# Induction Motor Problem, How to find Rc

• Engineering

## Homework Statement

A three phase 220 V , 60 Hz, six poles 10 hp induction motor has the following circuit. Parameters on a per basis referred to the stator:
Rs = 0.334 Ω,
R’R = 0.147 Ω,
Xs = 0.498 Ω
X’R = 0.224 Ω ,
Xm = 12.6 Ω
The rotational losses including the core losses amount to 262 W and may be assumed constant. For a slip of 2.8 % determine (a) The line current and the power factor (b) The shaft torque and output horsepower (c) The efficiency

## The Attempt at a Solution

To get the line current, I need the equivalent impedance. But how do I get Rc (core resistance) ?, Is their a way to get it or is it just neglected because it is very large ?

## Answers and Replies

NascentOxygen
Staff Emeritus
Science Advisor
the core losses amount to 262 W and may be assumed constant.
You're wondering how to involve this detail in your calculations?

You're wondering how to involve this detail in your calculations?

For question a), it says determine the line current, so IL = Vphase/Z, but they haven't given Rc. They also said that the core losses and rotational losses amount to 262W

NascentOxygen
Staff Emeritus
Science Advisor
Can you express these fixed losses as their equivalent in line current and then add that in after you have done the other calculations?

N.B., I'm not sure this is the accepted approach, but it seems a valid approximation.

Can you express these fixed losses as their equivalent in line current and then add that in after you have done the other calculations?

N.B., I'm not sure this is the accepted approach, but it seems a valid approximation.

Not sure how to do that, I have this equation

IL = Vphase/Zeq

Where Zeq would be the equivalent impedance of the circuit, but then I would have to neglect Rc

NascentOxygen
Staff Emeritus
Science Advisor
Not sure how to do that, I have this equation

IL = Vphase/Zeq

Where Zeq would be the equivalent impedance of the circuit, but then I would have to neglect Rc
Once you have taken account of Rc losses separately, you can then neglect Rc in the equivalent circuit. Which equivalent circuit are you using?