If full load slip is 2%, then starting torque as a percentage of full load torque is 18% of full load torque

Q.  A 3 phase induction motor in a short circuit current is equal 3 times of the full load current. If the full load slip is 2%, then the starting torque as a percentage of full load torque is
- Published on 29 Oct 15

a. 6% of full load torque
b. 18% of full load torque
c. 36% of full load torque
d. None of these

ANSWER: 18% of full load torque

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    Discussion

  • Manashi kalita   -Posted on 18 Mar 19
    Tstart/Tfull= (Ist/Ifl)^2*Sf
    = 3^2*.02
    =.18
    =18% of full load

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