Induction Machines - Power + Efficiency, Non-Sinusodial Current

Thread Starter


Joined Sep 13, 2010
Hello everyone,

This is from a lab which we recently preformed at school and I'm confused about a few parts.

The first question relates to the first figure. (See figure attached)

When we preformed Experiment 2 and observed the current and voltage waveforms on the scope the current waveform was non-sinusoidal, but I'm not entirely sure why.

Is it because the current going through the core shunt branch is about 40-50% of the rated current? Could this be why it's magnitude is almost 50%?

My second question relates to power and efficiency. (See second figure attached)

The first thing I am unsure of is how I calculated the stator copper losses.

I wrote,

\(P_{cu}(S) = 3I^{2}R_{s} = 23.97W\)

but I can't see to reproduce that number.

I was confused if I should use the \(R_{s}\) I calculated in the first part of the experiment(See first figure) or if I should calculate the value of \(R_{s}\) again using the table at the top of the second figure.

(i.e. \(R_{s} = \frac{V_{s}}{I_{s}} = \frac{117.7V}{0.88A}\)

Can someone clarify?

After this is sorted out we are asked to verify and confirm the above readings for the power, and thus the efficiency.

I attempted to calculate the rotor and stator losses using the values found in the first figure attached.

Note that we also have the rotational losses estimated in the second figure attached.

Are the calculations I've made for the rotor and stator losses correct? If so, how I reproduce the power and efficiency I've measured?

Thanks again!



Joined Mar 6, 2009
Magnetizing current can be quite significant when compared to the FLC in induction machines - 50% isn't unusual for smaller machines. Also the presence of harmonics (e.g. 3rd) from the non-linear magnetic circuit will contribute to the non-sinusoidal current waveform.

Thread Starter


Joined Sep 13, 2010
Can you offer any clarification as to reproducing input and output powers and in turn the efficiency?

I'm still having trouble with that.

Did I compute the stator and rotor losses correctly?


Joined Mar 6, 2009
It's hard to make much out of the various numbers that presents a consistent picture.

Regarding the 23.97W...

Per phase you have about 8W (~23.97/3)

The stator resistance is 12.15Ω.

To get 8W in 12.15Ω requires I=√(8/12.15)=0.811 A

So you must have used the rotor current of 0.811 A to calculate the total 23.97 W over the 3 phases. I guess that's consistent in the sense that the stator current of 0.88A would include the magnetizing current which you might not want to include in the stator resistance loss evaluation. It all depends on the equivalent circuit model you are expected to adopt. I'm also assuming the stator to rotor winding ratio is 1:1

Using just the stator Cu loss to find rotational losses is somewhat confusing. I would lump the stator & rotor Cu losses to find the total Cu loss - which would notionally be
P_Cu_loss=3*0.811^2*(12.15+5.15)=3*11.38=34.14W. This would make the rotational losses Prot=123.66-34.14=89.52W.

Perhaps the rotational losses alluded to in the lab notes might be better annotated as the air gap power.

Most IM linear models give the air gap power as 3*I^2*R_rotor/S. In this case this evaluates to 3*(0.811)^2*5.15/0.046=221W which is way out when compared with actual measurements. Hence I'm curious but not too surprised with the inconsistency..
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