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Discuss the power in $AC$ circuit containing only an inductor.
Solution
The current reaches its maximum value later than the voltage by one-fourth of a period $\frac{\mathrm{T}}{4}=\frac{\pi / 2}{\omega}$
The instantaneous power supplied to the inductor is,
$\mathrm{P}_{\mathrm{L}}=\mathrm{IV}=\mathrm{I}_{\mathrm{m}} \sin \left(\omega t-\frac{\pi}{2}\right) \times \mathrm{V}_{\mathrm{m}} \sin (\omega t)$
$=-\mathrm{I}_{\mathrm{m}} \mathrm{V}_{\mathrm{m}} \cos (\omega t) \sin (\omega t)$
$=-\frac{\mathrm{I}_{\mathrm{m}} \mathrm{V}_{\mathrm{m}}}{2} \sin (2 \omega t)$
The average power over a complete cycle is,
$\mathrm{P}_{\mathrm{L}}=\left\langle-\frac{\mathrm{I}_{\mathrm{m}} \mathrm{V}_{\mathrm{m}} \sin (2 \omega t)}{2}\right\rangle$
$=-\frac{\mathrm{I}_{\mathrm{m}} \mathrm{V}_{\mathrm{m}}}{2}(\sin (2 \omega t))=0$
Since the average of $\sin (2 \omega t)$ over a complete cycle is zero. Thus, the average power supplied to an inductor over one complete cycle is zero. Figures explains it in detail.
