How it works
Electrical power P = V × I × PF (single-phase) or √3 × V × I × PF (three-phase). Rearranging for current: I = P ÷ (V × PF) or I = P ÷ (√3 × V × PF). Entering kW multiplied by 1,000 converts to watts for the division.
Worked example
10 kW at 230 V single-phase, PF 0.9: I = 10,000 ÷ (230 × 0.9) = 48.3 A. Same load on 400 V three-phase: I = 10,000 ÷ (1.732 × 400 × 0.9) = 16.0 A — three-phase uses roughly one-third the current for the same power.
Assumptions & limitations
Uses RMS voltage and the real power formula for steady-state full-load current. Starting current for induction motors is typically 5–7× full-load amps — size circuit breakers and cable for starting conditions, not just running current. Results assume a balanced, sinusoidal load.