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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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In an $LCR$ circuit as shown below both switches are open initially. Now switch $S1$ is closed, $S2$ kept open. ($q$ is charge on the capacitor and $ \tau = RC$ is capacitive time constant). Which of the following statement is correct?

 

$\begin {array} {1 1} (A)\;At\: t = \tau,\: q = \large\frac{CV}{2} & \quad (B)\;At\: t = 2\tau, \: q=CV ( 1-e^{-2}) \\ (C)\;At\: t = 2\tau, \: q=CV ( 1-e^{-1}) & \quad (D)\;At\: t = \large\frac{\tau}{2}, \: q=CV ( 1-e^{-2}) \end {array}$

 

Can you answer this question?
 
 

1 Answer

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In a DC circuit the inductance is ineffective and so the circuit reduces to an RC circut.
q = CV ($1-e^t/\tau$)
$\tau=RC$
the standard formula for the variation of charge on a capacitor with time.
by sustituting the value of $t=\tau$ we get the answer
Ans : (B)

 

answered Feb 8, 2014 by thanvigandhi_1
edited Sep 22, 2014 by thagee.vedartham
 

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