In an LCR circuit as shown below both switches are open initially. Now switch $\mathrm{S}_{1}$ is closed, $\mathrm{S}_{2}$ kept open. ( q is charge on the capacitor and $\tau=\mathrm{RC}$ is capacitive time constant). Which of the following statement is correct?

(B) At $t=2 \tau, q=\operatorname{CV}\left(1-e^{-2}\right)$

(A) At $\mathrm{t}=\tau, \mathrm{q}=\mathrm{CV} / 2$

(C) $\mathrm{At} t=\frac{\tau}{2}, \mathrm{q}=\mathrm{CV}\left(1-\mathrm{e}^{-1}\right)$

(D) Work done by the battery is half of the energy dissipated in the resistor.

JEE Main · 2013 · 07 Apr 2013 · Offline · Current Electricity

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

τ = RC

is capacitive time constant). Which of the following statement is correct?

Answer: (B)

A capacitor charging through a resistor builds up its charge as

q = C V (1 - e^{- t / τ})

, where the time constant

τ = R C

sets the pace of the growth, so the charge at any instant is just a matter of reading off the exponent. Putting

t = 2 τ

makes the exponent

- 2 τ / τ = - 2

, and the charge carried by the capacitor at that moment is

q = C V (1 - e^{- 2})

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option