Convert watts to amps, amps to watts, find volts or use Ohm's law, for DC, single-phase and three-phase. Runs on your device.
For a DC circuit, amps are watts divided by volts: I = P ÷ V. A 2,300 W heater on 230 V draws 2,300 ÷ 230 = 10 A, and the same 2,300 W on 120 V draws 19.167 A. For a single-phase AC circuit the power factor (PF) enters the formula: I = P ÷ (V × PF). With 2,300 W, 230 V and a PF of 0.8 the current is 2,300 ÷ (230 × 0.8) = 12.5 A, and the apparent power is 230 × 12.5 = 2,875 VA. The box under the answer repeats every step with your numbers, and the table shows the current at 120, 230 and 240 V.
For a balanced three-phase AC load the voltage is the line-to-line value and the formula gains √3, about 1.732: I = P ÷ (√3 × V × PF). A 5,000 W load at 400 V and a PF of 0.9 draws 5,000 ÷ (1.732 × 400 × 0.9) = 8.019 A in each line. Going the other way, 10 A at 400 V and PF 0.9 is 1.732 × 400 × 10 × 0.9 = 6,235.38 W. The power factor is a number above 0 and up to 1 that compares real power with apparent power. It is 1 for a plain heater, and the label of a motor or a power supply may give a lower figure. The page does not guess it for you.
Amps to watts is the same relation turned around: P = V × I for DC, or P = V × I × PF for single-phase AC. 10 A at 120 V with a PF of 0.9 is 120 × 10 × 0.9 = 1,080 W. Volts mode finds V = P ÷ I, so 2,300 W at 10 A is 230 V. The Ohm's law mode takes two of volts, amps and ohms and returns the others plus power. 230 V and 10 A give R = 230 ÷ 10 = 23 Ω and P = 230 × 10 = 2,300 W; 230 V across 46 Ω gives 5 A and 1,150 W. Ohm's law holds for resistive loads, so it is shown for DC only.
The results describe an ideal, balanced circuit with the numbers you type. They do not include cable length, voltage drop, motor start-up current, harmonics, efficiency, a safety margin or any local code. They are an estimate, not a basis for choosing wire sizes, fuses or circuit breakers; for that, a qualified electrician applies the rules for your country. Voltage presets are only starting points: supply voltages differ between countries and buildings, so type the one on your equipment's label if it differs.
Divide watts by volts for DC: 2,300 W ÷ 230 V = 10 A. For single-phase AC also multiply the volts by the power factor: 2,300 ÷ (230 × 0.8) = 12.5 A. For three-phase, the volts term is √3 × V × PF.
For DC or a load with a power factor of 1, 1,000 ÷ 120 = 8.333 A. At 230 V the same 1,000 W is 4.348 A. With a power factor of 0.9 at 120 V single-phase it is 1,000 ÷ (120 × 0.9) = 9.259 A.
P = V × I for DC, P = V × I × PF for single-phase AC and P = √3 × V × I × PF for three-phase. 10 A at 120 V and a PF of 0.9 gives 1,080 W.
For a balanced three-phase load, total power works out to √3, about 1.732, times the line-to-line voltage times the line current times PF. So I = P ÷ (√3 × V × PF) with V line-to-line. 5,000 W at 400 V and PF 0.9 is 8.019 A.
Use the one on the equipment's label. A resistive load such as a heater is 1. For others a figure like 0.8 or 0.9 is common, but the page cannot know it. At 2,300 W and 230 V, PF 1 gives 10 A and PF 0.8 gives 12.5 A.
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