question_answer

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?                         [JEE MAIN 2013]

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

B) \[At\text{ }t=\tau ,\text{ }q=CV/2\]

C) \[~At\text{ }t=2\tau ,\text{ }q=CV(1-{{e}^{-2}})\]

D) \[At\text{ }t=\frac{\tau }{2},q=CV(1-{{e}^{-1}})\]

Correct Answer: C

Solution :

[c] Case − 1

Its normal RC circuit

\[{{W}_{bat}}=C{{V}^{2}}\]

\[U=\frac{1}{2}C{{V}^{2}}\]

\[H={{W}_{bat}}-U=\frac{1}{2}C{{V}^{2}}\]

\[\Rightarrow \] \[({{W}_{ba}})=2(H)\]

So (1) is wrong.

\[q=CV(1-{{e}^{-t/\alpha }})\]\[\Rightarrow \]At \[t=\alpha ,\]

at \[t=2\alpha ,\text{ }q=cv(1-{{e}^{2}})\]        3 is correct

\[q=CV(1-{{e}^{-1}})\]            2 is wrong.

At \[t=\frac{\alpha }{2}\]

\[q=CV\left( 1-{{e}^{-1/2}} \right)\] \[\Rightarrow \]            (4) is wrong.