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A 10kW, 50 Hz, 3-phase induction motor develops the rated torque at 1440rpm. If the load torque is reduced to half, then the power output that can now be obtained is?(a) 5 kW(b) 5.3 KW(c) 4.6 kW(d) 8 kWI have been asked this question by my school teacher while I was bunking the class.I want to ask this question from Circle Diagrams of Induction Machines in portion Induction Machines of Electrical Machines

Answer» RIGHT option is (a) 5 kW

To EXPLAIN I WOULD SAY: S2 = (1/2)*0.04 = 0.02

P = (10000*60*2*3.14*1470)/(2*2*3.14*1460*60)

P = 5.03 kW.


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