--- title: "Torque-slip curve and max torque condition" subject: "Electrical Machine" chapter: 5 key: "semester_4/electrical_machine:torque-slip-max-torque-condition" source: https://www.bctengineer.com/q/semester_4/electrical_machine/torque-slip-max-torque-condition site: BCT Engineering --- # Torque-slip curve and max torque condition Electrical Machine · Chapter 5: Three Phase Induction Machines - Asked in: 2073 Bhadra, 5+3 marks; 2071 Magh, 3+3 marks; 2070 Bhadra, 8 marks; 2068 Magh, 8 marks (asked 4 times) - Video: [torque slip characteristics of 3 phase induction motor | torque speed characteristics |motor control](https://www.youtube.com/watch?v=g4jfjVvgFrs) — Mind of Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machine/torque-slip-max-torque-condition - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-5 ## Question Explain the torque-slip characteristics of an induction motor. Show the condition for which the maximum torque develops in the induction motor. ## Short answer Torque is proportional to sR2/(R2²+(sX2)²). Maximum torque occurs when slip s_m = R2/X2, i.e., rotor resistance equals rotor reactance at that slip, and T_max = 3E2²/(2ω_s X2). ## Answer #### Torque equation In a 3-phase induction motor, the rotor current at slip s is I₂ = sE₂ / √(R₂² + (sX₂)²), where E₂ and X₂ are the rotor emf and reactance per phase at standstill. The torque is proportional to Φ I₂ cos φ₂, which gives T = (3 / ω_s) × s E₂² R₂ / (R₂² + (sX₂)²) newton-metre where ω_s = 2πN_s/60 is the synchronous angular speed. #### Torque-slip characteristic ```text T | | Tmax | ******* | * ***** | * ****** | * ***** Tst | * | * |* +---------+------------------+--> s 0 s_m 1 (N = Ns) (N = 0) ``` 1. **Low slip (s small)**: sX₂ << R₂, so T ∝ s. The curve is a straight line; this is the normal working region (stable). 2. **Maximum torque** at s = s_m. 3. **High slip (s large)**: R₂ << sX₂, so T ∝ 1/s. The curve is a rectangular hyperbola, and the motor is in the unstable region. 4. At s = 1 (standstill) the torque is the starting torque T_st = (3/ω_s) E₂² R₂ / (R₂² + X₂²). 5. At s = 0 (synchronous speed), the torque is zero. #### Condition for maximum torque Put dT/ds = 0. Writing T = K s R₂ / (R₂² + s²X₂²): d/ds [s / (R₂² + s²X₂²)] = 0 gives (R₂² + s²X₂²) − 2s²X₂² = 0 R₂² = s²X₂² **s_m = R₂ / X₂** That is, the torque is maximum when the rotor resistance equals the rotor reactance at that slip (R₂ = s_mX₂). Substituting: **T_max = 3E₂² / (2 ω_s X₂)** T_max is independent of the rotor resistance, and proportional to the square of the supply voltage. Also T / T_max = 2 s s_m / (s² + s_m²). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At what slip does maximum torque occur? A. s_m = R2/X2 B. s_m = 1 C. s_m = 0 D. s_m = X2/R2 Answer: A. Setting dT/ds = 0 gives R2 = s X2. (Key point: Maximum torque condition) 2. What is the torque behaviour at low slip? A. T is constant B. T ∝ s, a straight line C. T ∝ 1/s D. T ∝ s^2 Answer: B. At low slip sX2 << R2 so T ∝ s. (Key point: Low-slip region) 3. At high slip, how does torque vary? A. T ∝ s B. T is constant C. T ∝ 1/s, rectangular hyperbola D. T increases Answer: C. With R2 << sX2, T ∝ 1/s; the region is unstable. (Key point: High-slip region) 4. On what does T_max depend? A. Starting torque only B. R2 only C. Rotor resistance and slip D. Supply voltage squared and X2, not on R2 Answer: D. T_max = 3E2^2/(2 ω_s X2). 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