Average Power Associated with an Inductor

For a purely inductive circuit, the current lags the applied emf by a phase angle of $\pi/2$: $e=e_0\sin\omega t$ and $i=i_0\sin\left(\omega t-\dfrac{\pi}{2}\right)=-i_0\cos\omega t$. The instantaneous power is $P=ei=-e_0i_0\sin\omega t\cos\omega t$.

Using the identity $\sin\omega t\cos\omega t=\dfrac12\sin2\omega t$, this becomes $P=-\dfrac12 e_0i_0\sin2\omega t$ -- a sinusoidal function of $2\omega t$ that is symmetric about zero, taking equal and opposite values over each successive quarter cycle. Averaging this over one full cycle of the original AC (which spans exactly one full cycle of $\sin2\omega t$ as well) gives EXACTLY ZERO, since $\langle\sin2\omega t\rangle=0$ over any whole number of cycles. So $P_{av}=0$ for a pure inductor: over the course of each cycle, the inductor absorbs energy from the source while building up its magnetic field (current increasing) and returns EXACTLY that same amount of energy back to the source while the field collapses (current decrea …