--- title: "Effect of five-times rotor resistance on 40 kW slip ring motor" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:slip-ring-40kw-rotor-resistance-5x" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/slip-ring-40kw-rotor-resistance-5x site: BCT Engineering --- # Effect of five-times rotor resistance on 40 kW slip ring motor Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2078 Chaitra, 8 marks - 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-40kw-rotor-resistance-5x - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question A 40 kW, 3-phase slip ring induction motor of negligible stator impedance runs at a speed of 0.96 times synchronous speed at rated torque. The slip at maximum torque is 4 times the full-load value. If the rotor resistance of the motor is increased by 5 times, determine: i) The speed, power output and rotor copper loss at rated torque. ii) The speed corresponding to the maximum torque. ## Short answer For constant torque, R2/s stays constant, so new slip = 5×0.04 = 0.2, speed = 0.8Ns, output = 33.33 kW, rotor Cu loss = 8.333 kW. Slip at maximum torque becomes 5×0.16 = 0.8, so speed at Tmax = 0.2Ns. ## Answer With negligible stator impedance, torque depends on $R_2/s$: $$ T = \frac{K s E_2^2 R_2}{R_2^2 + (sX_2)^2} = \frac{K E_2^2 (R_2/s)}{(R_2/s)^2 + X_2^2} $$ So **for the same torque, $R_2/s$ stays constant**, i.e. $s \propto R_2$. Also $s_m = R_2/X_2 \propto R_2$. **Given:** rated output 40 kW at $N = 0.96 N_s$, so $s_1 = 0.04$. $s_{m1} = 4 \times 0.04 = 0.16$. New rotor resistance $R_2' = 5R_2$. Assumption: 40 kW is the mechanical power developed at rated torque (mechanical losses neglected). #### Original condition $$ \begin{aligned} P_{ag} &= \frac{P_m}{1 - s_1} = \frac{40}{0.96} = 41.667\ \text{kW} \\ P_{cu2} &= s_1 P_{ag} = 0.04 \times 41.667 = 1.667\ \text{kW} \end{aligned} $$ #### (i) Rated torque with $5R_2$ $$ \begin{aligned} s_2 &= 5 s_1 = 5 \times 0.04 = 0.20 \\ N_2 &= (1 - 0.20) N_s = 0.80\, N_s \end{aligned} $$ Torque and synchronous speed are unchanged, so air-gap power $P_{ag} = T\omega_s$ is unchanged $= 41.667$ kW. $$ \begin{aligned} P_{out} &= (1 - s_2) P_{ag} = 0.8 \times 41.667 = 33.33\ \text{kW} \\ P_{cu2} &= s_2 P_{ag} = 0.2 \times 41.667 = 8.333\ \text{kW} \end{aligned} $$ #### (ii) Speed at maximum torque $$ \begin{aligned} s_{m2} &= 5 s_{m1} = 5 \times 0.16 = 0.80 \\ N_{m2} &= (1 - 0.80) N_s = 0.20\, N_s \end{aligned} $$ (The maximum torque value itself is unchanged.) | Quantity | Original $R_2$ | $5R_2$ | |---|---|---| | Slip at rated torque | 0.04 | 0.20 | | Speed at rated torque | $0.96N_s$ | $0.80N_s$ | | Power output | 40 kW | 33.33 kW | | Rotor Cu loss | 1.667 kW | 8.333 kW | | Slip at $T_{max}$ | 0.16 | 0.80 | | Speed at $T_{max}$ | $0.84N_s$ | $0.20N_s$ | **Answer: (i) speed $= 0.8N_s$, output $= 33.33$ kW, rotor Cu loss $= 8.333$ kW; (ii) speed at maximum torque $= 0.2N_s$ (slip 0.8).** For example, for a 4-pole 50 Hz machine ($N_s = 1500$ rpm) these speeds would be 1200 rpm and 300 rpm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With negligible stator impedance, what happens to slip for the same torque when R2 is increased? A. Slip stays the same B. Slip is proportional to 1/R2 C. Slip is proportional to R2 because R2/s must stay constant D. Slip is proportional to R2^2 Answer: C. T depends on R2/s, so for unchanged torque s scales with R2. (Key point: Constant torque rule) 2. Original slip is 0.04. What is the speed at rated torque with 5R2? A. 0.96 Ns B. 0.80 Ns C. 0.20 Ns D. 0.84 Ns Answer: B. s2 = 5 x 0.04 = 0.20, so N = 0.8 Ns. (Key point: Slip at rated torque with 5R2) 3. Pag is 41.667 kW and unchanged. What are the output and rotor copper loss at s = 0.2? A. Output 40 kW, loss 8.333 kW B. Output 33.33 kW, loss 8.333 kW C. Output 33.33 kW, loss 1.667 kW D. Output 8.333 kW, loss 33.33 kW Answer: B. Pag = T x ws is unchanged since torque is unchanged; Pout = (1 - 0.2)Pag and Pcu2 = 0.2 Pag. (Key point: Output and rotor loss) 4. Original sm = 0.16. What is the speed at maximum torque with 5R2? A. 0.84 Ns B. 0.16 Ns C. 0.80 Ns D. 0.20 Ns Answer: D. sm is proportional to R2, so sm2 = 0.8 and N = (1 - 0.8)Ns = 0.2 Ns. 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