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?     AIEEE Solevd Paper-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 :

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.