Prove that an ideal capacitor in an AC circuit does not dissipate…

Question

Prove that an ideal capacitor in an AC circuit does not dissipate power

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Answer

In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of $\frac{\pi}{2} rad$. Here, for $e=e_0 \sin \omega t_{\text {, we have, }} i=i_0 \sin \left(\omega t-\frac{\pi}{2}\right)$
Instantaneous power, $P=$ ei
$=\left(e_0 \sin \omega t\right)\left[i_0\left(\sin \omega t \cos \frac{\pi}{2}+\cos \omega t \sin \frac{\pi}{2}\right)\right]$
$=-e_0 i_0 \sin \omega t \cos \omega t$ as $\cos \frac{\pi}{2}=0$ and $\sin \frac{\pi}{2}=1$.
Average power over one cycle, $P_{a v}$
$ P_{\text {av }}= \frac{\text { work done in one cycle }}{\text { time for one cycle }}$
$= \frac{\int_0^T P d t}{T}$
$= \frac{\int_0^T\left[e_0 i_0 \cos \phi \sin ^2 \omega t \pm e_0 i_0 \sin \phi \sin \omega t \cos \omega t\right) d t}{T}$
$= \frac{e_0 i_0}{T}\left[\cos \phi \int_0^T \sin ^2 \omega t d t \pm \sin \phi \int_0^T \sin \omega t \cos \omega t d t\right]$
$\text { Now, } \int_0^T \sin { }^2 \omega t d t=\int_0^T\left(\frac{1-\cos 2 \omega t}{2}\right) d t$
$= \int_0^T \frac{1}{2} d t-\int_0^T \frac{\cos 2 \omega t}{2} d t=\frac{T}{2}-\frac{1}{2}\left(\frac{\sin 2 \omega t}{2 \omega}\right)_0^T$
$= \frac{T}{2}-\frac{1}{4 \omega}(\sin 2 \omega T-\sin 0)$
$= \frac{T}{2}-\frac{1}{4 \omega}\left[\sin 2\left(\frac{2 \pi}{T}\right) T-0\right]$
$= \frac{T}{2}-\frac{1}{4 \omega}[0-0]=\frac{T}{2} $
Also,
$ \int_0^T \sin \omega t \cos \omega t d t=\frac{1}{2} \int_0^T \sin 2 \omega t d t=\frac{1}{2}\left[\frac{-\cos 2 \omega t}{2 \omega}\right]_0^T$
$=-\frac{1}{4 \omega}\left[\cos 2\left(\frac{2 \pi}{T}\right) T-\cos 0\right]=-\frac{1}{4 \omega}[1-1]=0 $
Hence, $P_{ av }=\frac{e_0 i_0}{T} \cos \phi \times \frac{T}{2}=\frac{e_0 i_0}{2} \cos \phi$
$=\frac{e_0}{\sqrt{2}} \cdot \frac{i_0}{\sqrt{2}} \cos \phi$
$= e _{ rms } i _{ rms } \cos \Phi= e _{ rms } i _{ rms }\left(\frac{R}{Z}\right)$, where the impedance $Z =\sqrt{R^2+\left(X_{ L }-X_{ C }\right)^2}$.
$\therefore P_{a v}=0$, i.e., the circuit does not dissipate power.