Ohm's Law Calculator — V, I, R & P

Voltage equals current times resistance: V = I × R. Rearranged, I = V ÷ R and R = V ÷ I. Combined with the power relationship P = V × I, those four quantities give twelve usable formulas — the "power wheel" — so any two known values yield the other two.

Divide: I = V ÷ R. A 120 V supply across a 10 Ω element draws 120 ÷ 10 = 12 amps, dissipating 1,440 W. That is a typical 1,500 W heating element — the arithmetic that ties element resistance to circuit loading.

Because power can be expressed in terms of whichever pair you know. P = V × I is the direct definition. Substituting Ohm's law gives P = I² × R (useful when you know current and resistance, as in conductor heating) and P = V² ÷ R (useful for a fixed-voltage load like a heating element). All three give the same answer.

For purely resistive AC loads, yes — heaters, incandescent lamps, resistance elements. Once inductance or capacitance is present (motors, transformers, ballasts, LED drivers) you must replace resistance with impedance and account for power factor, because voltage and current are no longer in phase. For those loads use the amps-to-watts or kVA calculators, which handle power factor properly.

Constantly, mostly for the P = I² × R case. Conductor heating and voltage drop are both I²R problems — that is why doubling the current quadruples the heat in a wire. It also underlies resistance testing: a megohmmeter reading, a ground-electrode resistance measurement, and a motor winding check are all Ohm's law applied to a known test voltage. For conductor runs specifically, see the Voltage Drop Calculator.

Because power in a resistance goes with the square of current: P = I² × R. Double the current and I² rises by a factor of four, so a conductor carrying 40 A dissipates four times the heat it did at 20 A — with the same resistance. This is the single most important consequence of Ohm's law for anyone sizing conductors, and the reason ampacity limits exist at all.