Question

The capacitor of capacitance $C$ can be charged (with the help of a resistance R) by a voltage source $V$, by closing switch $S_{1}$ while keeping switch $S_{2}$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_{2}$ and opening $S_{1}$.
Initially, the capacitor was uncharged. Now, switch $S_{1}$ is closed and $S_{2}$ is kept open. If time constant of this circuit is $r$, then

• A. after time interval $T$, charge on the capacitor is $CV / 2 .$
• B. after time interval $2 T$, charge on the capacitor is $CV (1$ $\left.-e^{-2}\right)$.
• C. the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged.
• D. after time interval $2 T$, charge on the capacitor is $CV (1$ $\left.-e^{-1}\right)$.

Answer 1

Answer: (B)

During charging, we get
$q=q_{0}\left(1-e^{-\frac{t}{\tau}}\right)$
where $q_{0}= CV =$ steady-state charge.
At $t =2 T$, we get
$q =q_{0}\left(1-e^{-2}\right)$