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 $2T$, charge on the capacitor is $CV(1$ $-e^{-2})$.
• 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 $2T$, charge on the capacitor is $CV(1$ $-e^{-1})$.

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 $t=2T$, we get
$q=q_0\left(1-e^{-2}\right)$