$A$ capacitor of capacitance $C$ is charged to a potential $V_1$. It is then connected to an ideal inductor of inductance $L$. When the potential across the capacitor drops to $V_2$,what is the current flowing through the inductor?

The frequency of oscillation of the current in the given circuit is:

In an oscillating $L-C$ circuit,the maximum charge on the capacitor is $Q$. What is the charge on the capacitor when the energy is stored equally between the electric and magnetic fields?

If $C$ and $L$ denote capacitance and inductance respectively,then the dimensions of $LC$ are

$A$ capacitor of capacitance $C$ is charged to a potential difference $V_0$. Now,this capacitor is connected to an ideal inductor. When $25\%$ of the energy of the capacitor is transferred to the inductor,what will be the potential difference across the capacitor?

Consider an $LC$ circuit,with inductance $L = 0.1 \ H$ and capacitance $C = 10^{-3} \ F$,kept on a plane. The area of the circuit is $1 \ m^2$. It is placed in a constant magnetic field of strength $B_0$ which is perpendicular to the plane of the circuit. At time $t = 0$,the magnetic field strength starts increasing linearly as $B = B_0 + \beta t$ with $\beta = 0.04 \ T \ s^{-1}$. The maximum magnitude of the current in the circuit is . . . . $mA$.