$A$ capacitor of capacitance $100 \ \mu F$ is charged to a potential of $12 \ V$ and connected to a $6.4 \ mH$ inductor to produce oscillations. The maximum current in the circuit would be: (in $A$)

$A$ charged $30\; \mu F$ capacitor is connected to a $27\; mH$ inductor. The initial charge on the capacitor is $6\; mC$. What is the total energy stored in the circuit initially? What is the total energy at a later time?

In the given circuit,when $S_1$ is closed,the capacitor $C$ gets fully charged. Then $S_1$ is kept open and $S_2$ is closed. Hence

$A$ capacitor of capacity '$C$' is charged to a potential difference of '$V_1$'. The plates of the capacitor are then connected to an ideal inductor of inductance '$L$'. The current through the inductor when the potential difference across the capacitor reduces to '$V_2$' is:

An $LC$ circuit contains a $20 \, mH$ inductor and a $50 \, \mu F$ capacitor with an initial charge of $10 \, mC$. The resistance of the circuit is negligible. Let the instant the circuit is closed be $t = 0$. At what time is the energy stored completely magnetic? $t = \dots \, ms$

$A$ $1 \, \mu F$ capacitor is charged to $50 \, V$. The charging battery is then disconnected and a $10 \, mH$ coil is connected across the capacitor so that $LC$ oscillations occur. What is the maximum current in the coil (in $, A$)? Assume that the circuit contains no resistance.