1

In a single phase meter measurement, the power is being measured by using the instantaneous power which is p(t) = v(t) * i(t). But as we need to sample the voltage and current to get the power, hence it need to be calculated by using this formula:

enter image description here

enter image description here

So is this formula can really measured the real active power without considering the phase angle between V and I? As we all know that active power, P = VIcosθ.

So what is the difference between apparent power measurement, S=VI and active power measurement if the P = VIcosθ is not being implemented?

JYelton
  • 32,302
  • 33
  • 134
  • 249
mynset
  • 13
  • 1
  • 5

1 Answers1

4

Instantaneous Volts multiplied by instantaneous Amps is true power and no compromise and when micros implement this they are doing it exactly the correct way.

VI x power factor may not be true power because hardly ever is the voltage waveform or the current waveform truly sinusoidal and only when V and I are truly sinusoidal can RMS Voltage multiplied by RMS current multiplied by power factor = v(t). i(t).

Here are several examples of voltage and current waveforms shifted in phase to each other: -

enter image description here

The first picture is when v and i are in phase and power is maximum. Note also the example when v and i differ by 90 degrees (bottom left) - power is zero (as you would expect) but, if there is any harmonic distortion, VICos(phi) will be inaccurate.

Andy aka
  • 434,556
  • 28
  • 351
  • 777