# How to show that imaginary part of complex power is reactive power?

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.

• 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. Dec 1 '19 at 1:13