Double the Current and the Cable Gets Four Times Hotter
Published 9/29/2026 · 3 min read · Everyday calculators
Daniel Okonkwo — Front-end developer and tech writer at OneKitly
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P = I²R, and the square is the whole story. A ten-metre extension lead in 1.5 mm² copper carries the current out and back, so twenty metres of conductor at 0.0175 ohm-millimetres-squared per metre gives 0.233 Ω. Draw 10 A through it and the lead itself turns 23.3 watts into heat along its length. Raise the load to 16 A — still within what a domestic socket offers in much of Europe — and the heat does not go up by 60 % but by 156 %, to 59.7 watts, because the current is squared while the resistance stays put. Sixty watts spread over ten metres of cable lying loose on a floor disperses; the same sixty watts inside a cable still wound on its drum has nowhere to go, which is why those drums carry two ratings and the coiled one is far lower.
Power lost as heat is current squared times resistance. Ten metres of 1.5 mm² extension lead has 0.233 Ω there and back: 23.3 W at 10 A, and 59.7 W at 16 A.
The tool's own default is a bar heater
Leave the fields at 2 A and 100 Ω and the answer is 400 watts, which is not a component but an appliance. The combination is a useful reminder of how the square behaves: in a signal circuit the resistances are large and the currents are in milliamps, so the product stays in the milliwatts; in a power circuit the currents are large and the resistances must be tiny for exactly this reason. Any time a calculation puts a substantial current through a substantial resistance, the result is a heating element whether or not that was the intention.
Thicker copper is the only lever that does not cost performance
Resistance is resistivity times length divided by cross-section, so the three ways to lose less heat are a shorter run, a thicker conductor or a smaller current — and the third usually means not doing the thing you wanted to do. Going from 1.5 mm² to 2.5 mm² on the same ten metres drops the resistance from 0.233 Ω to 0.140 Ω and the heat at 10 A from 23.3 W to 14 W, a 40 % reduction for nothing but a heavier cable. This is also why running two shorter leads instead of one long one is not an improvement if they end up in series: the lengths add, and so does the heat.
| Current | Heat in the cable |
|---|---|
| 5 A | 5.8 W |
| 10 A | 23.3 W |
| 16 A | 59.7 W |
Worked with our own calculator
Power dissipation calculator
Given
- Current (A)
- 2
- Resistance (Ω)
- 100
Result
- Power dissipated (W)
- 400
These figures are produced by the calculator below, not typed in by hand — they are recomputed whenever the tool changes.
Run it on your own figures →Frequently asked questions
- Is this the same as voltage drop?
- They are two readings of the same loss. The voltage drop is I × R — 2.33 volts at 10 A on this cable — and the power lost is that drop times the current, which is the same as I²R. The drop is what the appliance at the far end notices, since it receives a lower voltage than the socket supplies; the power is what the cable notices, since it has to get rid of it. Long runs are usually limited by the drop and hot enclosures by the power, and a cable can be acceptable on one and not the other.
- Why does the resistance count the cable twice?
- Because current has to return. A ten-metre lead contains at least two conductors of ten metres each, and the same current flows out along one and back along the other, so the loop is twenty metres of copper. A resistance calculator asked for "length" will give the one-way figure unless you feed it the round trip, and forgetting to double it halves every answer that follows — the heat, the drop and the margin against the cable's rating.
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