--- title: "Torque-slip curve and rotor resistance" subject: "Electrical Machine" chapter: 5 key: "semester_4/electrical_machine:torque-slip-rotor-resistance-effect" source: https://www.bctengineer.com/q/semester_4/electrical_machine/torque-slip-rotor-resistance-effect site: BCT Engineering --- # Torque-slip curve and rotor resistance Electrical Machine · Chapter 5: Three Phase Induction Machines - Asked in: 2070 Magh, 8 marks; 2072 Magh; 2072 Asoj (asked 3 times) - Video: [effect of rotor resistance variation on torque slip characteristics | torque speed characteristics](https://www.youtube.com/watch?v=-KGIlyWVJAg) — 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-rotor-resistance-effect - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-5 ## Question Explain the torque-slip characteristics of a 3-phase induction motor. Show the condition for which the maximum torque develops in the induction motor. Discuss the effect of variation of rotor resistance on this maximum torque. ## Short answer Maximum torque occurs at slip s_m = R2/X2 and is independent of rotor resistance. Increasing rotor resistance shifts the peak to higher slip, raising starting torque; when R2 = X2, starting torque equals maximum torque. ## Answer #### Torque-slip characteristic For a 3-phase induction motor, the torque at slip s is T = (3 / ω_s) × s E₂² R₂ / (R₂² + (sX₂)²) N·m where E₂, R₂ and X₂ are the standstill rotor emf, resistance and reactance per phase, and ω_s is the synchronous speed in rad/s. ```text T | | Tmax | ******* | * ***** | * ****** | * ***** Tst | * | * |* +---------+------------------+--> s 0 s_m 1 (N = Ns) (N = 0) ``` - At s = 0, T = 0. For small s, T ∝ s (stable working region, nearly linear). - T rises to T_max at s = s_m, then falls as T ∝ 1/s (unstable region) down to the starting torque T_st at s = 1. - Normal full-load slip is only 2 to 5 %, in the stable linear region. #### Condition for maximum torque Set dT/ds = 0: (R₂² + s²X₂²) − 2s²X₂² = 0, so R₂² = s²X₂² **s_m = R₂ / X₂**, and **T_max = 3E₂² / (2ω_sX₂)** Maximum torque occurs when the rotor resistance per phase equals the rotor reactance at that slip. #### Effect of rotor resistance - T_max does not depend on R₂, but the slip at which it occurs (s_m = R₂/X₂) increases in proportion to R₂. - Higher R₂ shifts the peak towards s = 1 (lower speed), and the **starting torque increases**. - If R₂ = X₂, then s_m = 1, and the starting torque equals the maximum torque. - If R₂ is increased beyond X₂, the starting torque begins to fall. ```text R2 small R2 larger R2 = X2 (low Tst) (higher Tst) (Tst = Tmax) T | ** T | ******* T | ****** | * ** | ** *** | *** |* ** | * | ** | ****** | * | ** | * |* |** +---------------> s +---------------> s +---------------> s 0 1 0 1 0 1 ``` | Rotor resistance | s_m | T_max | T_st | |---|---|---|---| | Low | small | same | low | | Medium | moderate | same | higher | | R₂ = X₂ | 1 | same | = T_max | This is why a slip-ring induction motor, with external resistance added to the rotor circuit, gives a high starting torque with a low starting current. The extra resistance is cut out gradually as the motor speeds up, to keep good efficiency and low slip. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At s_m, the rotor resistance equals what? A. Rotor reactance at that slip (s_m X2) B. Zero C. Stator resistance D. Twice X2 Answer: A. R2 = s_m X2. (Key point: Condition) 2. What happens to T_max when R2 is increased? A. It becomes zero B. It stays the same C. It decreases D. It increases Answer: B. T_max = 3E2^2/(2ω_s X2) is independent of R2. (Key point: Effect on T_max) 3. What happens to s_m when R2 increases? A. It stays fixed B. It becomes negative C. It increases in proportion; the peak shifts toward lower speed D. It decreases Answer: C. s_m = R2/X2. (Key point: Effect on s_m) 4. When does starting torque equal maximum torque? A. When R2 = 2X2 B. When s = 0 C. When R2 = 0 D. When R2 = X2 Answer: D. Then s_m = 1, so the peak occurs at standstill. 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