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A 10 kW, 400 V, 3-phase, 4-pole, 50 Hz induction motor develops the rated output at rated voltage with its slip rings shorted. The maximum torque is twice the full load torque occurs at the slip of 10% with zero external resistance in the rotor circuit. The speed at the full load torque will be?(a) 1460 rpm(b) 1400 rpm(c) 1360 rpm(d) 1470 rpmThe question was asked in an interview.Question is taken from Operating (or Performance) Characteristics of Induction Motors topic in section Induction Machines of Electrical Machines

Answer»

The correct CHOICE is (a) 1460 rpm

To EXPLAIN I WOULD say: 1/2 = 2/((0.1/sf)+(sf/0.1))

s = 2.68%

rotor speed = (1-0.0268)=1460rpm.



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