Induction Motor Slip
Also known as rotor slip · slip percentage
Worked example: 1750 rpm against 1800 rpm → 2.78% slip. 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. Change any unit and the answer follows.
Where Induction Motor Slip fits
- Motor calculations5 formulas
See also
Synchronous Speed from Frequency and Poles: Both use synchronous speed.Motor Torque from Power and Speed: Both use rotor speed.Synchronous speed and slip →
UniversityCircuits & Electrical Power
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Induction Motor Slip explained
An induction motor's rotor must lag the rotating field: if it ever caught up, the field would stop sweeping past the rotor bars, no voltage would be induced, and no torque would exist. That lag is slip. A 4-pole motor on 60 Hz has a synchronous speed of 1800 rpm; a nameplate reading of 1750 rpm means s = 100 × 50/1800 ≈ 2.8%, entirely typical for a healthy machine at full load.
Slip is a free load indicator: it is very nearly proportional to torque, so a tachometer reading tells you how hard the motor is working without a single meter lead. At no load slip falls under 1%; at breakdown torque it may hit 20%. Rising slip on a familiar machine means added mechanical load, low voltage, or a broken rotor bar. Note that this solver treats rpm as a frequency, so answers appear in hertz unless you pick rpm from the table.
Induction Motor Slip formula
- = Slip (%)
- = Synchronous speed (rpm)
- = Rotor speed (rpm)
Cite this page
formula.expert. (2026, October 16). Induction Motor Slip Formula & Calculator. https://formula.expert/solvers/motor-slip-percent
https://formula.expert/solvers/motor-slip-percent