Timeline for How to show that imaginary part of complex power is reactive power?
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Dec 1, 2019 at 1:13 | comment | added | ad2004 | If you consider $cos(\omega t + \phi) = cos(\omega t)cos(\phi) -sin(\omega t)sin(\phi)$, you can see that $cos(\phi)$ is associated with the "in-phase" component of the current (in-phase with the voltage) and $-sin(\phi)$ is associated with the out-of-phase (by $\pi/2$) term. The in-phase, I believe, corresponds with the real power and the out-of-phase corresponds with the "reactive" power. The "reactive power" should have a time average of zero because it doesn't correspond to any net energy transfer. I hope this helps. | |
Nov 30, 2019 at 20:13 | comment | added | praveen kr | I understand the part related to dissipated power which gives the non zero average. What I am confused about is the "reactive power" which has zero time average. It is found in many places that the imaginary part of the complex power gives the reactive power but I do not understand the reasoning behind this. | |
Nov 29, 2019 at 20:59 | history | answered | ad2004 | CC BY-SA 4.0 |