--- title: "Starting, no-load and full-load current of slip ring motor" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:slip-ring-0p8-4ohm-400v-ratio4" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/slip-ring-0p8-4ohm-400v-ratio4 site: BCT Engineering --- # Starting, no-load and full-load current of slip ring motor Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2073 Bhadra, 8 marks - Video: [Simple problems on 3-phase induction motors: Slip, Rotor Frequency(Part-I)](https://www.youtube.com/watch?v=H0SaAq7sMN0) — 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-0p8-4ohm-400v-ratio4 - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question A 4-pole, 50 Hz 3-φ slip ring induction motor has star connected stator and rotor windings. The rotor winding has resistance 0.8 Ω and reactance of 4 Ω per phase at standstill. The emf induced between slip rings at standstill is 400 V. The stator to rotor turn ratio is 4. The motor runs at 1490 rpm at no-load and 1300 rpm at full-load. Calculate: i) Starting current ii) No-load current iii) Full load current ## Short answer Slips: start s=1, no-load (1490 rpm) s=0.00667, full-load (1300 rpm) s=0.1333. Rotor phase EMF = 400/√3 V. Rotor currents: 56.6 A start, 1.92 A no-load, 32.0 A full-load. Stator current (ratio 4:1): 14.15 A, 0.48 A, 8.01 A. ## Answer **Given:** 4 poles, 50 Hz, star stator and rotor, $R_2 = 0.8\ \Omega$, $X_2 = 4\ \Omega$ per phase at standstill, emf between slip rings at standstill $= 400$ V, stator/rotor turns ratio $= 4$, $N_0 = 1490$ rpm, $N_{fl} = 1300$ rpm. Rotor emf per phase at standstill: $$ E_2 = \frac{400}{\sqrt{3}} = 230.94\ \text{V}, \qquad N_s = \frac{120 \times 50}{4} = 1500\ \text{rpm} $$ Rotor current at slip $s$: $$ I_2 = \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}} $$ Stator current (load component, neglecting no-load current) $I_1 \approx I_2/4$. #### (i) Starting current ($s = 1$) $$ \begin{aligned} I_{2,st} &= \frac{230.94}{\sqrt{0.8^2 + 4^2}} = \frac{230.94}{4.0792} = 56.61\ \text{A} \\ I_{1,st} &\approx \frac{56.61}{4} = 14.15\ \text{A} \end{aligned} $$ #### (ii) No-load current ($N = 1490$ rpm) $$ \begin{aligned} s_0 &= \frac{1500 - 1490}{1500} = 0.006667 \\ I_{2,0} &= \frac{0.006667 \times 230.94}{\sqrt{0.8^2 + (0.006667 \times 4)^2}} = \frac{1.5396}{0.80044} = 1.923\ \text{A} \\ I_{1,0} &\approx \frac{1.923}{4} = 0.481\ \text{A} \end{aligned} $$ #### (iii) Full-load current ($N = 1300$ rpm) $$ \begin{aligned} s_{fl} &= \frac{1500 - 1300}{1500} = 0.13333 \\ I_{2,fl} &= \frac{0.13333 \times 230.94}{\sqrt{0.8^2 + (0.13333 \times 4)^2}} = \frac{30.792}{0.96148} = 32.03\ \text{A} \\ I_{1,fl} &\approx \frac{32.03}{4} = 8.01\ \text{A} \end{aligned} $$ | Condition | Slip | Rotor current | Stator (load) current | |---|---|---|---| | Starting | 1 | 56.61 A | 14.15 A | | No load | 0.00667 | 1.923 A | 0.481 A | | Full load | 0.1333 | 32.03 A | 8.01 A | **Answer (rotor currents): starting 56.61 A, no load 1.92 A, full load 32.03 A** (stator side $\approx$ 14.15 A, 0.48 A and 8.01 A, ignoring the magnetizing current). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The emf between slip rings is 400 V with a star rotor. What is E2 per phase? A. 400 V B. 230.94 V C. 692.8 V D. 133.3 V Answer: B. E2 = 400/sqrt3 = 230.94 V. (Key point: Rotor emf per phase) 2. Which formula gives the rotor current at slip s? A. I2 = E2/sqrt(R2^2 + X2^2) at all slips B. I2 = s E2/sqrt(R2^2 + (sX2)^2) C. I2 = E2/(R2 + s X2) D. I2 = s E2/R2^2 Answer: B. At slip s the rotor emf is sE2 and reactance sX2, so I2 = sE2/sqrt(R2^2+(sX2)^2). (Key point: Rotor current formula) 3. For Ns = 1500 rpm, what are the no-load and full-load slips at 1490 and 1300 rpm? A. 0.0067 and 0.1333 B. 0.0133 and 0.1 C. 0.0067 and 0.2 D. 0.01 and 0.133 Answer: A. 10/1500 = 0.00667 and 200/1500 = 0.1333. (Key point: Slip values) 4. How is the stator current found from the rotor current, with turns ratio 4? A. I1 = 4 I2 B. I1 = I2/4 C. I1 = I2/16 D. I1 = I2 Answer: B. Currents transform inversely to turns, so I1 = I2/4 (starting 14.15 A). 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