What this converter does
This converter turns line current in amps into real power in kilowatts for single-phase and three-phase AC circuits. Enter the voltage and power factor, pick the phase, and read the power — or swap the arrow to go from kW back to amps. It updates as you type.
Three-phase power uses the line-to-line voltage and a √3 factor; single-phase drops the √3. Power factor scales the result: real power is always current × voltage × PF. For apparent power, see the kVA to kW converter.
The units it covers
The power depends on three inputs beyond the current — the phase, the voltage and the power factor.
View all units & their values
| Unit | Symbol | Value | Mainly used |
|---|---|---|---|
| Line current | A | I | What the conductor carries |
| Real power | kW | P | The load, in kilowatts |
| Voltage | V | V | Line-to-line for three-phase |
| Power factor | PF | cos φ | Ratio 0–1, motor ≈ 0.8 |
The formula
Power is current times voltage times power factor, with √3 for three-phase:
kW = √3 × V × A × PF ÷ 1000 — drop √3 for single-phaseWhere:
- A = line current in amps
- V = supply voltage (line-to-line for 3-phase)
- PF = power factor, 0 to 1
- √3 = 1.732, the three-phase factor
Worked example
A 180 A three-phase load at 400 V and 0.8 PF. Find the power.
kW = √3 × V × A × PF ÷ 10001.732 × 400 × 180 × 0.8 ÷ 1000 = 99.8 kWThe 180 A three-phase feeder delivers about 100 kW at these settings.
The units in this example
The amperes flowing in each line conductor. Multiplied by voltage and power factor, it gives the real power the load consumes.
- kW = V × A × PF ÷ 1000 — 1ph
- kW = √3 × V × A × PF ÷ 1000 — 3ph
- 1 A at 230 V, PF 1 = 0.23 kW
- √3 = 1.732
The useful power the load consumes. For a given current it rises with voltage and power factor, and with the √3 factor on three-phase supplies.
- 1 A, 230 V, 1ph, PF 1 = 0.23 kW
- 1 A, 400 V, 3ph, PF 0.8 = 0.554 kW
- A = kW × 1000 ÷ (V × PF) — 1ph
- 1 kW = 1,000 watts