Question

In an $LCR$ circuit as shown in figure, both switches are open initially. Now switch $S_1$ is closed, $S_2$ kept open. $(q$ is charge on the capacitor and 4$\tau =RC$ is capacity time constant).
Which of the following statement is correct?

• A. At $t=\frac{\tau }{2},q=CV(1-e^{-1})$
• B. Work done by the battery is half of the energy dissipated in the resistor
• C. At $t=\tau ,q=CV/2$
• D. At $t=2\tau .q=CV(1-e^{-2})$

Answer 1

The correct option is D At $t=2\tau .q=CV(1-e^{-2})$

At switch $S_1$ is closed and switch $S_2$ is kept open. Now capacitor is charging through a resistor $R$. Charge on a capacitor at any time $t$ is
$q=q_0(1-e^{-t/\tau }=CV(1-e^{-1/t})$
[As $q_0=CV]$
At $t=\frac{\tau }{2},q=CV(1-e^{-\tau /2\tau }=CV(1-e^{1/2})$
At $t=\tau ,q=CV(1-e^{-\tau /\tau }=CV(1-e^{-1})$
At $t=2\tau ,q=CV(1-e^{-2\tau /\tau })=CV(1-e^{-2})$.