Q.When an AC source is connected to an ideal inductor, show that the average power supplied by the source over a complete cycle is zero.

[!TLDR] For a pure inductor, $i=-i_0\cos\omega t$ so $P=ei=-e_0i_0\sin\omega t\cos\omega t=-\frac12 e_0i_0\sin2\omega t$, whose average over a full cycle is zero since $\langle\sin2\omega t\rangle=0$. [!ANSWER] $P_{av}=0$ for an ideal inductor over a complete cycle.

Let the source emf be $e=e_0\sin\omega t$. For a purely inductive circuit, the current lags the emf by $\pi/2$: $i=i_0\sin\left(\omega t-\dfrac{\pi}{2}\right)=-i_0\cos\omega t$. The instantaneous power delivered by the source is $P=ei=(e_0\sin\omega t)(-i_0\cos\omega t)=-e_0i_0\sin\omega t\cos\omega t$. Using the identity $2\sin\omega t\cos\omega t=\sin2\omega t$, this becomes $P=-\dfrac12 e_0i_0\sin2\omega t$.\n\nThe average power over one complete cycle (period $T=2\pi/\omega$) is $P_{av}=\dfrac{1}{T}\displaystyle\int_0^T P\,dt=-\dfrac{e_0i_0}{2T}\displaystyle\int_0^T\sin(2\omega t)\,dt$. Since $\sin2\omega t$ is itself a sinusoid that completes exactly two full oscillations within one period T (because its own angular frequency is $2\omega$), its integral over the full interval $0$ to $T$ is exactly zero -- the positive and negative half-cycles of $\sin2\omega t$ cancel perfectly. Hence $P_{av}=0$. Physically, this reflects that the inductor absorbs energy from the source while the current is building up (storing it in the magnetic field) and returns exactly that same energy back to the source while the current subsequently falls -- so there is no NET transfer of energy over a full cycle, even though the instantaneous power is nonzero (and alternates in sign) throughout. [!ANSWER] $P_{av}=0$: the average power supplied by the source to an ideal inductor over a complete cycle is exactly zero.

Write the instantaneous power $P=ei$ using the $\pi/2$-lag current for a pure inductor, simplify using the double-angle identity, and integrate over one full period, noting $\langle\sin2\omega t\rangle=0$.

Trying to average $P=ei$ term by term without first using the double-angle identity, or incorrectly assuming $\langle\sin\omega t\cos\omega t\rangle$ needs a separate, more complicated calculation rather than simply recognising it as $\frac12\sin2\omega t$, itself an ordinary zero-mean sinusoid.