A 500 hp, 6-pole, 3-phase, 440 V, 50 Hz induction motor runs at 950 rpm at full load. The full-load slip and the number of cycles the rotor voltage makes per minute are:

A) 10% and 150
B) 10% and 125
C) 5% and 150
D) 5% and 125

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✅ Answer:

➡️ C) 5% and 150

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🧠 Detailed Solution

◆ Step 1: Synchronous Speed

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

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

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◆ Step 2: Full-load Slip

$$s=\frac{N_s-N_r}{N_s}=\frac{1000-950}{1000}=0.05=5\mathrm{\%}$$

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◆ Step 3: Rotor Frequency

$$f_r=s\times f=0.05\times 50=2.5\text{Hz}$$

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◆ Step 4: Rotor Voltage Cycles per Minute

➡️ Rotor voltage completes $f_r$ cycles per second

$$\text{Cycles per minute}=f_r\times 60=2.5\times 60=150$$

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❌ Why Other Options Are Incorrect

◆ A) 10%, 150 → Slip is wrong
◆ B) 10%, 125 → Both values incorrect
◆ D) 5%, 125 → Rotor cycles calculation wrong

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🔑 Key Points to Remember

◆ Slip: $s=\frac{N_s-N_r}{N_s}$
◆ Rotor frequency: $f_r=sf$
◆ Rotor cycles/min = $f_r\times 60$
◆ Rotor frequency is always very low at full load

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🎯 Final Conclusion

➡️ Full-load slip = 5%
➡️ Rotor voltage cycles = 150 cycles/min ✔️