Single-Phase Real Power with Power Factor

P=VI PFP = V I \, \text{PF}
sig figs

Worked example: 240 V, 20 A, 0.95 PF → 4560 W. Press Try an example to run it live, then adjust anything.

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See also

Three-Phase Real Power: Both use real power and line current.Power Factor from Real and Apparent Power: Both use real power and power factor.Power-Factor Correction kvar: Both use real power and power factor.Delta Line and Phase Current: Both use line current.Voltage Drop, Three Phase: Both use line current.Phase Angle from Power Factor: Both use power factor.
2 more relatedThree-Phase Motor Full-Load Current: Both use power factor.Transformer Efficiency from Core and Copper Losses: Both use power factor.
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Single-Phase Real Power with Power Factor explained

PIVPF

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 formula

P=VI PFP = V I \, \text{PF}
Where
  • PP= Real power (W)
  • VV= Line voltage (V)
  • II= Line current (A)
  • PF\text{PF}= Power factor
Revised 16 October 2026
Cite this page

formula.expert. (2026, October 16). Single-Phase Real Power with Power Factor Calculator. https://formula.expert/solvers/single-phase-real-power-pf