The slip of an induction motor normally does not depend on


1️⃣ Rotor speed
2️⃣ Synchronous speed
3️⃣ Shaft torque
4️⃣ Core-loss component


✅ Answer (Detailed Solution Below)

✔️ Option 4: Core-loss component


📖 Detailed Solution


🔷 Definition of Slip

Slip of an induction motor is defined as:

s=NsNrNs

Where:
➤ Ns = Synchronous speed
➤ Nr = Rotor speed

Slip is usually expressed as percentage:

%s=NsNrNs×100

🔍 Dependence of Slip


✔️ 1️⃣ Rotor Speed

From the slip formula:

s(NsNr)

If rotor speed changes → slip immediately changes.

➡ Therefore, slip directly depends on rotor speed.


✔️ 2️⃣ Synchronous Speed

Synchronous speed:

Ns=120fP

Since Ns appears in the slip equation, slip depends on synchronous speed.


✔️ 3️⃣ Shaft Torque

In the normal operating region:

Ts

When shaft load increases:

➤ Rotor slows down
➤ Slip increases
➤ Rotor current increases
➤ Torque increases

Thus, slip is indirectly dependent on shaft torque.


❌ 4️⃣ Core-Loss Component

Core losses consist of:

➤ Hysteresis loss
➤ Eddy current loss

These losses:

✔ Affect efficiency
✔ Affect power factor
✔ Increase heating

❌ Do NOT appear in slip equation
❌ Do NOT control rotor speed directly

Therefore, slip is independent of core-loss component.


📊 Summary Table

ParameterDoes Slip Depend On It?
Rotor speed✅ Yes
Synchronous speed✅ Yes
Shaft torque✅ Yes (indirectly)
Core-loss component❌ No

⭐ Important Exam Concept

✔ Slip depends on relative speed between rotor and stator field
✔ Slip increases with load
✔ Slip is independent of core losses


📌 One-Line Exam Statement ⭐

Slip of an induction motor depends on rotor speed, synchronous speed, and load torque — but not on the core-loss component.


📝 Final Conclusion

Slip does not depend on core-loss component

👉 Correct Answer: Option 4

Post a Comment (0)
Previous Post Next Post