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Question

A 3-phase, 4 pole, 50 Hz induction motor has rotor copper losses equal to 300 W. The torque developed by this motor at 4% slip will be _____ N-m.
  1. 47.75

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Solution

The correct option is A 47.75
We know, Pcu=sPg .... (copper loss in terms of air gap power)

Pg=Pcus=3000.04=7500 W

Now, Pm=Pg(1s)=7500(10.04)=7200 W

Here synchronous speed, Ns=120fP=120×504=1500 rpm

Torque developed = Pgωs=75002π×150060=47.75 Nm

Alternatively, torque can be calculated as

T=Pmωs(1s)=72002π×150060(10.04)
=47.75 Nm

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