--- title: "Max torque independent of rotor resistance" subject: "Electrical Machine" chapter: 5 key: "semester_4/electrical_machine:torque-slip-max-torque-independent-rr" source: https://www.bctengineer.com/q/semester_4/electrical_machine/torque-slip-max-torque-independent-rr site: BCT Engineering --- # Max torque independent of rotor resistance Electrical Machine · Chapter 5: Three Phase Induction Machines - Asked in: 2074 Bhadra, 6 marks - Video: [Derivation for condition of Maximum torque and Maximum Torque of 3phase induction motor](https://www.youtube.com/watch?v=fy4CJAxCAAA) — Eee Apple - 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-independent-rr - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-5 ## Question Explain the torque-slip characteristics of a three phase induction motor. Starting with the expression for torque as a function of slip, show that the value of maximum torque is independent of rotor resistance. ## Short answer Torque T = (3/ω_s) s E2² R2/(R2² + s²X2²). Differentiating gives maximum torque at slip s_m = R2/X2. Substituting s_m yields T_max = 3E2²/(2 ω_s X2), which contains no R2 term, so maximum torque is independent of rotor resistance; R2 only shifts the peak position. ## Answer A three-phase induction motor develops torque because of the interaction of the air-gap flux with the rotor current. Its torque changes with slip, and the plot of torque against slip (or speed) is the torque-slip characteristic. #### Torque-slip characteristics ```text T | Tmax | *** | * * | * * | * * | * * |*Tst * T_fl +--|--------|-----------|---|--> s s=1 s_m 0.05 0 (standstill) (near sync) ``` - **s = 0 (synchronous speed):** rotor emf and current are zero, so T = 0. - **Small slip (0 to about 5 %):** the rotor reactance sX₂ is negligible, so I₂ ∝ s and T ∝ s. The curve is nearly a straight line. This is the stable, normal running region. - **Slip s_m (about 10 to 25 %):** torque reaches its maximum value (breakdown or pull-out torque). - **Beyond s_m up to s = 1:** sX₂ dominates, T ∝ 1/s, and the torque falls. This is the unstable region. At s = 1 the torque is the starting torque. #### Torque expression The rotor emf per phase at slip s is E₂ᵣ = sE₂ and the reactance is sX₂, where E₂ and X₂ are standstill values. I₂ = sE₂ / √(R₂² + s²X₂²) cos φ₂ = R₂ / √(R₂² + s²X₂²) Torque T = P₂ / ω_s, where the rotor air-gap power is P₂ = 3 E₂ I₂ cos φ₂ (= 3I₂²R₂/s). This gives **T = (3 / ω_s) · s E₂² R₂ / (R₂² + s²X₂²)**, with ω_s = 2πN_s/60. #### Condition for maximum torque Differentiate and put dT/ds = 0: ```text d/ds [ s R2 / (R2² + s²X2²) ] = 0 (R2² + s²X2²) - s(2 s X2²) = 0 R2² - s²X2² = 0 s_m = R2 / X2 ``` Put s = s_m = R₂/X₂ in the torque equation: ```text Tmax = (3/ws) · E2² · (R2/X2) · R2 / (R2² + R2²) = (3/ws) · E2² · (R2²/X2) / (2 R2²) Tmax = 3 E2² / (2 ws X2) ``` **Result:** T_max depends only on the supply voltage (through E₂, since E₂ ∝ V), the standstill reactance X₂ and the synchronous speed. R₂ does not appear, so the maximum torque is independent of rotor resistance. R₂ only decides the slip s_m = R₂/X₂ at which T_max occurs. Raising R₂ (external resistance in a slip-ring motor) moves the peak toward s = 1 without changing its height, and when R₂ = X₂ the peak occurs at starting (s_m = 1). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which expression is correct? A. T = (3/ω_s) s E2^2 R2/(R2^2 + s^2X2^2) B. T = (3/ω_s) s E2 R2/(R2 + sX2) C. T = s E2^2 X2/(R2^2 + s^2X2^2) D. T = (3/ω_s) E2^2 R2/s Answer: A. Rotor input over synchronous speed gives this form. (Key point: Torque expression) 2. Setting dT/ds = 0 gives? A. R2 = 0 B. R2^2 = s^2X2^2 C. s = 1 D. X2 = 0 Answer: B. The derivative numerator reduces to R2^2 - s^2X2^2. (Key point: Maximum condition) 3. Why is T_max independent of R2? A. X2 cancels B. R2 is zero C. Substituting s_m = R2/X2 cancels R2 D. R2 is constant Answer: C. The result 3E2^2/(2ω_sX2) contains no R2. (Key point: Independence) 4. What does raising R2 do? A. Removes the peak B. Raises the peak C. Lowers the peak D. Moves the peak toward s = 1 without changing height Answer: D. Only s_m shifts. 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