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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Electromagnetic Induction

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 ?

$\begin {array} {1 1} (A)\;At \: t=\tau, \: q=\large\frac{CV}{2} \\ (B)\;At \: t=2\tau, \: q=CV(1-e^{-2}) \\ (C)\;At \: t=\large\frac{\tau}{2},\normalsize \: q=CV(1-e^{-1}) \\ (D)\;Work\: done \: by\: the \: battery \: is \: half\: of \: the \: energy\: dissipated \: in \: the \: resistor \end {array}$

 

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1 Answer

Charge on the capacitor at any time 't' is
$ q = CV (1-e^{\large\frac{-t}{t}})$
at $ t = 2\tau$
$ q = CV (1-e^{-2})$
Ans : (B)
answered Feb 15, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1
 

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