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. |
You need to login to perform this action.
You will be redirected in
3 sec