Question 340: AC fundamentals

Question

In the given circuit, when $S_1$ is closed, the capacitor gets fully charged. Now $\mathrm{S}_1$ is open and $\mathrm{S}_2$ is closed. Then
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m3s8ww8l/7c7484fc-6c23-4332-9d54-76f0b261695b/dbbf8650-a88a-11ef-8324-df9cf7a574c9/file-jaoe38c1m3s8ww8m.png?format=png)

Accepted Answer

Maximum energy in capacitor = Maximum energy in inductor
$\begin{array}{ll} 
& \frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}^2 \
	herefore \quad & \mathrm{I}^2=\frac{\mathrm{C}}{\mathrm{~V}} \mathrm{~V}^2 \
	herefore \quad & \mathrm{I}=\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{~L}}}
\end{array}$

Answer

there is no exchange of energy between L and C .

Answer

the current in the circuit is in the same direction.

Answer

the instantaneous current in the circuit may be $\mathrm{V}\left(\sqrt{\frac{\mathrm{C}}{\mathrm{L}}}\right)$.

Answer

the energy stored in the circuit is purely in the form of magnetic energy.

AC fundamentals

In the given circuit, when S_1 is closed, the capacitor gets fully charged. Now \mathrm{S}_1 is open and \mathrm{S}_2 is closed. Then

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