An 8-pole alternator runs at 750 rpm. It supplies power to a 6-pole induction motor which has a full-load slip of 3%. The full-load speed of the motor is

February 13, 2026

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1️⃣ 1,050 rpm

2️⃣ 970 rpm
3️⃣ 960 rpm
4️⃣ 1,250 rpm

✅ Answer (Detailed Solution Below)

✔️ Option 2: 970 rpm

📖 Detailed Solution

🔷 Step 1: Calculate Frequency of Alternator

Frequency formula for alternator:

f = \frac{P N}{120}

Where:
➤ P = 8 poles
➤ N = 750 rpm

f = \frac{8 \times 750}{120}

f = \frac{6000}{120}

f = 50 Hz

✔ Supply frequency = 50 Hz

🔷 Step 2: Calculate Synchronous Speed of Motor

Motor is 6-pole.

N_{s} = \frac{120 f}{P}

N_{s} = \frac{120 \times 50}{6}

N_{s} = 1000 rpm

✔ Synchronous speed of motor = 1000 rpm

🔷 Step 3: Calculate Full-Load Speed

Slip formula:

N_{r} = N_{s} (1 - s)

Given:

s = 3 % = 0.03

N_{r} = 1000 (1 - 0.03)

N_{r} = 1000 \times 0.97

N_{r} = 970 rpm

🎯 Final Full-Load Speed = 970 rpm

❌ Why Other Options Are Incorrect

❌ 1️⃣ 1,050 rpm

Speed cannot exceed synchronous speed (1000 rpm).

❌ 3️⃣ 960 rpm

This corresponds to 4% slip, not 3%.

❌ 4️⃣ 1,250 rpm

Not possible for 6-pole, 50 Hz motor.

⭐ Important Exam Points

✔ Alternator frequency: $f = \frac{P N}{120}$
✔ Motor synchronous speed: $N_{s} = \frac{120 f}{P}$
✔ Rotor speed: $N_{r} = N_{s} (1 - s)$
✔ Rotor always runs below Ns

📌 Key One-Line Exam Statement ⭐

An 8-pole alternator running at 750 rpm generates 50 Hz supply, and a 6-pole induction motor with 3% slip runs at 970 rpm.

📝 Final Conclusion

Alternator frequency = 50 Hz
Motor synchronous speed = 1000 rpm
With 3% slip:

F u l l - l o a d s p e e d = 970 r p m

👉 Correct Answer: Option 2 (970 rpm)

Tags: Induction Machine MCQ