A 6-pole, 50 Hz wound rotor induction motor develops 100 V across slip rings at rated supply (open-circuited rotor). If the rotor is driven at 1000 rpm opposite to stator field, the frequency of voltage across slip rings is:


A) zero
B) 50 Hz
C) 100 Hz
D) 200 Hz


Answer:

➡️ C) 100 Hz


🧠 Detailed Theory + Solution


◆ Step 1: Synchronous Speed


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

◆ Step 2: Rotor Speed Condition

➡️ Rotor is driven at 1000 rpm in opposite direction

So:


N = -1000 \, \text{rpm}

◆ Step 3: Slip Calculation


s = \frac{N_s - N}{N_s}
= \frac{1000 - (-1000)}{1000}
= \frac{2000}{1000} = 2

◆ Step 4: Rotor Frequency

➡️ Rotor (slip-ring) frequency is:


f_r = s f

f_r = 2 \times 50 = 100 \, \text{Hz}

◆ Step 5: Key Insight 🚨

➡️ Rotor running opposite → relative speed doubles
➡️ Slip becomes greater than 1 (s = 2)
➡️ Rotor frequency increases accordingly


Why Other Options Are Incorrect

◆ A) zero → Only at synchronous speed same direction
◆ B) 50 Hz → Only at standstill
◆ D) 200 Hz → Overestimation (double counting error)


🔑 Key Points to Remember

◆ Rotor frequency:
◆ Opposite rotation ⇒ slip can be > 1
◆ At standstill →
◆ At synchronous speed →


🎯 Final Conclusion

➡️ Slip = 2 due to opposite rotation
➡️ Rotor frequency = 100 Hz ✔️

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