We've just started AC circuit analysis in class, and the terms that I mentioned above popped up all at once. I'm new to the idea of phasors(I usually visualize them as arrows making a certain angle with another reference arrow), and I can visualize V and I, but how does power be represented as a phasor?

The phasor representation is $$S=P+jQ$$ and what actually are these terms? I know P is the average part. What exactly does Q represent, the RMS reactive power? And what is the relation between these terms and instantaneous power? Is S here instantaneous power?

P.S: Sorry about so many trivial/stupid questions. The things kinda made sense individually but I can't seem to connect the dots.


1 Answer 1


First of all consider this picture: -

enter image description here

It paints 4 scenarios all of which apply to this question: -

  • Top left shows voltage and current in phase and the multiplication of voltage and current waveforms produces a power waveform. Note that the power waveform has twice the frequency of the voltage and current hence, it's not usually meaningful to represent it as a phasor (and certainly not useful if you have other phasors representing voltage and current). The average power is P.

  • Top right shows voltage leading current by 60 degrees - note that the average power (P) has reduced but, the power waveform has the same shape, frequency and overall peak-to-peak magnitude. In this scenario we can start to realize what the apparent power is. Basically, S is half the distance between peak and trough of the power waveform and, in fact S = P in the top left scenario.

  • Bottom left is when a purely reactive load is connected to a voltage source. The power waveform has an average value of 0 watts (P = 0) but, apparent power (S) remains the same as the previous two examples.

  • Bottom right is likely to be encountered when a motor turns into a generator and pushes real power (P) back to the voltage source. In other words, there is real power (P) but it is negative.

So, you should now understand real power (P) and apparent power (S). To understand Q we use the power triangle: -

enter image description here

Picture of power triangle from here.

Where the angle theta is the phase angle between voltage and current waveforms.

What exactly does Q represent, the RMS reactive power?

RMS power is a misnomer; RMS voltage x RMS current x power factor = power (not RMS power).

And what is the relation between these terms and instantaneous power?

Instantaneous power is any point on the red power curves shown in the top diagram in 4 scenarios and equals instantaneous volts x instantaneous amps.

  • \$\begingroup\$ This! Wow, what an amazing explanation! I just knew the formulas, but no one ever explained what the terms P, S, Q actually meant. Thanks! Also, last question. How did they get from this, to phasor representation of power (as VI*)? \$\endgroup\$
    – user_9
    Jan 31, 2020 at 12:54
  • \$\begingroup\$ ^Should have been more specific-- how come it is P+jQ, rather than just say, S=P+Q? I know the math(since it's due to j involved in C and L reactance, but how does it make sense intuitively? Like, some explanation from the graph above maybe? \$\endgroup\$
    – user_9
    Jan 31, 2020 at 12:57
  • \$\begingroup\$ Phasor representation of power has no real meaning to me; only phasor representations of voltage, current and impedance make sense but, if you have some example to show? P + jQ means that \$S = \sqrt{P^2+Q^2}\$. Do you follow that last bit (Pythagoras)? \$\endgroup\$
    – Andy aka
    Jan 31, 2020 at 13:41
  • \$\begingroup\$ Oh yeah, I get what the complex number means and all. I'm just asking if there's a reason why it's in complex form in the first place. For example, why not reactive power= (apparent minus average) or something like that. It's stupid, I know, but I was just wondering if I could visualize via the graph in your answer as to why complex numbers enter the picture. \$\endgroup\$
    – user_9
    Jan 31, 2020 at 14:34
  • 1
    \$\begingroup\$ Hmm maybe there's a way. I'll have a think. \$\endgroup\$
    – Andy aka
    Jan 31, 2020 at 15:42

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