I'm designing AC power consumption meter device for apartment based on popular MCU and current/voltage transformers. Currently I successfully managed to measure current and voltage in each phase line individually. There're lots of mentioning 3-phase power formula, like this:

P (kW) = (I (Amps) × V (Volts) × PF × 1.732) ÷ 1,000

I even found that PF is basically phase shift between current and voltage sine.

The only thing I don't get is which current and voltage to use here - I have three of each. Is it average for three or some kind of RMS or what?

Will greately appreciate any advice here.

  • \$\begingroup\$ What are the voltages and currents you get? The voltages should be the same. The currents and pf will vary if the load is not balanced. If it is balanced, any voltage, any current and any pf, since they will all be the same. \$\endgroup\$ Commented Nov 25, 2022 at 21:14
  • \$\begingroup\$ Is your meter measuring line to neutral voltages or line to line? \$\endgroup\$
    – user57037
    Commented Nov 25, 2022 at 23:51

2 Answers 2


Measuring 3-phase power. How to calculate?

The simplest way is probably this; the two wattmeter method: -

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Image from Two Wattmeter Method of Power Measurement and, I mention this because, it is a true 3-phase measurement of power into a 3-phase load. There is also the three wattmeter method: -

enter image description here

Image from Measurement of Three Phase Power: Three Wattmeter Method and, again I'm showing old fashioned wattmeter methods because, if you want accuracy despite non-linear loads or asymmetrical voltage supplies you have to look at what a wattmeter does.

A wattmeter is an analogue multiplier of the voltage and current waveforms. In other words, it's \$v \times i\$ with no worries about power factor. If you try and calculate power using power factor then you have to measure phase angle between \$v\$ and \$i\$ and, the massive problem with this, is that a typical load may be somewhat non-linear (i.e. not wholly resistive) therefore, using zero-crossing detectors (for example) for estimating phase angle, is a bit useless and, gives rise to significant errors.

In other words, your formula assumes a balanced and linear load with a voltage source that is balanced both in amplitude and 120° phase angles. It's folly to think you might be able to do this for practical loads except a kettle and, I don't think they make three phase kettles. Sure there are 3-phase heating elements and these'll be OK but, in the main, you'll be disappointed with V.I.Cos(φ).

But, if your supply is both amplitude and phase balanced and feeds a linear load then you can use that method.

Going back to \$v \times i\$, you can employ modern digital techniques and sample at around 20 times the line frequency as a minimum to avoid current waveform harmonics contributing too much of an error. If you sample and multiply at a higher rate then even better.

Then you accumulate the \$v \times i\$ sampled value over (say for instance) 1 second and divide by the samples per second. Use this method in the equivalent of a two or three wattmeter method.

  • \$\begingroup\$ Wow! Thanks for dtailed explanation. \$\endgroup\$
    – x_arrange
    Commented Nov 25, 2022 at 22:15
  • \$\begingroup\$ From what I understand so far typical apartment's load can not ever be balanced because you can plug anything into any wall socket. So even voltage may remain the same on all threee lines the currents differ completely. I guess my choice would be three-wattmeters. And I'll also try to add power factor measure to each of them \$\endgroup\$
    – x_arrange
    Commented Nov 25, 2022 at 22:26
  • \$\begingroup\$ @x_arrange you don't need to add power factor measurements to compute power with the methods I've shown and described and linked. Power factor is a red herring for real practical power measurement. Nobody uses it because it just does not work unless you have perfect balanced sinewaves (amplitude and phase) and perfectly linear loads. Forget it as an option to help you with power measurement. \$\endgroup\$
    – Andy aka
    Commented Nov 25, 2022 at 22:58
  • \$\begingroup\$ @x_arrange What should I do when someone answers my question? \$\endgroup\$
    – Andy aka
    Commented Nov 26, 2022 at 9:48

You can just purchase one of the many three phase wattmeter and power analyzer ICs and add the required external PTs and CTs, and interface it to a microcontroller to log and display the values. Or you can purchase dedicated power analyzers that display RMS voltage, current, power, power factor, phase angle, and total (and peak) watt-hours. Some even allow entry of cost/kWh to obtain total cost.

You can, of course, design something like this using a microcontroller with six simultaneous ADCs to compute instantaneous voltage and current, and multiply them to obtain power, and average the readings over a period of time for true-RMS and power, and you can compute power factor using W/VA, or phase angle by comparing zero crossings of voltage and current,

It is easiest and most accurate to read current in each phase and phase-to-neutral voltage, but you can derive the third voltage or current from two currents or two voltages, although it may be less accurate for unbalanced loads and distortion.

  • \$\begingroup\$ Thanks. MCU with 6 ADCs is exactly what i'm building. Looks like indeed it's the easiest and most accurate way \$\endgroup\$
    – x_arrange
    Commented Nov 26, 2022 at 7:45

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