How is it in uk about reactive power charges ? Here up to 20% is free for consumers, but businesses need to either compensate or pay for kVar:s.

As it can be measured as difference between apparent and real power would like to know how accurate the current algorithms are in calculating apparent power. So far a lot of info about real power calculation is made avail here so guess its fairly accurate.

## Re: True and reactive power

There is no charge for VAr to domestic users in the UK. Only real power is charged.

To properly measure var, you need to phase shift voltage by 90° and then do the same calculation as you do for real power. Reactive power (when I was taught it was called "imaginary power" - a much better description because it is VA resolved on the imaginary axis) is not the difference between apparent power and real power - remember the vector diagrams:

P

^{2}+ VAR^{2}= VA^{2}The calculations are essentially as accurate as the calibration of the sensors and the resolution of the ADC. There will be small rounding errors in the maths but these will be orders of magnitude smaller.

## Re: True and reactive power

Was thinking more from the standpoint of using Vrms and Irms to calculate PF. As there is some delay in reducing the bias voltage by various filters this promotes some error in the currently available scripts.

Alternatively one could measure the max AD volts point for a CT and get a phase angle compared to maximum Volts AD.

Not all the devices are fully inductive (90 degrees) so dont know if that would work ?

## Re: True and reactive power

You need to look at the definition of power factor. The text book equation power factor = cosφ only applies to a text book perfect sine wave. The

has nothing to do with phase angles, nor timing differences, and is the ratio of real power to apparent power, which is exactly as emonLib calculates it. Therefore it works for real world voltage and current waveforms.definitionBut you originally wanted reactive power - var - did you not? The only way to measure that accurately, as I wrote above, is (assuming the voltage phasor is the reference and aligned with the real axis) to resolve the current vector into the real and imaginary components. And to do that, you need a reference shifted 90 ° from the real axis.