Current depends on the circuit and the kind of power
A watt measures real electrical power, while an ampere measures current. Dividing watts by volts gives current directly for a DC load. Alternating-current equipment also needs a power-factor term because real power can be smaller than voltage multiplied by current. This converter offers DC, single-phase AC and a balanced three-phase AC model. Enter real electrical input power, not mechanical output from a motor or heat output from a heat pump. A conversion from output to input would need efficiency or another performance factor.
The three current equations
DC: I = P ÷ V; single-phase: I = P ÷ (V × PF); balanced three-phase: I = P ÷ (√3 × V × PF)Single-phase voltage is the RMS voltage across the load. Three-phase voltage is the line-to-line RMS value, and the power entry is total real power across all three phases. The displayed three-phase current is the line current for that balanced model. AC power factor must be greater than zero and no more than one. DC mode ignores the power-factor field and uses one. Apparent power is real power divided by power factor and is reported in volt-amperes for AC. The same power and voltage quantities are used in metric and imperial modes.
Examples that show the power-factor effect
A 1,200 watt DC load on a 240 volt supply draws 1,200 ÷ 240 = 5 amps in the ideal steady model. The same real power on a single-phase 240 volt AC supply at a power factor of 0.8 gives 1,200 ÷ (240 × 0.8) = 6.25 amps. For a balanced three-phase load consuming 6,000 watts at 400 volts line-to-line and a power factor of 0.8, line current is 6,000 ÷ (√3 × 400 × 0.8), approximately 10.83 amps. Using 230 volts in place of the required 400 volt line-to-line value would give the wrong answer for that example.
Use nameplate data consistently
Equipment may list input watts, volt-amperes, rated current or output capacity; those are not interchangeable. Use a stated operating power factor when available rather than assuming all AC loads behave like resistive heaters. Electronic and motor loads can change their draw with operating conditions, and startup peaks are outside this steady calculation. The current result is not a breaker, cable or protective-device recommendation. Circuit protection and conductor suitability require the equipment instructions and a complete installation check. Zero input power returns zero current; zero supply voltage is rejected. Keep voltage-drop calculations separate when the cable route causes the equipment terminal voltage to differ from the source voltage.
Frequently asked questions
How many amps is 1200 watts at 240 volts?
For DC or single-phase AC at power factor one, the result is five amps. A lower AC power factor increases the current for the same real power.
Does three-phase use line-to-line voltage?
Yes. Enter total real power for all three balanced phases and their line-to-line RMS voltage. The result is line current.
What power factor should I enter?
Use equipment data for the operating condition. The tool does not infer power factor from watts or choose a default for a particular appliance.
Can I use motor output watts?
Not directly. Divide mechanical output by efficiency to find real electrical input, then use the relevant voltage and power factor.
Can the result choose my circuit breaker?
No. It converts steady power to current. Startup behaviour, continuous loading, fault protection and applicable installation requirements need separate assessment.
Last reviewed: 2026-10-03 · How we calculate