Question 398: AC fundamentals

Question

For the circuit shown below, instantaneous current through inductor '$\mathrm{L}$' and capacitor '$\mathrm{C}$' is respectively.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1ltx4pzse/2447e164-8459-45ee-91d1-43ff766a5abf/38c4bff0-e540-11ee-b8cb-9be8d18c4bd5/file-1ltx4pzsf.png?format=png)

Accepted Answer

The current through the inductor lags behind the applied emf by $\frac{\pi}{2}$ and the current through the capacitor leads the current by $\frac{\pi}{2}$
$\begin{aligned}
& 	herefore \mathrm{i}_{\mathrm{L}}=\frac{\mathrm{e}_0}{\omega \mathrm{L}} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}ight)=-\frac{\mathrm{e}_0}{\omega \mathrm{L}} \cos \omega \mathrm{t} \
& 	ext { and } \mathrm{i}_{\mathrm{c}}=\mathrm{e}_0 \omega \mathrm{C} \sin \left(\omega \mathrm{t}+\frac{\pi}{2}ight)=\mathrm{e}_0 \omega \mathrm{C} \cos \omega \mathrm{t}
\end{aligned}$

Answer

$\frac{-\mathrm{e}_0}{\omega \mathrm{L}} \cos \omega \mathrm{t} ; \mathrm{e}_0 \omega \mathrm{c} \cos \omega \mathrm{t}$

Answer

$\frac{-\mathrm{e}_0}{\omega \mathrm{L}} \sin \omega t ; \frac{\mathrm{e}_0}{\omega \mathrm{C}} \cos \omega \mathrm{t}$

Answer

$\frac{\mathrm{e}_0 \mathrm{C}}{\mathrm{L}} \cos \omega \mathrm{t}; \frac{\mathrm{e}_0 \mathrm{~L}}{\mathrm{C}} \sin \omega \mathrm{t}$

Answer

$\frac{-\mathrm{e}_0 \mathrm{C}}{\mathrm{L}} \sin \omega \mathrm{t} ; \frac{\mathrm{e}_0 \mathrm{~L}}{\mathrm{C}} \cos \omega \mathrm{t}$

AC fundamentals

For the circuit shown below, instantaneous current through inductor '\mathrm{L}' and capacitor '\mathrm{C}' is respectively.

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