I am trying to obtain a wattage measurement from an air compressor using only an amperage measurement from one of the conductors, and the nominal voltage stated on the machine.

The purpose of the measurement is to understand how much power this machine is drawing in order to calculate approximate electricity costs.

It is a three-phase, 575 V air compressor. When I place my current meter clamp over one of the conductors I get a reading of 16 A. The measurement is taken between the circuit breaker and the machine.

To get watts, would the formula be: (16 A × 575 V)/√3?

This would get me the wattage for one phase; would the total wattage be the result from the formula above multiplied by 3?

  • 1
    \$\begingroup\$ Isn’t 3 phase power calculated as cube root(V I cos phi)? \$\endgroup\$
    – Solar Mike
    Commented Feb 27, 2023 at 21:21
  • \$\begingroup\$ @SolarMike power is calculated as the average of the instantaneous and continuous multiplication of v and i. power factor cannot be used as a means of calculating power from Vrms and Irms due to current distortion harmonics. \$\endgroup\$
    – Andy aka
    Commented Feb 27, 2023 at 21:40

2 Answers 2


\$ VA = 3V_{phase}I_{phase} \$

\$ I_{phase} = \frac{I_{line}}{\sqrt{3}} \$

\$ P(W) = VA\cdot cos(\phi) \$

So, what you did is correct. Except, you will be measuring VA not Watts. You need to consider the phase angle between the two. A Wattmeter will do this automatically.

When the compressor is heavily loaded, the angle will be small. And VA is always more than Watts. So, your method is an approximation, it may be close enough for your purposes.

When the compressor is lightly loaded, the phase angle can be large, and the VA can be much more than the Watts.


I am trying to obtain a wattage measurement from an air compressor using only an amperage measurement from one of the conductors...

That measurement tells you the amperage in one of the three phases.

...and the nominal voltage stated on the machine

That doesn't tell you much about the true voltage applied. Only the true applied voltage can be used for accurate power measurements (unless you are happy with guesswork).

Even a single phase current measurement and a single phase voltage measurement doesn't given the full picture of total three-phase power consumed.

  • Neither does a single phase current and voltage RMS measurement tell you anything about power factor.

  • Neither does it take into account load or supply imbalances.

  • Neither does it properly deal with harmonic distortion in the current.

The simplest way to measure 3-phase power is the two wattmeter method: -

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Image from here. Here are some other links to help you understand: -


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