--- title: "Max-torque speed and efficiency of 400 V slip ring motor" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:slip-ring-400v-190v-600w-1380rpm" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/slip-ring-400v-190v-600w-1380rpm site: BCT Engineering --- # Max-torque speed and efficiency of 400 V slip ring motor Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2070 Magh, 8 marks - Video: [(English)EM ||Question 7.7 || Induction Motor || Power|| Torque|| Speed|| Losses|| Efficiency](https://www.youtube.com/watch?v=jz7w_qH0FE0) — 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_machines_i/slip-ring-400v-190v-600w-1380rpm - 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 star 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. The stator to rotor turn ratio is 1:1. The motor consumes 600 watts at no load. Calculate: i) Speed at which the motor develops the maximum torque. ii) Efficiency of the motor when it is running at 1380 rpm. ## Short answer Voltage ratio K = 190/400 = 0.475; rotor referred R2' = 0.2659 Ω, X2' = 1.33 Ω. Max torque slip sm = 0.0689, speed 1397 rpm. At 1380 rpm (s=0.08), input 20.89 kW, output 16.106 kW, efficiency 77.1%. ## Answer **Given:** $V_L = 400$ V (star), 4 poles, 50 Hz, slip-ring voltage at standstill $= 190$ V (star), $Z_1 = 0.5 + j2.5\ \Omega$, $R_2 = 0.06\ \Omega$, $X_2 = 0.3\ \Omega$, no-load input $= 600$ W, $N = 1380$ rpm. **Assumption:** the stated 1:1 turns ratio does not agree with 400 V / 190 V, so the measured voltages are used for the effective ratio $K = 190/400$. Rotor values are referred to the stator, the approximate equivalent circuit is used, and the 600 W no-load loss is taken as constant (core + mechanical) loss. #### Rotor values referred to stator $$ \begin{aligned} K &= \frac{190}{400} = 0.475, \qquad K^2 = 0.225625 \\ R_2' &= \frac{0.06}{0.225625} = 0.2659\ \Omega, \qquad X_2' = \frac{0.3}{0.225625} = 1.3296\ \Omega \\ X &= X_1 + X_2' = 2.5 + 1.3296 = 3.8296\ \Omega \end{aligned} $$ #### (i) Speed at maximum torque $$ \begin{aligned} s_m &= \frac{R_2'}{\sqrt{R_1^2 + X^2}} = \frac{0.2659}{\sqrt{0.25 + 14.666}} = \frac{0.2659}{3.8621} = 0.06886 \\ N_m &= (1 - 0.06886) \times 1500 = 1396.7\ \text{rpm} \end{aligned} $$ #### (ii) Efficiency at 1380 rpm $$ s = \frac{1500 - 1380}{1500} = 0.08, \qquad \frac{R_2'}{s} = \frac{0.2659}{0.08} = 3.324\ \Omega $$ $$ \begin{aligned} V_{ph} &= \frac{400}{\sqrt{3}} = 230.94\ \text{V} \\ Z &= (0.5 + 3.324) + j3.8296 = 3.824 + j3.830 = 5.412\ \Omega \\ I_1 &= \frac{230.94}{5.412} = 42.67\ \text{A} \end{aligned} $$ $$ \begin{aligned} P_{in} &= 3I_1^2(R_1 + R_2'/s) = 3 \times 42.67^2 \times 3.824 = 20\,890\ \text{W} \\ P_{cu1} &= 3 \times 42.67^2 \times 0.5 = 2731\ \text{W} \\ P_{ag} &= 3 \times 42.67^2 \times 3.324 = 18\,158\ \text{W} \\ P_{cu2} &= sP_{ag} = 0.08 \times 18\,158 = 1453\ \text{W} \\ P_m &= 18\,158 - 1453 = 16\,706\ \text{W} \\ P_{out} &= P_m - 600 = 16\,106\ \text{W} \\ \eta &= \frac{16\,106}{20\,890} = 0.771 = 77.1\% \end{aligned} $$ Loss check: $2731 + 1453 + 600 = 4784$ W $= 20\,890 - 16\,106$. **Answer: (i) maximum torque at about 1397 rpm ($s_m = 0.0689$); (ii) efficiency at 1380 rpm $\approx 77.1\%$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this problem the effective turns ratio K is found from: A. K = 190/400, the ratio of measured rotor to stator voltage B. K = 1, from the stated 1:1 ratio C. K = 400/190 D. K = 0.06/0.3 Answer: A. The stated 1:1 ratio contradicts the measured voltages, so K = 190/400 = 0.475 is used. (Key point: Effective ratio) 2. What is s_m and the corresponding speed? A. s_m = 0.0689, about 103 rpm B. s_m = 0.0689, about 1397 rpm C. s_m = 0.2659, about 1101 rpm D. s_m = 0.08, 1380 rpm Answer: B. s_m = R2'/sqrt(R1^2 + X^2) = 0.2659/3.8621 = 0.0689, so N = (1 - 0.0689) x 1500 = 1397 rpm. (Key point: Slip at maximum torque) 3. What is the slip at 1380 rpm and R2'/s? A. s = 0.92 and R2'/s = 0.289 ohm B. s = 0.08 and R2'/s = 0.0213 ohm C. s = 0.08 and R2'/s = 3.324 ohm D. s = 0.12 and R2'/s = 2.2 ohm Answer: C. s = (1500 - 1380)/1500 = 0.08, so R2'/s = 0.2659/0.08 = 3.324 ohm. (Key point: Slip at 1380 rpm) 4. Which losses make up the difference between input and output at 1380 rpm? A. Only stator copper and rotor copper B. Only the 600 W no-load loss C. Rotor copper and core loss, with stator copper ignored D. Stator copper, rotor copper and the 600 W constant loss Answer: D. 2731 + 1453 + 600 = 4784 W = 20 890 - 16 106 W, which gives efficiency 77.1%. 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