What is the watts to amps formula?

For DC and AC single-phase, I = W ÷ (V × PF). For AC three-phase, I = W ÷ (√3 × V × PF), where PF is the power factor and √3 ≈ 1.732. For purely resistive loads PF is 1.0.

Why does power factor appear here but not in kVA to amps?

Watts is real power, which is only the component doing work. For inductive loads the line current is larger than watts alone implies, so you must divide by the power factor to recover the actual current. kVA is already apparent power, so it needs no PF.

What power factor should I use?

Use 1.0 for resistive loads like heaters and incandescent lamps. Motors and electronic ballasts typically run 0.7 to 0.9; use the nameplate value when available. Guessing too high underestimates the current and can undersize the breaker.

Which voltage for three-phase?

Use the line-to-line voltage (208, 240, 480, or 600 V). The √3 in the formula accounts for the three-phase relationship, so the line-to-neutral voltage would give the wrong answer.

Can I use this to size a breaker?

It gives the operating current. For continuous loads apply the NEC 125% factor and round up to the next standard breaker, and confirm the conductor ampacity with any temperature or bundling derating separately.

Watts to Amps Calculator

Turning a wattage rating into real line current

Equipment is often rated in watts, but to size a breaker or conductor you need amps. The conversion depends on the voltage and, for AC loads, the power factor. Watts is real power — only the part doing useful work — so for inductive loads such as motors the actual line current is higher than the wattage alone suggests. This calculator covers DC, single-phase, and three-phase loads and includes the power factor term so the result reflects the true current the circuit must carry.

How it works

For DC and AC single-phase loads the current is:

I = W / (V × PF)

For AC three-phase loads the line current is:

I = W / (√3 × V × PF)

with V the line-to-line voltage for three-phase, PF the power factor (1.0 for resistive loads, lower for motors), and √3 ≈ 1.732. DC has no power factor, so the tool fixes it at 1.0 in that mode. Dividing by the power factor is what corrects for the reactive current an inductive load draws on top of its real-power current.

Example and notes

A 3600 W resistive water-heater element at 240 V single-phase draws 3600 ÷ (240 × 1.0) = 15 A. A 3600 W motor at 240 V with a 0.85 power factor draws 3600 ÷ (240 × 0.85) ≈ 17.6 A — the same watts but more current because of the reactive component. Always use the nameplate power factor for motors; assuming 1.0 underestimates current and risks undersizing the conductor. For continuous loads, multiply the result by 1.25 before picking the overcurrent device and check the conductor ampacity with any derating applied.