--- title: "Slip for maximum torque and full-load output of slip ring motor" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:slip-ring-400v-delta-star-190v" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/slip-ring-400v-delta-star-190v site: BCT Engineering --- # Slip for maximum torque and full-load output of slip ring motor Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2078 Baisakh, 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-400v-delta-star-190v - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question A 400 V, 4-pole, 50 Hz, 3 phase, slip ring induction motor has a delta connected stator winding and a star connected rotor winding. At standstill the voltage between the two slip rings is 190 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 maximum torque of 150 N-m. Calculate: (i) Slip at which motor develops the maximum torque. (ii) Torque, power output at full load, given that full load slip is 0.04. ## Short answer Slip at maximum torque is 0.1226. Full-load torque is 91.14 N·m, and power output at 1440 rpm is approximately 13.74 kW. ## Answer **Given:** $V_L = 400$ V, stator in **delta** (so stator phase voltage $= 400$ V), rotor in star with 190 V between slip rings at standstill, $Z_1 = 0.5 + j2.5\ \Omega$, $R_2 = 0.06\ \Omega$, $X_2 = 0.3\ \Omega$, $T_{max} = 150$ N·m, 4 poles, 50 Hz, full-load slip $= 0.04$. #### Step 1: Turns ratio and stator impedance referred to rotor $$ \begin{aligned} E_{2,ph} &= \frac{190}{\sqrt{3}} = 109.70\ \text{V}, \qquad E_{1,ph} = 400\ \text{V} \\ K &= \frac{109.70}{400} = 0.27424, \qquad K^2 = 0.075208 \\ R_1' &= K^2 R_1 = 0.075208 \times 0.5 = 0.0376\ \Omega \\ X_1' &= K^2 X_1 = 0.075208 \times 2.5 = 0.1880\ \Omega \\ X &= X_1' + X_2 = 0.1880 + 0.3 = 0.4880\ \Omega \end{aligned} $$ #### (i) Slip at maximum torque $$ \begin{aligned} s_m &= \frac{R_2}{\sqrt{R_1'^2 + X^2}} = \frac{0.06}{\sqrt{0.0376^2 + 0.4880^2}} = \frac{0.06}{0.48947} = 0.1226 \end{aligned} $$ (Speed at maximum torque $= 1500(1 - 0.1226) \approx 1316$ rpm.) #### (ii) Full-load torque and power $$ T = \frac{3}{\omega_s}\frac{V'^2 (R_2/s)}{(R_1' + R_2/s)^2 + X^2}, \qquad T_{max} = \frac{3}{\omega_s}\frac{V'^2}{2\left(R_1' + \sqrt{R_1'^2 + X^2}\right)} $$ At $s = 0.04$: $R_2/s = 0.06/0.04 = 1.5\ \Omega$. $$ \begin{aligned} \frac{T_{fl}}{T_{max}} &= \frac{(R_2/s)\cdot 2\left(R_1' + \sqrt{R_1'^2 + X^2}\right)}{(R_1' + R_2/s)^2 + X^2} \\ &= \frac{1.5 \times 2(0.0376 + 0.48947)}{(1.5376)^2 + 0.4880^2} \\ &= \frac{1.5 \times 1.05414}{2.6024} = 0.6076 \end{aligned} $$ $$ T_{fl} = 0.6076 \times 150 = 91.14\ \text{N·m} $$ Full-load speed $N = (1 - 0.04) \times 1500 = 1440$ rpm: $$ \begin{aligned} \omega &= \frac{2\pi \times 1440}{60} = 150.80\ \text{rad/s} \\ P_{out} &= T_{fl}\,\omega = 91.14 \times 150.80 = 13\,744\ \text{W} \approx 13.74\ \text{kW} \end{aligned} $$ (This is the gross mechanical power developed; mechanical losses are not given.) **Answer: (i) $s_m = 0.1226$; (ii) full-load torque $= 91.14$ N·m, power output $\approx 13.74$ kW.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Stator is delta at 400 V and rotor star with 190 V between slip rings. What is K (rotor/stator phase voltage ratio)? A. 190/400 = 0.475 B. (190/sqrt3)/400 = 0.2742 C. 190/(400 sqrt3) x sqrt3 = 0.475 D. 400/(190/sqrt3) = 3.65 Answer: B. Delta stator phase voltage is 400 V and star rotor phase emf is 190/sqrt3 = 109.7 V, so K = 0.2742. (Key point: Turns ratio) 2. How is the stator impedance referred to the rotor? A. Multiply R1 and X1 by K B. Divide by K C. Multiply by K^2 D. Leave unchanged Answer: C. Impedance transforms with the square of the turns/voltage ratio, so R1' = K^2 R1 and X1' = K^2 X1. (Key point: Referring stator impedance) 3. With stator impedance included, which formula gives sm? A. sm = R2/X2 B. sm = R2/sqrt(R1'^2 + (X1' + X2)^2) C. sm = R2/(R1' + X1') D. sm = (R1' + R2)/X Answer: B. The stator impedance adds to the rotor, giving sm = R2/sqrt(R1'^2 + X^2) = 0.1226. (Key point: Slip at maximum torque) 4. How is the full-load torque found using Tmax = 150 N-m? A. Tfl = Tmax x (1 - s) B. Tfl = Tmax x s C. Tfl = Tmax/2 D. Tfl = Tmax x ratio of the torque expressions at s = 0.04 to the Tmax expression (0.6076) Answer: D. The common factor 3V'^2/ws cancels in the ratio, giving Tfl/Tmax = 0.6076 and Tfl = 91.14 N-m. 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