Single-Phase Real Power with Power Factor

P=VIPFP = V I \, \text{PF}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

On DC, watts are simply volts times amps. On AC the current can lag or lead the voltage, and only the in-phase component delivers energy; the power factor cos φ is the bookkeeping for that. A 240 V single-phase welder pulling 20 A at 0.95 PF consumes 240 × 20 × 0.95 = 4560 W, not the 4800 VA its supply cable and breaker must actually carry.

Resistive loads — heaters, incandescent lamps, kettles — sit at PF = 1, which is why the clamp-meter product matches the wattmeter on those and not on a motor. Watch out for modern electronics: a switching power supply may draw badly distorted current, giving a total power factor well below the displacement cos φ this formula assumes, so the meter reading and the calculation drift apart.

Single-Phase Real Power with Power Factor
P=VIPFP = V I \, \text{PF}
Where
  • PP= Real power
  • VV= Line voltage
  • II= Line current
  • PF\text{PF}= Power factor