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 $L$ open. If time constant of this circuit is $r$, then___

• A. after time interval $r$ charge on the capacitor is $CV/2$
• B. after time interval $2r$, 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 $2r$, charge on the capacitor is CV $(1-e^{-1})$

Answer 1

The correct option is B after time interval $2r$, charge on the capacitor is CV $(1-e^{-2})$

$Q=Q_0(1-e^{-t/T})$
$Q=$ CV $(1-e^{-2})$ after time interval 2r.