If the full-load speed of a 6-pole, 50 Hz induction motor is 950 rpm, what is its half-load speed nearly equal to?

A) 1000 rpm
B) 450 rpm
C) 1900 rpm
D) 975 rpm

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✅ Answer:

➡️ D) 975 rpm

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🧠 Detailed Solution

◆ Step 1: Find Synchronous Speed

$N_s=\frac{120f}{P}$

$$N_s=\frac{120\times 50}{6}=1000\text{rpm}$$

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◆ Step 2: Calculate Full-load Slip

$$s_{FL}=\frac{N_s-N}{N_s}=\frac{1000-950}{1000}=0.05(5\mathrm{\%})$$

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◆ Step 3: Key Concept (Very Important 🚨)

➡️ In induction motor:
Slip ∝ Load torque (approximately)

So at half load:

$$s_{HL}\approx \frac{1}{2}\times s_{FL}=0.025$$

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◆ Step 4: Find Half-load Speed

$$N_{HL}=N_s(1-s_{HL})=1000\times (1-0.025)$$$$N_{HL}=975\text{rpm}$$

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❌ Why Other Options Are Incorrect

◆ A) 1000 rpm → Only at no-load (slip ≈ 0)
◆ B) 450 rpm → Completely unrealistic
◆ C) 1900 rpm → Greater than synchronous speed (impossible)

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🔑 Key Points to Remember

◆ Induction motor always runs slightly below synchronous speed
◆ Slip is directly proportional to load (approx.)
◆ Half load ⇒ Slip halves ⇒ Speed increases slightly
◆ Speed change is small, not drastic

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🎯 Final Conclusion

➡️ At half load, slip reduces → speed increases slightly

Half-load speed ≈ 975 rpm ✔️