The rotor frequency of a 3-phase, 5 kW, 400 V, 50 Hz, 4-pole slip-ring induction motor is 25 Hz. The speed of the motor when connected to 400 V, 50 Hz supply will be:


A) 1,500 rpm
B) 1,000 rpm
C) 750 rpm
D) Zero

✅ Answer:
C) 750 rpm

🧠 Detailed Solution

🔹 1. Given Data:
Supply frequency → f = 50 Hz
Rotor frequency → fr = 25 Hz
Number of poles → P = 4


🔹 2. Calculate Synchronous Speed (Ns):

Ns = (120 × f) / P

Ns = (120 × 50) / 4 = 1500 rpm


🔹 3. Calculate Slip (s):

fr = s × f

s = fr / f = 25 / 50 = 0.5

👉 Slip = 50%


🔹 4. Calculate Motor Speed (N):

N = Ns (1 − s)

N = 1500 × (1 − 0.5) = 1500 × 0.5 = 750 rpm


❌ Why Other Options Are Wrong

A) 1500 rpm ❌ → This is synchronous speed (only at zero slip)
B) 1000 rpm ❌ → Incorrect slip assumption
D) Zero ❌ → Motor is running, not at standstill


📌 Key Points

• Rotor frequency = Slip × Supply frequency
• Slip = Rotor frequency / Supply frequency
• Higher rotor frequency → Higher slip → Lower speed
• At slip = 0.5 → Speed becomes half of synchronous speed


🎯 Final Conclusion

✔ Motor speed = 750 rpm
✔ Correct Option → C

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