I want to know how energy meters (Electro-static or electronic) measure Energy (let's assume power, ignore time) when there are harmonics present.

I know what are harmonics, how are they generated and how do they affect actual power. I read that harmonics power (in most cases) is in the opposite direction of Actual Power. That means if the utility is providing power to a consumer who has a dominant non-linear load, and in-between them is an energy meter (plz ignore class, type, etc.). So energy meter would record less power. That is understandable.

I tested this scenario like this, I applied a linear load to meter and recorded power of ~2.3KW (230V x 10A x 1.0 P.F). But when I applied Phase-fired (90 degrees) waveform which contains odd harmonics, the power recorded by meter was ~1.1KW given the same set of V, I and P.F.

I know meter records power accurately because you can see from the waveform that rms have been decreased (as each half-cycle is 50% clipped to zero). But there lies my question.

If meters are recording less power in the presence of harmonics, then isn't it a loss of utility and benefit of consumers? While consumer is responsible for generating harmonics and is getting profit on top of it?

Should meter record actual power while filtering out the harmonics?

To me, it sounds really confusing, and I tried to search for utilities and meter manufacturer point of view, and they seem to agree that meter should consider harmonics effect while measuring power as it would be beneficial to both? How?

I tried to raise my question as much as I could find the write words, I hope that people here would understand, and help me find the answer.

Thank You

  • \$\begingroup\$ You contradict yourself. You say 1.1KW given the same set of V, I and P.F then you say the waveform that rms have been decreased (as each half-cycle is 50% clipped to zero) and those statements are contradictory making this question difficult to answer. Power meters (you might find) find it quite easy to compute real true power in the presence of all manner of harmonics and distortions. They use a very simple mathematical relationship that is pretty difficult to mess up. \$\endgroup\$
    – Andy aka
    Commented Jan 16, 2020 at 8:29
  • \$\begingroup\$ That's exactly what I said, I know meters measures power minus the power lost in harmonics, my real question was, should it work like that? You said there are meters who measure true power? What is true power? V&I*P.F, but that V/I gets clipped due to harmonics (half-wave or phase fired), so the RMS V and I are reduced so as the power. The meters you mentioned, do they filter our harmonics before measuring power? \$\endgroup\$ Commented Jan 16, 2020 at 12:01
  • 1
    \$\begingroup\$ True power 1. True power 2. Power factor is never computed when measuring true power; power factor is just a convenient number that tells you about the load. Meters will measure all power including any power transmitted by harmonics. RMS values are never used in meters to calculate power; it's more fundamental than that. \$\endgroup\$
    – Andy aka
    Commented Jan 16, 2020 at 12:37

1 Answer 1


There are electromagnetic energy meters with rotating aluminum discs and electronic energy meters, but I have never heard of an "electro-static" energy meter.

Electromagnetic energy meters measure real power and ignore harmonics because they measure power as the speed of a small motor that has a synchronous speed that is proportional to frequency. In effect, they can not measure harmonic volt-amperes because they can not tour at more than one speed at a time. There are torques developed due to harmonics, but some tend to drive the disc in the same direction as the fundamental and some in the opposite direction. The torques and counter-torques do not match, but they apparently don't interfere with the proportionality between speed and real power.

Electronic energy meters work by calculating instantaneous power at short sampling intervals. They multiply instantaneous voltage by instantaneous current. The result is integrated to calculate average power. If the voltage waveform is sinusoidal, no real power is transmitted and the result of the calculation is the fundamental power. If the voltage waveform is distorted, there is real power transmitted at both the fundamental and the harmonic frequencies and that is correctly included in the calculation.

If the objective is to calculate the power separately for each harmonic, the individual harmonic voltages need to be determined. I believe that fourier transform methods can do that.

I hope that provides a satisfactory intuitive explanation. Perhaps someone else can provide a rigorous explanation.

  • \$\begingroup\$ Electro-static/electronic/digital meters are the same, i think, though i edited it in my question. You said: "If the voltage waveform is distorted, there is real power transmitted at both the fundamental and the harmonic frequencies and that is correctly included in the calculation.", thats what i was looking for, meters based on ADC's would definitely measure power with harmonics. It is a kind of non-technical question, that meters installed at customer premises by utility would measure less power because there are always harmonics (non-linear)? Is this normally happens? \$\endgroup\$ Commented Jan 16, 2020 at 12:08
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    \$\begingroup\$ Energy measured by utilities that use electromechanical meters is less than the actual energy transferred to the extent that the voltage at the meter is distorted. However, to a great extent, that energy is harmful rather than beneficial to the customer. Harmonic energy causes additional heat to be dissipated in motors and transformers. Harmonic energy is only used beneficially to the extent that it is used in resistance heaters and incandescent lights. Utilities have an incentive to charge for customer harmonic watt-hours and total volt-ampere hours. Modern utility meters make that possible. \$\endgroup\$
    – user80875
    Commented Jan 16, 2020 at 15:54
  • \$\begingroup\$ thanks, that explains, so some places, utilities are concerned with harmonic power. In my region, only Low P.F penalties are present but no such concern for harmonics. \$\endgroup\$ Commented Jan 17, 2020 at 5:44
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    \$\begingroup\$ It is likely that the utilities calculation of power factor is pf = (real power perhaps including harmonic power) / (total VARs including harmonics). That is relatively easy to determine and can reasonably be used to estimate the basic economic impact on the utilities of low power factor and harmonics. \$\endgroup\$
    – user80875
    Commented Jan 17, 2020 at 13:54

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