--- title: "Slip ring motor max-torque slip and full-load output" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:slip-ring-380v-180v-60nm-starting" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/slip-ring-380v-180v-60nm-starting site: BCT Engineering --- # Slip ring motor max-torque slip and full-load output Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2077 Chaitra, 8 marks - Video: [Numerical problems on 3-phase induction motors: Maximum Torque & Corresponding slip](https://www.youtube.com/watch?v=14-83DFgA4Q) — Gargi Basu - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/slip-ring-380v-180v-60nm-starting - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question A 380 V, 4-pole, 50 Hz, 3 phase, slip ring induction motor has a star connected stator winding and a star connected rotor winding. At standstill the voltage between the two slip rings is 180 V. The stator impedance is 0.5+j2.5 ohm. The rotor resistance and reactance at standstill are 0.06 ohm and 0.3 ohm respectively and it develops a starting torque of 60 Nm. Calculate: (i) Slip at which the motor develops the maximum torque. (ii) Torque, power output at full load, given that full load slip is 0.05. ## Short answer Slip at maximum torque is 0.0691. Full-load torque is 375.6 N·m, and power output at 1425 rpm is approximately 56.04 kW. ## Answer **Given:** $V_L = 380$ V (stator star), rotor star with 180 V between slip rings at standstill, $Z_1 = 0.5 + j2.5\ \Omega$, $R_2 = 0.06\ \Omega$, $X_2 = 0.3\ \Omega$, $T_{st} = 60$ N·m, 4 poles, 50 Hz, full-load slip $= 0.05$. **Method:** refer the stator impedance to the rotor side, then use torque ratios (the voltage cancels), with $T_{st}$ as the reference. #### Step 1: Turns ratio and referred impedance Both windings are in star, so the line-voltage ratio equals the phase ratio: $$ \begin{aligned} K &= \frac{180}{380} = 0.47368, \qquad K^2 = 0.22438 \\ R_1' &= K^2 R_1 = 0.22438 \times 0.5 = 0.1122\ \Omega \\ X_1' &= K^2 X_1 = 0.22438 \times 2.5 = 0.5609\ \Omega \\ X &= X_1' + X_2 = 0.5609 + 0.3 = 0.8609\ \Omega \end{aligned} $$ #### (i) Slip at maximum torque $$ \begin{aligned} s_m &= \frac{R_2}{\sqrt{R_1'^2 + X^2}} = \frac{0.06}{\sqrt{0.1122^2 + 0.8609^2}} = \frac{0.06}{0.8682} = 0.0691 \end{aligned} $$ (Corresponding speed $= 1500(1 - 0.0691) \approx 1396$ rpm.) #### (ii) Full-load torque and power Torque at any slip: $$ T = \frac{3}{\omega_s}\cdot\frac{V'^2\,(R_2/s)}{(R_1' + R_2/s)^2 + X^2} $$ At $s = 0.05$: $R_2/s = 0.06/0.05 = 1.2\ \Omega$. $$ \begin{aligned} \frac{T_{fl}}{T_{st}} &= \frac{R_2/s}{R_2}\cdot\frac{(R_1' + R_2)^2 + X^2}{(R_1' + R_2/s)^2 + X^2} \\ &= \frac{1.2}{0.06}\times\frac{(0.1722)^2 + 0.8609^2}{(1.3122)^2 + 0.8609^2} \\ &= 20 \times \frac{0.7709}{2.4631} = 6.259 \end{aligned} $$ $$ T_{fl} = 6.259 \times 60 = 375.6\ \text{N·m} $$ Full-load speed $N = (1 - 0.05) \times 1500 = 1425$ rpm: $$ \begin{aligned} \omega &= \frac{2\pi \times 1425}{60} = 149.23\ \text{rad/s} \\ P_{out} &= T_{fl}\,\omega = 375.6 \times 149.23 = 56\,044\ \text{W} \approx 56.04\ \text{kW} \end{aligned} $$ (Gross mechanical power; mechanical losses are not given.) **Answer: (i) $s_m = 0.0691$; (ii) full-load torque $\approx 375.6$ N·m, power output $\approx 56.04$ kW.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Both windings are star-connected. What is K? A. 180/(380/sqrt3) B. 180/380 = 0.4737 C. 380/180 D. (180 sqrt3)/380 Answer: B. With both in star the line-voltage ratio equals the phase ratio, so K = 180/380. (Key point: Turns ratio for star-star) 2. K^2 = 0.2244. What are R1' and X1' ? A. 0.1122 Ω and 0.5609 Ω B. 0.2369 Ω and 1.184 Ω C. 0.5 Ω and 2.5 Ω D. 2.228 Ω and 11.14 Ω Answer: A. R1' = 0.2244 x 0.5 = 0.1122 and X1' = 0.2244 x 2.5 = 0.5609 Ω. (Key point: Referred stator impedance) 3. Which expression gives sm and what is its value? A. R2/X2 = 0.2 B. R2/sqrt(R1'^2 + X^2) = 0.0691 C. (R1' + R2)/X = 0.2 D. R2/R1' = 0.53 Answer: B. X = X1' + X2 = 0.8609 Ω, so sm = 0.06/0.8682 = 0.0691. (Key point: Slip at maximum torque) 4. How is the full-load torque found from Tst = 60 N-m? A. Tfl = Tst x s B. Tfl = Tst x (R2/s)/R2 x [(R1'+R2)^2 + X^2]/[(R1'+R2/s)^2 + X^2] = 6.259 Tst C. Tfl = Tst/(1 - s) D. Tfl = Tst x sm Answer: B. Taking the ratio of the general torque equation at s and at s = 1 cancels the voltage and gives 6.259, so Tfl = 375.6 N-m. 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