Question 240: Electromagnetic Induction

Question

Consider the following circuit. By keeping $\mathrm{S}_1$ closed, the capacitor is fully charged and then $S_1$ is opened and $S_2$ is closed, then

Accepted Answer

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

Answer

At time $\mathrm{t}=0$, the energy stored in the circuit is purely in the form of magnetic energy.

Answer

At $\mathrm{t}>0$, there is no exchange of energy between $L$ and $C$.

Answer

At any time $\mathrm{t}>0$, the current in the circuit is in the same direction.

Answer

At any time $\mathrm{t}>0$, the instantaneous current in the circuit may be $\mathrm{V} \sqrt{\frac{\mathrm{c}}{\mathrm{L}}}$

Electromagnetic Induction

Consider the following circuit. By keeping \mathrm{S}_1 closed, the capacitor is fully charged and then S_1 is opened and S_2 is closed, then

Choose the correct answer: