Question

After the capacitor gets fully charged, $S_1$ is opened and $S_2$ is closed so that the inductor is connected in series with the capacitor. Then,

• A. at $t=0$, energy stored in the circuit is purely in the form of magnetic energy
• B. at any time $t>0$, current in the circuit is in the same
• C. at $t>0$, there is no exchange of energy between the inductor and capacitor
• D. at any time $t>0$, instantaneous current in the circuit may $v\sqrt{\frac{C}{L}}$

Answer 1

The correct option is D at any time $t>0$, instantaneous current in the circuit may $v\sqrt{\frac{C}{L}}$

Energy stored in the circuit at t=0 is magnetic as well as electrostatic($\frac{1}{2}C{V}^2$). Current in the circuit is not the same

For t>0

$Q=Q_0\sin (\omega t+\delta )$

$\frac{dQ}{dt}=i=Q_0\omega \cos (\omega t+\delta )$

$i=CV\times \frac{1}{\sqrt{CL}}$ as $\omega t+\delta =0$

$i=V\sqrt{\frac{C}{L}}$