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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 equal to twice the full load torque occurs at the slip of 10% with zero external resistance in the rotor circuit. The slip at the full load torque will be?(a) 2.7%(b) 5%(c) 3.7%(d) 10%The question was asked in an interview for job.My enquiry is from Operating (or Performance) Characteristics of Induction Motors in section Induction Machines of Electrical Machines

Answer» CORRECT answer is (a) 2.7%

To EXPLAIN I would SAY: Tfl/Tem = 2/((sm/sf)+ (sf/sm))

1/2 = 2/((0.1/sf)+(sf/0.1))

s^2-0.4s+0.01=0

s=0.0268.


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