Suppose I have a voltage across an element $V(t) = A \cos(\omega t)$ and the current through it given by $I(t) = B \cos(\omega t + \phi)$. The instantaneous power is
$$P(t) = V(t)I(t) = AB \cos(\omega t)\cos(\omega t + \phi).$$ A simple trigonometric identity reduces this to $$P(t) = \frac{AB}{2}(\cos(\phi) + \cos(2\omega t + \phi)).$$ The first term is just the average power dissipated in the element and the second term is the power that goes back and forth in the element, which as per my understanding, is the reactive power.

Now in the phasor form the power is defined as $$\frac{1}{2}\tilde{V}\tilde{I}^* = \frac{AB}{2} e^{-j\phi} = \frac{AB}{2}(\cos(\phi) - j \sin(\phi)).$$ Clearly the real part here gives the average power dissipated. But I do not understand why the imaginary part equals the reactive power.


I believe that some of the discrepancy is due to the fact that the instantaneous power equation that you obtained in the first part of your question needs to be time averaged over one period (i.e. integrated over one period and divided by the period). This will yield the real power (i.e. that which is actually dissipated in the load) I believe. I hope this helps.

  • $\begingroup$ 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. $\endgroup$
    – praveen kr
    Nov 30 '19 at 20:13
  • $\begingroup$ 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. $\endgroup$
    – ad2004
    Dec 1 '19 at 1:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.