Torque-slip characteristics of induction motors

Why it matters

Whether a motor can start a loaded conveyor, ride through a voltage dip or survive an overload depends on its torque–slip curve. Starting torque, maximum (breakdown) torque and the slip at which it occurs decide motor selection, and the same curve explains why rotor resistance is added for starting and why low voltage makes motors stall and overheat.

Key ideas

The torque equation. From the approximate equivalent circuit (magnetising branch moved to the terminals), per-phase voltage V1, stator R1, rotor R2' and total leakage reactance X = X1 + X2' (all referred to stator): T = 3·V1²·(R2'/s) / {ωs·[(R1 + R2'/s)² + X²]}. In many GATE problems R1 is neglected, which gives clean closed forms.

Shape of the curve (R1 neglected)

• Low slip (normal running, s ≪ s_m): R2'/s dominates, so T ≈ 3·V1²·s/(ωs·R2'). Torque is proportional to slip — the curve is almost a straight line, and speed falls only slightly as load rises.
• High slip (s ≫ s_m): X dominates, so T ≈ 3·V1²·R2'/(ωs·X²·s). Torque is inversely proportional to slip — a rectangular hyperbola.
• Maximum (breakdown, pull-out) torque occurs where R2'/s = X (maximum power transfer to R2'/s), i.e. at s_m = R2'/X.
• Tmax = 3·V1²/(2·ωs·X) — independent of rotor resistance, proportional to V1², inversely proportional to leakage reactance.
• Starting torque is the value at s = 1.

Stable and unstable regions. Between s = 0 and s_m, dT/ds > 0: if load increases, the motor slows, slip rises, torque rises and a new equilibrium is reached — stable. Between s_m and s = 1, dT/ds < 0: an increase in slip reduces torque, so the motor decelerates further and stalls — unstable for most loads. Full-load slip is normally well inside the stable region, and Tmax is typically 2–3 times full-load torque.

Effect of rotor resistance. Increasing R2' moves s_m = R2'/X to the right without changing Tmax. Starting torque therefore rises until R2' = X (then Tst = Tmax), and falls if R2' is increased further. A slip-ring motor exploits this: external resistance for starting, shorted out for running (to keep running slip and rotor copper loss low). Deep-bar and double-cage rotors get the same effect automatically, because rotor frequency is high at start.

Effect of supply voltage. Every torque value scales with V1². A 10 % voltage drop cuts Tmax and Tst by about 19 %; at a given load torque the motor runs at higher slip and draws more current, so it overheats.

Other regions. For s < 0 (rotor driven above Ns) the machine generates (induction generator, e.g. wind turbines). For s > 1 (rotor driven against the field) it brakes — plugging.

Formulas

• T = 3·V1²·(R2'/s) / {ωs·[(R1 + R2'/s)² + X²]} — torque (N·m); V1 per-phase voltage (V), ωs = 2π·Ns/60 (rad/s), X = X1 + X2' (Ω).
• s_m = R2' / √(R1² + X²) ≈ R2'/X — slip at maximum torque (R1 neglected in the second form).
• Tmax = 3·V1² / (2·ωs·X) — maximum torque, R1 neglected (N·m).
• T / Tmax = 2·s·s_m / (s² + s_m²) — Kloss relation (R1 neglected).
• Tst / Tmax = 2·s_m / (1 + s_m²) — starting to maximum torque ratio.
• s_m = s·(k + √(k² − 1)) with k = Tmax/T at slip s — finds s_m from an overload ratio.
• T ∝ V1² at a given slip.

Worked examples

Example 1 (standard): maximum and starting torque. Given: 400 V, 50 Hz, 4-pole, star-connected motor; per phase R2' = 0.3 Ω, X1 + X2' = 1.5 Ω; R1 and the magnetising branch neglected.

1. V1 = 400/√3 = 230.9 V; ωs = 2π × 1500/60 = 157.08 rad/s.
2. s_m = R2'/X = 0.3/1.5 = 0.2 (speed at Tmax = 1200 rpm).
3. Tmax = 3·V1²/(2·ωs·X) = 3 × 53333/(2 × 157.08 × 1.5) = 339.5 N·m.
4. Starting torque: Tst = 3·V1²·R2'/(ωs·(R2'² + X²)) = 3 × 53333 × 0.3/(157.08 × (0.09 + 2.25)) = 130.6 N·m.
5. Check with the ratio: Tst/Tmax = 2 × 0.2/(1 + 0.04) = 0.385; 0.385 × 339.5 = 130.6 N·m, which agrees.

Example 2 (GATE level): from overload capacity to starting torque. Given: a motor's maximum torque is 2.5 times its full-load torque, and full-load slip is 4 %. Neglect R1. Find the slip at maximum torque, the ratio of starting torque to full-load torque, and the factor by which rotor resistance must be raised for maximum torque at starting.

1. With k = Tmax/Tfl = 2.5 at s = 0.04: s_m = s·(k + √(k² − 1)) = 0.04 × (2.5 + 2.291) = 0.192.
2. Tst/Tmax = 2 × 0.192/(1 + 0.192²) = 0.370.
3. Tst/Tfl = 0.370 × 2.5 = 0.92 — this motor cannot start a full-load torque that is constant from standstill.
4. For Tst = Tmax we need s_m = 1, i.e. R2' increased by 1/0.192 = 5.2 times (possible only with a slip-ring rotor).

Common mistakes

• Using line voltage in the per-phase torque formula for a star-connected motor.
• Omitting ωs, which gives "torque" in watts instead of N·m.
• Writing R2' instead of R2'/s in the numerator.
• Thinking more rotor resistance raises Tmax; it only moves where Tmax occurs.
• Taking the smaller root when solving for s_m from the Kloss relation; the root less than s is the running-region solution, s_m is the larger one.
• Assuming torque is proportional to V (it is V²).

For GATE IN

Common numericals: Tmax and s_m from equivalent-circuit data, Tst/Tmax and Tfl/Tmax via the Kloss relation, the new torque or slip after a voltage change (T ∝ V² and, at low slip, T ∝ s·V²), and the rotor resistance for a required starting torque. Conceptual MCQs ask which quantities change with R2' and V1.

Quick check

1. Supply voltage falls by 20 %. By what factor does Tmax change?
2. R2' = 0.2 Ω, X = 1 Ω. At what slip does maximum torque occur?
3. Does increasing rotor resistance change maximum torque?
4. In the low-slip region, how does torque vary with slip?

Answers: 1. 0.8² = 0.64. 2. s_m = 0.2. 3. No, only the slip at which it occurs. 4. Approximately linearly (T ∝ s).