Resistor Power Dissipation
Computes power dissipated by a resistor from V*I or V^2/R.
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Resistor power: P = R·I² = V²/R = V·I
The power dissipated by an ohmic resistor comes straight out of Ohm's law combined with the definition of electrical power: P = V·I = R·I² = V²/R, measured in watts. In a pure resistor, every bit of energy you push in leaves as heat. Example: run 10 mA through a 100 Ω resistor and it dissipates P = 100 · (0.01)² = 0.01 W = 10 mW, which sits comfortably inside a standard 1/4 W (250 mW) part. A 0.1 Ω current-sense resistor carrying 3 A is a different story: P = 0.1 · 9 = 0.9 W, so you want at least a 1 W or 2 W package. Common ratings run 1/8 W, 1/4 W, 1/2 W, 1 W, 2 W and up.
Applications: sizing, derating, current sense, LED drivers
The point is to size resistors properly so they don't burn out, and most engineers leave a 2× safety margin on dissipation as a matter of habit. It matters a lot in current-sense design (shunt resistors) and LED driver ballasts, but also in snubber networks, voltage dividers that carry real current, and power resistors that might need a heatsink or forced airflow. When the power gets high, you'll usually see wirewound or thick-film resistors rated anywhere from 5 to 500 W.
FAQ
Why a 2× margin? The power rating on the datasheet assumes free-air convection at 25 °C. Real boards run hotter than that, so doubling the headroom keeps the resistor under its derating curve and helps it last longer.
RMS or peak? With sine waves, plug in the RMS values of V and I. P = V_rms · I_rms gives you the true average dissipation. Use peak values instead and you'll overestimate by 2×.
Does the formula apply to AC? For a pure resistor, yes. Once the load has a reactive component, switch to P = V·I·cos(φ), which gives you the resistive (active) part of the power.
What about temperature? Resistance drifts as things heat up (that's the TCR, given in ppm/°C). In precision circuits, reach for low-TCR metal film parts and keep operation well under the rated power.
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