The maximum possible speed of a 3-phase squirrel cage induction motor running at a slip of 4% is

1️⃣ 2,880 rpm
2️⃣ 3,000 rpm
3️⃣ 1,440 rpm
4️⃣ 960 rpm

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✅ Answer (Detailed Solution Below)

✔️ Option 1: 2,880 rpm

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📖 Detailed Solution

🔹 Step 1: Understand Maximum Possible Speed

The maximum possible rotor speed occurs when:

✔️ Supply frequency is standard (50 Hz)
✔️ Number of poles is minimum

The smallest practical number of poles in a 3-phase motor is:

$$P=2$$

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🔹 Step 2: Calculate Synchronous Speed

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

Substitute:

$$N_s=\frac{120\times 50}{2}$$$$N_s=3000\text{ rpm}$$

So, maximum synchronous speed = 3000 rpm

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🔹 Step 3: Use Slip Formula

Slip is defined as:

$$s=\frac{N_s-N_r}{N_s}$$

Given:

$$s=4\mathrm{\%}=0.04$$

Rearranging:

$$N_r=N_s(1-s)$$$$N_r=3000(1-0.04)$$$$N_r=3000\times 0.96$$$$N_r=2880\text{ rpm}$$

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🎯 Final Rotor Speed = 2,880 rpm

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❌ Why Other Options Are Incorrect

❌ 2️⃣ 3,000 rpm

🔹 3000 rpm is synchronous speed
🔹 Rotor never reaches synchronous speed
🔹 Slip would be zero → No torque

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❌ 3️⃣ 1,440 rpm

🔹 This corresponds to 4-pole motor at 50 Hz with slip
🔹 Not the maximum possible speed

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❌ 4️⃣ 960 rpm

🔹 This corresponds to 6-pole motor
🔹 Lower synchronous speed

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⭐ Important Exam Points

✔️ Maximum synchronous speed occurs with minimum poles
✔️ Rotor speed = Ns(1 − s)
✔️ Rotor can never reach Ns
✔️ Slip must exist for torque production

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📌 Key One-Line Exam Statement ⭐

For a 50 Hz, 2-pole motor running at 4% slip, the maximum possible rotor speed is 2,880 rpm.

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📝 Final Conclusion

📍 The maximum possible speed of a 3-phase squirrel cage induction motor with 4% slip occurs in a 2-pole, 50 Hz motor, giving a rotor speed of:

$$\boxed{{\displaystyle 2880\text{ rpm}}}$$

👉 Correct Answer: 2,880 rpm (Option 1)