Short notes

Slip Torque Characteristics Short Notes

Plot the torque of a 415 V induction motor against slip from s = 1 (standstill) to s = 0 (synchronous speed) and you get the familiar curve that every EEE student has to sketch in the exam hall — steep linear rise near s = 0, a peak at slip sm, then a fall back toward zero at starting if R2 is small.

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How it works

Torque T = (3V1²R2/s) / [ωs × ((R1 + R2/s)² + (X1+X2)²)]. For the simplified case neglecting stator impedance, T ∝ sR2/(R2² + s²X2²). Maximum torque Tmax = 3V1²/[2ωs(X1+X2)] — independent of rotor resistance R2. The slip at maximum torque sm = R2/X2, so increasing R2 (possible in slip-ring motors by adding external resistance in steps of, say, 0.5 Ω per phase) shifts the peak torque to higher slip without reducing its magnitude. Starting torque Tst = T|s=1 = 3V1²R2/[ωs(R2²+X2²)]; maximum starting torque occurs when R2 = X2.

Key points to remember

The ratio Tst/Tmax = 2sm/(1 + sm²) — at sm = 0.1 this ratio is only about 0.198, meaning starting torque is much less than maximum torque for typical squirrel-cage motors. The curve has two regions: stable (s < sm) where motor self-corrects load changes, and unstable (s > sm) where a load increase causes further speed drop and eventual stall. Voltage affects torque as T ∝ V1²; a 10% voltage dip reduces torque by 19%, which is why motors struggle during grid sags. This curve must be drawn accurately with both Tst and Tmax labelled.

Exam tip

Every Anna University model paper asks you to derive the condition for maximum torque (dT/ds = 0 leading to sm = R2/X2) — work through the differentiation at least once so you can reproduce it in the exam.

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