Question

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

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

Answer 1

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

Charge on capacitor at time t is
$q=CV(1-e^{-t/\tau })$
At $t=2\tau$ $q=CV(1-e^{-2})$