A 6-pole, 3-phase alternator running at 1000 rpm supplies power to an 8-pole, 3-phase induction motor. The rotor current frequency is 2 Hz. The speed at which the motor operates is:

A) 1000 rpm
B) 960 rpm
C) 750 rpm
D) 720 rpm

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✅ Answer:

➡️ D) 720 rpm

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🧠 Detailed Solution

◆ Step 1: Find Supply Frequency (from Alternator)

$f=\frac{PN}{120}$

$$f=\frac{6\times 1000}{120}=50\text{Hz}$$

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◆ Step 2: Synchronous Speed of Induction Motor

$$N_s=\frac{120\times 50}{8}=750\text{rpm}$$

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◆ Step 3: Calculate Slip

$$s=\frac{f_r}{f}=\frac{2}{50}=0.04=4\mathrm{\%}$$

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◆ Step 4: Rotor Speed

$$N_r=N_s(1-s)=750(1-0.04)$$$$N_r=750\times 0.96=720\text{rpm}$$

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❌ Why Other Options Are Incorrect

◆ A) 1000 rpm → Alternator speed, not motor speed
◆ B) 960 rpm → Not possible (greater than synchronous for this motor)
◆ C) 750 rpm → This is synchronous speed, not actual speed

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🔑 Key Points to Remember

◆ Alternator frequency: $f=\frac{PN}{120}$
◆ Rotor frequency relation: $f_r=sf$
◆ Motor speed: $N_r=N_s(1-s)$
◆ Rotor always runs below synchronous speed

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🎯 Final Conclusion

➡️ Motor operates at:
720 rpm ✔️