with all three appliances running, and state whether the cord's 15 A rating is exceeded.
Read the worked answer
Total power = 1200 + 700 + 40 = 1940 W.
Current = P / V = 1940 ÷ 100 = 19.4 A.
The cord is rated at 15 A, so it is overloaded by 19.4 − 15.0 = 4.4 A, about 29% above its rating.
Reasonable assumptions to state: the supply is close to 100 V, the appliances behave as resistive loads, and the rating is a continuous rating, not a brief peak. A 29% overload is not marginal - it is the kind of overload that heats insulation over minutes and softens it over months.
2. Explain with a calculation why the cord becomes warm, and why the heater element reaches a much higher temperature even though the same current passes through both.
Read the worked answer
Power dissipated in the cord: P = I²R = 19.4² × 0.10 ≈ 38 W.
That is spread along 5 m of thin wire inside plastic insulation: about 7.5 W per metre, with very little surface area and no airflow. 38 W is small compared with 1200 W, but it is trapped, so the copper and insulation warm up. There is also a voltage drop of V = IR = 19.4 × 0.10 = 1.9 V along the cord, so the appliances receive slightly less than 100 V.
The heater element has resistance R = V²/P = 100² ÷ 1200 ≈ 8.3 Ω and dissipates the full 1200 W in a short coil of high-melting-point alloy designed to run hot and to radiate heat. Same current, very different resistance and very different design: the cord is meant to carry current, the element is meant to get hot.
Judgement: the discoloured plug is evidence that heat damage has already begun at a contact point, which has higher resistance than the wire itself.
3. Judge the two proposals quantitatively and recommend what the teacher should actually do.
Read the worked answer
Proposal 1 - a 20 m cord of the same type: resistance is proportional to length, so R becomes 0.10 × (20/5) = 0.40 Ω. Dissipation becomes 19.4² × 0.40 ≈ 150 W - four times worse, in a cord that is also more likely to be coiled and to trap heat. Reject.
Proposal 2 - unplug the 40 W lamp: power becomes 1900 W, current 19.0 A, still well above 15 A. The lamp was never the problem; removing it changes almost nothing. Reject.
Recommendation: never run the heater and the kettle on the same cord at the same time. Heater alone = 1200 W → 12.0 A; heater plus lamp = 1240 W → 12.4 A, both within rating. Kettle alone = 700 W → 7.0 A. Put the kettle on a second socket on a different circuit. Replace the discoloured plug or cord, because scorching shows the contact has already been damaged and a damaged contact has higher resistance and gets hotter still.