Power Dissipation
Heat dissipated in a component, and the power rating it needs.
P = I² × R — this energy leaves the component as heat.
Needs 5 W, so the next standard size up is 5 W. A part run at its exact rating gets extremely hot.
Note that power goes as the square of current, so doubling the current quadruples the heat. A resistor rated exactly at its dissipation will still work but runs hot enough to discolour boards and drift in value, which is why derating is standard practice rather than caution.
Power dissipated as heat is P = I²R, or V²/R, or simply V × I — whichever two quantities you know. Half an amp through 10 Ω dissipates 2.5 W, which with the usual 2× derating means specifying at least a 5 W resistor.
Three forms of one equation
All three expressions come from combining P = V × I with Ohm's law, so they always agree. Which one to use depends on what you happen to know: substitute V = IR to get P = I²R, or I = V/R to get P = V²/R.
The form worth internalising is P = I²R, because the square makes the behaviour non-obvious. Doubling the current through a component quadruples the heat. A circuit that runs comfortably at 0.5 A becomes a fire risk at 2 A even though the current has only gone up fourfold — the heat has gone up sixteenfold.
Rating is not the same as dissipation
A resistor's power rating is the point at which it survives, not the point at which it is comfortable. A 1/4 W resistor dissipating exactly 0.25 W will work, but it runs hot enough to discolour the board beneath it, drift measurably in value, and shorten its own life.
Standard practice is to specify a part rated at least twice the calculated dissipation, and more where airflow is poor or the part sits near other hot components. That is why this calculator applies a headroom factor before rounding up to a standard size.
Common through-hole resistor ratings are 1/8, 1/4, 1/2, 1, 2, 3, 5, and 10 W. Above that you are into wirewound or heatsinked parts.
Worked example: 0.5 A through 10 Ω
Calculate the heat, then choose a part that can shed it:
- 1 Pick the right form. Knowing current and resistance, use P = I²R.
- 2 Square the current. 0.5² = 0.25. This is where the non-linearity bites.
- 3 Multiply by the resistance. 0.25 × 10 Ω = 2.5 W of heat.
- 4 Apply headroom. 2.5 W × 2 = 5 W minimum rating.
- 5 Round to a standard size. 5 W is itself a standard value, so specify a 5 W resistor.
- 6 Sanity-check the voltage. V = IR = 5 V across the part, and 5 V × 0.5 A = 2.5 W — the three forms agree.
Dissipation and the rating to specify
Using 2× headroom and the standard through-hole series. Note how quickly current escalates the requirement.
| Current | Resistance | Dissipation | Minimum rating |
|---|---|---|---|
| 10 mA | 1 kΩ | 0.1 W | 1/4 W |
| 50 mA | 100 Ω | 0.25 W | 1/2 W |
| 100 mA | 100 Ω | 1 W | 2 W |
| 0.5 A | 10 Ω | 2.5 W | 5 W |
| 1 A | 10 Ω | 10 W | Above 10 W |
| 2 A | 1 Ω | 4 W | 10 W |
Where the heat actually matters
The classic place beginners get caught is a current-limiting resistor in a high-current path — an LED driver at 20 mA is trivial, but the same circuit arrangement at 1 A is a completely different component. Voltage regulators are the other: a linear regulator dropping 12 V to 5 V at 1 A dissipates 7 W, which needs a heatsink and is exactly why switching regulators exist.
Two effects compound as things get hot. Copper's resistance rises with temperature, so a hot conductor dissipates slightly more for the same current. And a resistor's value drifts, which in a precision circuit shows up as measurement error long before anything is at risk of failing.
One practical note on derating: a resistor's rating usually assumes free air at room temperature. Mounted flat against a board, packed among other components, or inside a sealed enclosure, its real capacity is meaningfully lower. When in doubt, choose the next size up — the cost difference is negligible and the failure mode is not.