--- title: "6-pole IM: torque at 4% slip" subject: "Electrical Machine" chapter: 5 key: "semester_4/electrical_machine:im-6pole-max-torque-200nm-850rpm" source: https://www.bctengineer.com/q/semester_4/electrical_machine/im-6pole-max-torque-200nm-850rpm site: BCT Engineering --- # 6-pole IM: torque at 4% slip Electrical Machine · Chapter 5: Three Phase Induction Machines - Asked in: 2073 Bhadra, 8 marks - Video: [(B)EM (5th ed) || Example 6.5 || Slip, Power, and Torque of Induction Motor](https://www.youtube.com/watch?v=z9j23rttgWY) — Electrical Engineering Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machine/im-6pole-max-torque-200nm-850rpm - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-5 ## Question A 6-pole, 50 Hz, three-phase induction motor has rotor resistance of 0.4 Ω/phase, maximum torque is 200 Nm at 850 rpm. Find (i) torque at 4% slip, and (ii) additional rotor resistance to get (2/3)rd of maximum torque at starting. ## Short answer Torque at 4% slip: T = 200 × 2×0.04×0.15/(0.04²+0.15²) = 99.6 N·m. For starting torque 2/3 of maximum, added rotor resistance = 0.619 Ω per phase (using s_m'=0.382, R=1.019 Ω). ## Answer **Given:** 6-pole, 50 Hz; R₂ = 0.4 Ω/phase; T_max = 200 N·m at 850 rpm. N_s = 120 × 50 / 6 = 1000 rpm. Slip at maximum torque: s_m = (1000 − 850)/1000 = 0.15. Since s_m = R₂/X₂, standstill reactance X₂ = R₂ / s_m = 0.4 / 0.15 = 2.667 Ω/phase. #### (i) Torque at 4 % slip Neglecting stator impedance, T / T_max = 2 s s_m / (s² + s_m²) ```text T/Tmax = 2 × 0.04 × 0.15 / (0.04² + 0.15²) = 0.0120 / (0.0016 + 0.0225) = 0.0120 / 0.0241 = 0.4979 T = 0.4979 × 200 = 99.6 N·m ``` **Torque at 4 % slip ≈ 99.6 N·m.** #### (ii) Extra rotor resistance for starting torque = (2/3) T_max At starting s = 1, with total rotor resistance R and slip at maximum torque s_m' = R / X₂: T_st / T_max = 2 s_m' / (1 + s_m'²) = 2/3 ```text 3 s_m' = 1 + s_m'² -> s_m'² - 3 s_m' + 1 = 0 s_m' = (3 ± √5)/2 = 0.382 or 2.618 ``` Taking the practical value s_m' = 0.382 (the peak occurs at a speed within the running range, and the smaller resistance is needed): R = s_m' X₂ = 0.382 × 2.667 = 1.019 Ω/phase External resistance per phase = 1.019 − 0.4 = **0.619 Ω**. (The other root, 2.618, would need R = 6.98 Ω, i.e. 6.58 Ω extra. It gives the same starting torque but is not used because it wastes more power and the maximum torque would then lie in the reverse region.) **Answer: (i) 99.6 N·m; (ii) about 0.62 Ω per phase added in the rotor circuit.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Ns = 1000 rpm and maximum torque occurs at 850 rpm. What is sm? A. 0.85 B. 0.15 C. 0.015 D. 0.4 Answer: B. sm = (1000 - 850)/1000 = 0.15. (Key point: Slip at maximum torque) 2. How is X2 found? A. X2 = R2 x sm = 0.06 ohm B. X2 = R2/sm = 0.4/0.15 = 2.667 ohm C. X2 = sm/R2 D. X2 = Tmax/R2 Answer: B. Since sm = R2/X2, X2 = R2/sm = 2.667 ohm. (Key point: Standstill reactance) 3. Which formula gives torque at slip s? A. T/Tmax = 2 s sm/(s^2 + sm^2), giving 99.6 N·m B. T = Tmax x s C. T/Tmax = s/sm, giving 53.3 N·m D. T/Tmax = (s^2+sm^2)/(2 s sm) Answer: A. T/Tmax = 0.012/0.0241 = 0.498, so T = 99.6 N·m. (Key point: Torque at 4 % slip) 4. For Tst = (2/3)Tmax, what equation is solved? A. 2 sm'/(1 + sm'^2) = 2/3, giving sm' = 0.382 B. sm' = 2/3 C. sm'^2 = 2/3 D. 2 sm' = 1 + sm'^2 Answer: A. At s = 1, the equation reduces to sm'^2 - 3sm' + 1 = 0; the practical root 0.382 gives R = 1.019 ohm, so 0.619 ohm extra. 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