Single-Phase Real Power with Power Factor
Worked example: 240 V, 20 A, 0.95 PF → 4560 W — press Try an example to run it live, then adjust anything.
Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Try different units for next level excitement!
Power factor →
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Single-Phase Real Power with Power Factor explained
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
- = Real power (W)
- = Line voltage (V)
- = Line current (A)
- = Power factor
Missing one of these? Work it out first, then come back
- Real power — Three-Phase Real Power, Power Factor from Real and Apparent Power
- Line voltage — Three-Phase Real Power, Three-Phase Apparent Power
- Line current — Three-Phase Real Power, Three-Phase Apparent Power
- Power factor — Three-Phase Real Power, Power Factor from Real and Apparent Power