A 3-phase, 4-pole, self-excited induction generator feeds a load at frequency f1 ​. When the load is partially removed, the frequency becomes f2 . The speed is maintained at 1500 rpm in both cases. Which statement is correct?

A) $f_1,f_2>50$ and $f_1>f_2$
B) $f_1<50$ and $f_2>50$
C) $f_1,f_2<50$ and $f_2>f_1$
D) $f_1>50$ and $f_2<50$

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✅ Answer:

➡️ C) $f_1,f_2<50$ and $f_2>f_1$

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🧠 Detailed Solution

◆ Step 1: Synchronous Speed

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

For 4 poles and 50 Hz:

$$N_s=\frac{120\times 50}{4}=1500\text{rpm}$$

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◆ Step 2: Key Concept of Induction Generator

➡️ For generator operation:

$$N>N_s$$

➡️ But here speed is fixed at 1500 rpm ≈ synchronous speed

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◆ Step 3: Important Insight (Very Critical 🚨)

➡️ In self-excited induction generator:

• Frequency depends on load + excitation (capacitors)

• Not strictly fixed like grid supply

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◆ Step 4: Effect of Load Reduction

→ Load ↓
→ Current drawn ↓
→ Magnetizing current ↓
→ Air-gap flux ↓

➡️ Slip magnitude decreases

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◆ Step 5: Frequency Relation

$$f\propto \text{air-gap flux and slip interaction}$$

➡️ With reduced load:
→ Slip reduces
→ Frequency increases

So:

$$f_2>f_1$$

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◆ Step 6: Absolute Value of Frequency

➡️ Since excitation is self-generated,
➡️ Frequency is generally less than rated (50 Hz)

$$f_1,f_2<50$$

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❌ Why Other Options Are Incorrect

◆ A) Both > 50 → Not possible in SEIG
◆ B) One <50, one >50 → Not consistent
◆ D) Opposite trend → Wrong

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🔑 Key Points to Remember

◆ SEIG frequency is not constant
◆ Depends on load and excitation
◆ Load ↓ ⇒ Slip ↓ ⇒ Frequency ↑
◆ Frequency generally < rated (50 Hz)

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🎯 Final Conclusion

➡️ Both frequencies are below 50 Hz and increase when load reduces

✔️ $f_1,f_2<50$ and $f_2>f_1$