Formula
DC: P (kW) = V × I ÷ 1000
AC single phase: P = V × I × PF ÷ 1000
AC three phase: P = √3 × VL-L × I × PF ÷ 1000
Worked example
A three-phase feeder carrying 20 A at 400 V, PF 0.8: P = 1.732 × 400 × 20 × 0.8 ÷ 1000 = 11.08 kW.
Reference table
Quick reference (PF = 0.8)
| Amps | kW @ 230 V (1φ) | kW @ 400 V (3φ) |
|---|---|---|
| 5 | 0.92 | 2.77 |
| 10 | 1.84 | 5.54 |
| 20 | 3.68 | 11.09 |
| 50 | 9.2 | 27.71 |
| 100 | 18.4 | 55.43 |
| 200 | 36.8 | 110.85 |
Where this shows up in the real world
Energy audits live in this direction: walk a building with a clamp meter, log amps per circuit, convert to kW, and suddenly you know which loads dominate the bill. Facility managers use it to find the 'mystery kilowatts' running at 2 a.m.; homeowners use it to learn that the old chest freezer in the garage costs more than its contents.
Common mistakes to avoid
Remember that a snapshot reading is not a profile — a water heater drawing 18 A for ten minutes an hour is a much smaller energy story than the same current running continuously. Pair the kW figure with run-time before drawing money conclusions, and use real power factor for motor loads, not the optimistic 1.0.
Frequently asked questions
Is this the power I'm billed for?
Utilities bill energy (kWh), which is this kW figure multiplied by hours of use. Some industrial tariffs also charge for peak demand in kW or kVA.
Why is my clamp-meter kW different?
A clamp meter alone measures current; converting to kW requires the actual voltage and power factor at that moment. A power analyzer measures all three.
What PF should I assume?
Measure it if you can. Otherwise: ~1.0 resistive, 0.8–0.9 for motor loads, often lower for lightly loaded motors.