--- title: "IM slip and starting current from torque ratios" subject: "Electrical Machine" chapter: 5 key: "semester_4/electrical_machine:im-starting-torque-1-25-max-2-5" source: https://www.bctengineer.com/q/semester_4/electrical_machine/im-starting-torque-1-25-max-2-5 site: BCT Engineering --- # IM slip and starting current from torque ratios Electrical Machine · Chapter 5: Three Phase Induction Machines - Asked in: 2070 Bhadra, 8 marks - Video: [IM18 - ex34p5 problem on ratio of torques](https://www.youtube.com/watch?v=2nuCA57bHLQ) — Pradeep Yemula - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machine/im-starting-torque-1-25-max-2-5 - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-5 ## Question A 3-phase, 50 Hz induction motor has starting torque which is 1.25 times full load torque and a maximum torque which is 2.5 times the full load torque. Neglecting stator resistance and rotational losses and assuming constant rotor resistance. Find i) slip at maximum torque ii) the slip at full load iii) the current at starting in per unit full load current ## Short answer Slip at maximum torque s_m = 0.268. Full-load slip s_fl = 0.0559. Starting current is 4.73 times full-load current. ## Answer **Given:** T_st = 1.25 T_fl, T_max = 2.5 T_fl. Stator resistance and rotational losses neglected, R₂ constant. Using T / T_max = 2 s s_m / (s² + s_m²), where s_m = R₂/X₂. #### (i) Slip at maximum torque At starting s = 1: T_st / T_max = 1.25 / 2.5 = 0.5 = 2 s_m / (1 + s_m²) ```text s_m² - 4 s_m + 1 = 0 s_m = (4 - √12)/2 = 2 - √3 = 0.268 ``` (The root 3.73 is rejected, since s_m is below 1 for a practical motor.) **Slip at maximum torque = 0.268.** #### (ii) Slip at full load T_fl / T_max = 1 / 2.5 = 0.4 = 2 s s_m / (s² + s_m²) ```text 0.4 s² - 2 s_m s + 0.4 s_m² = 0 s² - 5 s_m s + s_m² = 0 s = s_m (5 - √21)/2 = 0.268 × 0.2087 = 0.0559 ``` (The other root, 0.268 × 4.79 = 1.28, is above 1 and rejected.) **Full-load slip = 0.0559 (5.59 %).** #### (iii) Starting current in per unit of full-load current Neglecting stator impedance and magnetising current, I ∝ E₂ / √((R₂/s)² + X₂²) with R₂ = s_m X₂: I ∝ 1 / √((s_m/s)² + 1) (X₂ common) ```text Starting: s = 1 : 1 / √(s_m² + 1) = 1 / 1.0353 = 0.9659 Full load: s = 0.0559 : 1 / √((s_m/s)² + 1) = 1 / √(4.791² + 1) = 1 / 4.894 = 0.2043 I_st / I_fl = 0.9659 / 0.2043 = 4.73 ``` **Starting current = 4.73 times full-load current (4.73 pu).** | Quantity | Result | |---|---| | Slip at T_max | 0.268 | | Full-load slip | 0.0559 | | I_st / I_fl | 4.73 | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which equation links torque and slip, neglecting stator resistance? A. T/Tmax = 2 s sm/(s^2 + sm^2) B. T/Tmax = s/sm C. T/Tmax = s x sm D. T/Tmax = (s^2+sm^2)/(2 s sm) Answer: A. This is the standard torque-slip ratio used for all three parts. (Key point: Torque ratio formula) 2. Tst/Tmax = 0.5 at s = 1 gives which equation and root? A. sm^2 - 4sm + 1 = 0; sm = 0.268 B. sm^2 - 2sm + 1 = 0; sm = 1 C. sm^2 - 4sm + 1 = 0; sm = 3.73 D. sm = 0.5 Answer: A. 2sm/(1+sm^2) = 0.5 gives sm = 2 - sqrt(3) = 0.268; 3.73 is rejected as above 1. (Key point: Slip at maximum torque) 3. Why is the root 1.28 rejected for full-load slip? A. It is below zero B. Slip above 1 is not a motoring operating point C. It is below sm D. It gives zero torque Answer: B. The valid root is 0.0559; the other root exceeds 1 and is not in the motoring range. (Key point: Full-load slip) 4. Starting current in pu of full-load current is found using which relation? A. I proportional to 1/sqrt((sm/s)^2+1), giving 4.73 B. I proportional to s, giving 17.9 C. I proportional to torque, giving 1.25 D. I equals Tst/Tfl Answer: A. I is proportional to E2/sqrt((R2/s)^2+X2^2); the ratio 0.9659/0.2043 = 4.73. (Key point: Starting current ratio) ## Related in Electrical Machine - [Slip at max torque and full load, starting current](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-slip-torque-ratios) — Electrical Machines, chapter 5 - [Slip for maximum torque and full-load output of slip ring motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aslip-ring-400v-delta-star-190v) — Electrical Machines I, chapter 5, asked 2× - [Starting, no-load and full-load current of slip ring motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aslip-ring-0p8-4ohm-400v-ratio4) — Electrical Machines I, chapter 5 - [Slip ring motor max-torque slip and full-load output](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aslip-ring-380v-180v-60nm-starting) — Electrical Machines I, chapter 5, asked 4× - [Slip ring motor: starting, no-load, full-load currents](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-4pole-slipring-1490-1300-currents) — Electrical Machines, chapter 5 - [Starting current and rotor resistance for 150 kW slip ring motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-150kw-3000v-starting-current-30a) — Electrical Machines I, chapter 5 ## Related in other subjects - [Single phase IM principle, torque-slip and starting](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asingle-phase-principle-torque-slip) — Electric Machines II, chapter 3 - [Induction motor plugging: braking current and torque](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ainduction-motor-plugging-braking-current) — Utilization of Electrical Energy, chapter 3 - [Slip ring motor: speed at max torque, resistance](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-slipring-max-torque-ratio) — Electrical Machines, chapter 5 - [6-pole induction motor: slip, losses, efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-4pole-975rpm-slip-losses) — Electrical Machines, chapter 5