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A fully charge capacitor of capacity $0.1\; \mu F$ is discharged through a resistor of resistance $10^6$ ohms. The time in which the potential falls to half its original value is $[\log_e2=0.6931]$

$(a)\;1.3862\;S \\(b)\;0.6931\;S \\ (c)\;1.414\;S \\ (d)\;1.732\;S $

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Charge at any instant.
$q=q_0e^{\large\frac{-t}{Rc}}$
$t= RC \log_e \bigg( \large\frac{q_0}{q}\bigg)$
Let t be the time after which the potential falls to half its original value.
$\bigg[V=\large\frac{V_0}{2}\bigg]$
$\large\frac{q_0}{q} =\frac{CV_0}{CV}=\frac{V_0}{V}$$=2$
$t= RC \log _e (2)=10^7 \times 0.1 \times 10^{-6}\;\log_e^2$
$\qquad= 0.6931\;S$
Hence b is the correct answer.
answered Mar 25, 2014 by meena.p
 

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