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Pull out torque of a squirrel cage induction motor occurs at that value of slip where rotor power factor equals
1. 0.5
2. 0.866
3. 0.707
4. 1

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Correct Answer - Option 3 : 0.707

Concept:

Pull-out torque:

  • It is the maximum torque that the synchronous motor can develop without pulling out of step
  • When the synchronous motor is loaded, the rotor falls back by some angle a is called a load angle
  • The stator and rotor magnetic rotate at synchronous speed in spite of some load is applied on the rotor
  • The synchronous motor developed maximum torque when the rotor falls back by an angle of 90°
  • When the load increases beyond its maximum rating, the rotor steps out of synchronism, and synchronous motor stop
     

Explanation:
In the case of the squirrel cage induction motor, we can’t add external resistance as the rotor bars are shorted through end rings.
The condition at which pull out torque achieved when: R2 = sX2
 ∴\(ϕ = {\tan ^{ - 1}}\left( {\frac{{s{X_2}}}{{{R_2}}}} \right) = 45^\circ \)

Power factor = cos ϕ = 0.707

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