# Induction motor rotor current

I don't want a solution; the solution is already given.

A 575 V, 60 Hz, 150 hp, 0.80 power factor lagging, 2-pole, 3-phase induction motor has the following equivalent circuit parameters as seen from the stator side:

Core losses = 600 W, friction/windage losses = 300 W, stray losses = 100 W The line current drawing by this induction motor is 150 A.

The rotor current of the given induction motor is 41 A by using the below formula as solved by the instructor:

However, if I try using the equivalent circuit in the figure below to find the rotor current:

Where is the mistake in the second approach?

• You need to show your working. Commented Jan 6, 2022 at 10:31
• @Andy aka Done.
– OMAR
Commented Jan 6, 2022 at 11:46
• In the statement of the problem, I believe that "150 hp" should be "150 A." The problem contains no value or basis for calculating Pcore, AKA Piron. I can revise the question to show the equivalent circuit if you are unable to do that. More later.
– user80875
Commented Jan 6, 2022 at 15:30
• I think both the first and the second approach are correct. You have been given the input voltage current and power factor. You have also been given the motor equivalent circuit and iron losses. Those two sets of parameters should agree with each other. I have not done the complete calculation of the first set from the second set, but I think they must not agree with each other. Also an induction motor with 29% stator copper and iron losses is an extraordinarily inefficient motor. An induction motor with only 1.5% FL slip is quite a bit more efficient than average in that respect.
– user80875
Commented Jan 6, 2022 at 17:38
• @ Andy aka: Here is the equivalent circuit in question with some differences in labeling and including the iron loss branch that is mentioned in my comment. i.sstatic.net/fu9Ud.jpg
– user80875
Commented Jan 6, 2022 at 17:56