The Warm Extension Cord
A homeowner plugs a 120-volt, 1200-watt space heater into a wall receptacle through a 100-foot 16-gauge extension cord. The cord's two conductors together have a resistance of 0.8 ohm, and the heater element has a constant resistance of 12 ohms. The receptacle supplies a steady 120 volts. The cord is rated 10 amps but feels distinctly warm after an hour of use. A 25-foot cord of the same gauge, also on hand, has a total conductor resistance of 0.2 ohm. The homeowner wants to know whether the heater is defective or the cord is the problem.
1. Using the 100-foot cord, find the actual current, the voltage across the heater, and the power the heater delivers.
Read the worked answer
The cord resistance is in series with the heater, so total resistance is 12 + 0.8 = 12.8 Ω.
- Current: I = V/R = 120 / 12.8 = 9.375 A
- Voltage at the heater: V = IR = 9.375 × 12 = 112.5 V
- Heater power: P = I²R = (9.375)² × 12 ≈ 1,055 W
So the heater runs about 145 W below its 1,200 W rating. Note also that 9.375 A is under the cord's 10 A rating - the cord is warm even though it is not overloaded. That is the key clue: the rating is not the whole story.
2. How much power is being dissipated in the cord itself, and why does that matter?
Read the worked answer
P_cord = I²R_cord = (9.375)² × 0.8 ≈ 70 W
That is roughly 7% of the heater's output, released as heat along 100 feet of cord rather than in the room where it is wanted. It matters for three reasons:
- The cord is doing the heating. Any cord coiled, bundled, or buried under a rug cannot shed that heat, so its surface temperature climbs well above what the amp rating alone would suggest.
- The lost 70 W shows up as a voltage drop, which is why the heater underperforms - the appliance is fine; the supply path is not.
- Losses scale with the square of current, so a slightly larger load on the same cord would push the dissipation up steeply, not gradually.
3. Repeat the calculation for the 25-foot cord and state the practical conclusion.
Read the worked answer
Total resistance = 12 + 0.2 = 12.2 Ω.
- Current: 120 / 12.2 = 9.836 A
- Heater voltage: 9.836 × 12 = 118.0 V
- Heater power: ≈ 1,161 W
- Cord loss: (9.836)² × 0.2 ≈ 19 W
Conclusion: swapping to the short cord raises the heater's output by about 106 W and cuts the cord's own dissipation from about 70 W to about 19 W - most of the heat problem disappears. The heater was never defective; the long, thin cord was the bottleneck. Practical judgement: use the shortest cord that reaches, keep it uncoiled, and if a long run is unavoidable, go to a heavier gauge rather than trusting the amp rating on the jacket.