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Discuss power in $AC$ circuit containing only capacitor.
Solution
The instantaneous power supplied to the capacitor is,
$p_{c}=\mathrm{IV}=\mathrm{I}_{\mathrm{m}} \cos (\omega t) \mathrm{V}_{\mathrm{m}} \sin (\omega t)$
$=\mathrm{I}_{\mathrm{m}} \mathrm{V}_{\mathrm{m}} \cos (\omega t) \sin (\omega t)$
$=\frac{\mathrm{I}_{\mathrm{m}} \mathrm{V}_{\mathrm{m}}}{2} \cdot(2 \cos \omega t \sin \omega t)$
$=\frac{\mathrm{I}_{\mathrm{m}} \mathrm{V}_{\mathrm{m}}}{2} \sin (2 \omega t)$
Average power,
$\mathrm{P}_{\mathrm{C}}=\left\langle\frac{\mathrm{I}_{\mathrm{m}} \mathrm{V}_{\mathrm{m}} \sin 2 \omega t}{2}\right\rangle=\frac{\mathrm{I}_{\mathrm{m}} \mathrm{V}_{\mathrm{m}}}{2}\langle\sin 2 \omega t\rangle$
Since $\sin (2 \omega t)=0$ over a complete cycle. This is explained by figure in detail.
$\therefore$ Average power $\mathrm{P}_{\mathrm{C}}=0$
