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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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