When an AC source is connected to an ideal inductor show that the…

Question

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.

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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 $e=e_0 \sin \omega t$, 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_{ 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 $
$= e_{ rms } i _{ rms } \cos \Phi= e _{ rms } i _{ rms }\left(\frac{R}{Z}\right) \text {, where the impedance } Z =\sqrt{R^2+\left(X_{ L }-X_{ C }\right)^2} .$
$\therefore P _{ av }=0$, i.e., the circuit does not dissipate power.

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