# Effect of load on power factor in induction motor

I got stuck in an MCQ question. Which is as following :

Q.1) power factor of a squirrel cage induction motor is :

(a) Low at heavy load only

(b) Low at light and heavy loads

(c) Low at light load only

(d) Low at rated load only

I am getting an answer from what I understand but answer is given some other option. telling which will bias opinions. So I would just like to knoe what your thoughts are. Also not long explanations are required (just tell what would happen to slip and why?)

• Google should be your friend here. Commented Mar 21, 2018 at 14:32
• No, you give us your answer and your understanding behind it, then we can help.
– user16324
Commented Mar 21, 2018 at 14:55
• Understand that real work efficiency (high pf) vs stored ( reactive ) no work is low pf. Commented Mar 21, 2018 at 15:18

## 1 Answer

This kind of question is really not very helpful to the student. The answer depends on how you define "low." If the desired power factor for the power distribution system is 0.95, anything below that could be considered to be "low." By that definition, the power factor of a squirrel cage induction motor is "low" at light and heavy loads. If you compare the power factor at rated load with the power factor at light loads, then the power factor is "low" at light loads only.

The following is a graph of power factor vs. percent of rated load for a particular 3-phase induction motor. The shape of the curve and the power factor at rated load vs. light load for this motor is generally typical.

• I just noticed the parenthetical comment about slip at the end. Slip is proportional to torque (percent load). Why would be the subject of another question.
– user80875
Commented Mar 21, 2018 at 15:28
• Your graph is correct but what I don't get is how slip is decreasing. I mean on one hand we know that p.f=R/[{R^2+(sX)^2}^0.5] which means slip shoud decrease to increase pf. But how can slip decrease with load. I mean isn't slip almost zero at no load conditions? Commented Mar 21, 2018 at 15:31
• I think you must be looking at an over-simplification of the equivalent circuit or something. I can't make any sense out of that. The reactive VA is mostly due to the magnetizing current (Xm branch of eq. ckt.). The stator and rotor leakage reactances (X1 & X2) also contribute. The only thing preventing slip from being zero at no-load is bearing friction and aerodynamic drag (windage). Study and ask individual questions about individual points. I will have time for more later.
– user80875
Commented Mar 21, 2018 at 15:56