The question and the answer is:

enter image description here

However, I think there is a mistake. Below is my solution:

Since ZRotor = 0.3 + j0.35 Ω

R2/s = 0.3 Ω

from the below equation, the maximum slip can be found:

enter image description here

s = 0.474

So, R2/s = 0.3/0.474 = 0.632Ω

To find the current

I = 120/(0.2+j0.25+j0.35+0.632)

The magnitude of the current = 117 A

Using the following equation to find the maximum power:

enter image description here

Pr = 3 * 117^2 * 0.632 = 25.95 kW

Is my approach correct?


1 Answer 1


Your calculation appears to correctly provide the maximum electrical power transferred to the rotor. However 0.474 of that power is dissipated as heat in the rotor and 0.526 of the power is converted to mechanical power. The method that you used is the same as presented by Fitzgerald, Kingsley, Umans, Electric Machinery 4nd ed. It doesn't say why the maximum is based on the maximum power transferred to the rotor rather than the maximum converted to mechanical power. It seems to be assumed to occur at the same slip. That is probably a good assumption for a "normal design" squirrel-cage motor, but it would probably not be a good assumption for a motor with a high rotor resistance.


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