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Physics Question on Alternating current

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

A

Att=τ2,q=CV(1−e−1)At t=\frac{\tau}{2}, q=CV\left(1-e^{-1}\right)

B

Work done by the battery is half of the energy dissipated in the resistor

C

Att=τ,q=CV/2At t=\tau, q=CV/2

D

Att=2τ,q=CV(1−e−2)At t=2\tau, q=CV\left(1-e^{-2}\right)

Answer

Att=2τ,q=CV(1−e−2)At t=2\tau, q=CV\left(1-e^{-2}\right)

Explanation

Solution

As switch S is closed and switch S is kept open. Now, capacitor is charging through a resistor R. Charge on a capacitor at any time t is q=q0(1−et/τ)=CV(1−e−t/τ)q=q_{0}\left(1-e^{t/\tau }\right)=CV\left(1-e^{-t/\tau }\right) [As q0=CV]\left[As\,q_{0}=CV\right] At t =τ2, q=CV(1−e−τ2/τ)=CV(1−e−1/2)=\frac{\tau }{2},\,q=CV\left(1-e^{-\tau 2/\tau }\right)=CV\left(1-e^{-1/2}\right) At t =τ, q=CV(1−e−τ/τ)=CV=(1−e−1)=\tau , \,q=CV\left(1-e^{-\tau/\tau}\right)=CV=\left(1-e^{-1}\right) At t =2τ, q=CV(1−e−2τ/τ)=CV(1−e−2)=2\tau,\,q=CV\left(1-e^{-2\tau/\tau}\right)=CV\left(1-e^{-2}\right)