AC power
Single-phase active power
- where
- active power (W)
- voltage (V)
- current (A)
- power factor
Explain this equation
Works out the real power, in watts, taken by a single-phase AC load. This is the power that does useful work or heating, and what the energy meter records.
V is the RMS supply voltage and I the RMS current; their product is the apparent power. PF is the power factor, a decimal from 0 to 1 showing what fraction of that is actually consumed. Inductive loads such as motors draw some current out of step with the voltage, and that part flows back and forth without doing work.
Multiplying V × I by PF strips out the non-working part. For a resistive heater PF is 1, so P is simply V × I.
230 V × 10 A at PF 0.8 gives 1840 W; the same current at unity would be 2300 W. To find current, rearrange to I = P / (V × PF).
Watch out: enter PF as a decimal, not a percentage. For distorted waveforms use the measured true power factor rather than cos φ.
Active power is the rate at which electrical energy becomes work or heat. Voltage multiplied by current gives apparent power, and power factor identifies the active proportion.
Write the unknown first. To find current, rearrange I = P/(V × PF); convert kW into W before using volts and amperes.
Enter your values to see the result.
Common mistake
Do not treat shaft output kW as electrical input kW or use PF as a percentage.
Limits
Use RMS voltage and current. The cos φ form assumes sinusoidal voltage and current; for distorted waveforms use measured true power factor, P/S.
Sources
The tools calculate the displayed equation. They do not automatically choose cable sizes, approve installations, supply current regulatory tables, certify equipment or replace inspection and testing.