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Regarding AC power feeding an impedance, is it possible to plot an example of V and I versus time for an inductive load R+jwL and especially the instantaneous power on the same plot where we can mark/paint three different areas for a power period where one area will show the active power the other reactive and the other apparent power?

Instead of using phasors, I would like to see in time the areas corresponding to three types of powers namely active reactive and apparent.

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  • \$\begingroup\$ Reactive and apparent power don't exist in the time domain. They only exist as average quantities. \$\endgroup\$
    – Hearth
    Mar 16, 2019 at 19:10
  • \$\begingroup\$ Yes thats why am asking the area per period. Energy per period. \$\endgroup\$
    – user1999
    Mar 16, 2019 at 19:11
  • \$\begingroup\$ Showing the the areas to be averged I mean \$\endgroup\$
    – user1999
    Mar 16, 2019 at 19:15
  • \$\begingroup\$ As for myself, there are may be a lot better answers on the internet, and might be redundant to answer it here. If you have a DSO, by multiplying the current and voltage at both channel you can see the plot, if the resulting parts of the power plot is positive then you have a load consuming power acting as active power, otherwise if it's negative the load is generating power acting as a reactive power. This video may somewhat give you some inspiration. \$\endgroup\$
    – Unknown123
    Mar 16, 2019 at 21:16

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I have done something like that with MS Excel. See below. The negative power shown represents energy returned to the source during each half cycle. An equal amount of energy in excess of that represented by the average power is received from the source during each half cycle. I am sure that something like this can be done to show more explicitly what is happening.

enter image description here

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  • \$\begingroup\$ That seems pretty explicit to me. Positive average implies real power. Power wave goes negative, implies reactive power. The waveform would be S (VA). \$\endgroup\$ Mar 17, 2019 at 3:41
  • \$\begingroup\$ That is the easy part (real power). What about reactive power? This paper (you can get via ieee xplore) describes a methodology to calculate the reactive power at each sample point for three-phase system using cross products. I've used it and is easy to calculate. “Harmonic and Reactive Power Compensation Based on the Generalized Instantaneous Reactive Power Theory for 3-Phase and 4-Wire Systems”, F.Z. Peng, G. W. Ott, D.J. Adams \$\endgroup\$ Mar 17, 2019 at 7:59
  • \$\begingroup\$ I’m thinking this means we can use present voltage sample, and current sample from 1/4 cycle earlier (quadrature relationship) to calculate present sample reactive. I’m sure we can figure something out. \$\endgroup\$ Mar 17, 2019 at 8:15
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Charles got the easy part for us - real power. Real power is just voltage times current at each instance (each sample point). To calculate the reactive power take the voltage samples from 1/4 cycle earlier (90 degrees) and multiply by the present current sample. I just verified that this is how EMTP-RV does the calculation with their q(t) meter.

Charles can probably modify his spreadsheet easily to do this - of course, there will be no reactive calculation until you have 1/4 cycle of data collected first. You must assume an operating frequency (e.g. 60Hz).

Here below are the results using a 120V source connected to an (R + jX) load of (10 + j10) ohms. enter image description here

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