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Resistor Power Dissipation Calculator

Power a resistor dissipates from current and resistance.

Result

0.55 W

How it is calculated

The tool computes the heat a resistor must dissipate with P = I² × R, where I is the current in amperes (A) and R the resistance in ohms (Ω); the answer is in watts (W), rounded to two decimals. Resistance cannot be negative. Because the current is squared, the sign of the current does not matter and the result is always zero or positive. The tool gives the power actually dissipated, not the rating to buy: you need a resistor whose rated power is comfortably higher, and the tool does not add a safety margin for you. It also assumes a steady (DC or RMS) current. For millamps, divide by 1000 first, for example 50 mA is 0.05 A.

Worked example

A 220 Ω resistor carries 50 mA, which is 0.05 A. Step 1: square the current: 0.05² = 0.0025. Step 2: multiply by the resistance: 0.0025 × 220 = 0.55. Result: 0.55 W. A ¼ W part would be overloaded; a 1 W resistor would give roughly a 2× margin, which is the comfortable choice for this load.

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Common mistakes

  • Entering the current in milliamps (50) instead of amps (0.05), which makes the power a million times too large.
  • Choosing a resistor whose rating equals the calculated power, leaving no margin, so it runs hot.
  • Entering the resistance in kΩ as if it were ohms, for example 2.2 for a 2.2 kΩ part.
  • Using the supply voltage divided by the resistance as the current when other parts in the circuit also drop voltage.
  • Applying the result to a pulsed or AC signal without using the RMS current.

About this calculator

Joule heating in a resistor follows P = I²R — power dissipated as heat rises with the square of current, so doubling the current quadruples the heat, not doubles it. The same power can equivalently be found from P = V²/R when voltage is known instead of current, or P = V×I when both voltage and current are known (this calculator takes current and resistance, so divide the voltage by R first if you only have V).

Electronics designers calculate this to pick a resistor whose printed power rating (commonly ⅛W, ¼W, ½W, 1W, or 2W for through-hole parts) comfortably exceeds the dissipated power, usually with at least a 2× safety margin, since a resistor run near its rated limit runs hot, drifts in value, and can fail. It's a routine check when sizing current-limiting resistors for LEDs or bleeder resistors in power supplies.

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Comments (1)

  • Editorial team

    A common real-world check is an LED current-limiting resistor. At 5 V with a 2 V LED drop and 20 mA, the resistor is 150 ohms and dissipates 0.06 W, so a 1/8 W part gives about the 2x margin the guide recommends. Because power grows with the square of current, a modest increase in LED current heats the resistor much more. What circuit are you sizing a resistor for?

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