Question

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 kW

I 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 1

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.