Four quantities describe an induction motor at full load: the current it draws, the speed it runs at, the slip between the rotating field and the rotor, and the torque at the shaft. They follow directly from the nameplate power, voltage, power factor, efficiency, poles and frequency, and they feed almost every downstream calculation — cable size, overload setting, starter rating and drive selection.
Reviewed: June 19, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: Wikipedia: Induction motor, IEC 60034.
The four motor formulas
P is the rated mechanical output; dividing by efficiency η gives the electrical input that sets the current. The constant 9550 = 60 × 1000 / (2π) converts kW and rpm directly to newton-metres.
Worked example — 11 kW pump motor
Scenario: 11 kW, 415 V, 3-phase, 4-pole, 50 Hz, PF 0.85, efficiency 90%, nameplate speed 1450 rpm.
That 20 A FLA drives the cable size, overload setting and starter choice — carry it into the motor starter sizing calculator and the cable-sizing calculator.
Frequently Asked Questions
Three-phase: FLA = (P ÷ efficiency) ÷ (√3 × V × PF), with P the rated output in watts. An 11 kW, 415 V, 0.85 PF, 90% motor: input 12,222 W, FLA ≈ 20 A.
Ns = 120 × frequency ÷ poles. A 4-pole 50 Hz motor: 1500 rpm; on 60 Hz, 1800 rpm. The rotor runs slightly slower.
Slip = (Ns − N) ÷ Ns. A 4-pole 50 Hz motor at 1450 rpm slips 3.3%. Induction motors typically slip 2–5% at full load.
T (N·m) = 9550 × kW ÷ rpm at the rated speed. 11 kW at 1450 rpm ≈ 72 N·m. The 9550 constant is 60 ÷ (2π) × 1000.
An induction motor needs relative motion between field and rotor to induce current and make torque. At synchronous speed there would be none, so it settles a few percent below, and that slip grows with load.