# Can Average Power of a 3-Phase AC motor be calculated from Mean Current & Voltage RMS from each winding?

If this is possible could a formula be provided. + Explaination.

FYI I am completely new to this site so please be kind. Could I ask that any criticism be constructive. Any questions/further clarity of the problem required please feel free to ask.

Can Average Power of a 3-Phase AC motor be calculated from Mean Current & Voltage RMS from each winding?

No it can't because the mean current taken on any phase is zero. That leaves you with just the RMS voltage and this is no indication of power.

If you want to definitively measure power on any AC machine you need to know certain things and these can be: -

1. RMS volts, RMS current and power factor (for each phase)
2. Instantaneous voltage and instantanous current over one full cycle of power (per phase)

Without power factor, RMS current and RMS voltage are meaningless. Even if you know the power factor, if the waveform of current is not sinusoidal (as it could easily be in a motor core that shows signs of saturation), then you will get an inaccurate answer.

The most accurate method is multiplying instantaneous values of voltages and current then averaging (or integrating) the result. This is how wattmeters work either analogue or digital. Below is a picture of two scenarios where the relative phase of current is changed: -

Hopefully you can see that if the phase angle of current is different to that of voltage then the average value of power reduces. In this next picture I've multiplied a sinewave voltage by the 3rd harmonic of current (usually present in an AC motor) and the resulting power is zero: -

This means that a harmonically distorted RMS current value cannot be reliably used to measure power because the third harmonic has no power content associated with it. As a side note, if the voltage waveform had a 3rd harmonic present (and the load was resistive) then the 3rd harmonic of the voltage would properly multiply with the third harmonic of resistor current to produce a real power.

• Cruel, but apt... I like it! Nov 11 '15 at 17:37
• Don't forget loss of power factor as well. Nov 11 '15 at 17:42
• @JonRB Did you intend that comment for me? Nov 11 '15 at 17:48
• @Andyaka no, the OP, w.r.t. determining electrical power from "mean" and RMS. it was either add additional "comments" or post another answer which seemed a waste as the majority is covered here Nov 11 '15 at 18:07

Assuming all three phases to be identical, three phase power $P=I_{RMS}V_{RMS}\sqrt{3}$. If all you have is the mean current, you'll need a way to get the RMS current. Assuming a sine wave, $I_{mean} = I_{pk}\frac{2}{\pi}$, and $I_{RMS} = \frac{I_{pk}}{\sqrt{2}}$. Solve and substitute, and $I_{RMS} = \frac{I_{mean}\pi}{2\sqrt{2}}$.

Of course, as Andy points out, the mean current of an actual sine wave is zero, and power factor is critical. The above assumes you have a rectified sine wave current, in phase with the voltage.

• To the OP: In AC systems, there is an important concept called power factor. Typically when driving an AC motor, the reactance of the motor causes the current and phase to be out of sync. This "out of sync" causes the actual electric power to be less than what you get using the formula. To correct for this, multiply the result of the formula by the power factor. Power factor is the cosine of the angle between peak current and peak voltage. Nov 11 '15 at 17:50