Joule Heating Calculator

Use our Joule heating calculator to compute the heat generated by a current-carrying cable.

Power654.5455 mW

Voltage12 V

Current54.5455 mA

Resistance220 Ω

Check: I²R654.5455 mWthe same power, computed a different way

Check: V²/R654.5455 mW

Against a 0.25 W rating261.8% of ratedover rating — the resistor will overheat

Rating usually recommended1.3091 Wtwice the dissipation is the common derating rule of thumb

Energy in one hour2.3564 kJ0.000655 kWh

At twice the voltage2.6182 Wfour times the power into a fixed resistance — power goes as the square of voltage

The formula

V = IR; P = VI = I²R = V²/R

Two of four, and the square

Voltage, current, resistance and power are linked such that any two determine the other two. That is the whole content of Ohm's law plus the power relation, and it is why this page takes any pair.

The relationship worth internalising is the square. Power goes as the square of either voltage or current, so doubling the supply voltage across a fixed resistor quadruples the heat. This is why a resistor that runs warm at 5 V will fail at 10 V, and why transmission lines run at high voltage — for a given power, higher voltage means lower current, and the I²R losses fall with the square.

On resistor ratings: a resistor at its rated dissipation is running as hot as the manufacturer will tolerate, not as hot as you should want. Common practice is to pick a part rated at twice the calculated dissipation, which keeps it appreciably cooler and considerably longer-lived.