A resistor turns the power it cannot pass into heat: P = I²R, which is the same as V × I or V² ÷ R. A 0.1 Ω current-sense resistor carrying 1 A burns 0.1 W, so you would fit a part rated 0.25 W or more for margin. Set your own below.
Inputs
Result
100 mW
dissipated as heat
Use a part rated 0.25 W or more (2x margin). Voltage across it: 0.100 V.
Smallest standard rating (2x margin)
0.25 W
The formula
Three forms of the same thing, depending on what you know: P = I²R if you have the current and the resistance, P = V × I if you have the voltage across it and the current, P = V² ÷ R from the voltage and resistance. For a current-sense or ballast resistor you usually know the current and the value, so I²R is the one to reach for.
Pick the rating with margin
A resistor’s power rating is the point where it sits near its maximum temperature in still air. Running one at its rating is asking for a hot, drifting, short-lived part, so the calculator recommends the smallest standard rating at twice the dissipation. Derate further if the resistor lives in a hot enclosure or next to other heat, and check the datasheet’s derating curve, which pulls the allowed power down as ambient temperature rises.
From a real board
Power dissipation is the whole design constraint for the OTD bn-02 DC electronic load, a bench tool that sinks a programmable current and burns it as heat in a resistor and a transistor (One Thousand Drones, bench-tools arc 2026). Sink 1 A from a 5 V supply and you are turning 5 W into heat that has to go somewhere, which is why a load like that is mostly heatsink. See the electronic-load course.
References
- Your resistor’s datasheet for the power rating and the temperature derating curve.
- One Thousand Drones. ESP32 DC electronic load (bn-02), power handling. Build the board.