Let $C$ be the capacitance of a capacitor discharging through a resistor $R$. Suppose $t_{1}$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $t_{2}$ is the time taken for the charge to reduce to one-fourth its initial value. The the ratio $t _{1} / t _{2}$ will be
AIEEE 2010, Diffcult
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$U\,\, = \,\,\frac{1}{2}C{V^2}\,\,\, \Rightarrow \frac{{{U_0}}}{2}\,\, = \,\,\frac{1}{2}C{V_0}^2{e^{ - 2{t_1}/RC}}\,$
$\, \Rightarrow \frac{1}{2}\,\, = \,\,{e^{ - 2{t_1}/RC}}\,\,\,\,[\because \,\,{U_0}\,\, = \,\,\frac{1}{2}C{V_0}^2]$
$\therefore - \frac{{2{t_1}}}{{RC}}\,\, = \,\,\ln 2\,\,\, \Rightarrow {t_1}\,\, = \,\,\frac{{RC\ln 2}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..........\,\,(i)\,$
અને $\frac{{{q_0}}}{4}\,\, = \,\,{q_0}{e^{ - {t_2}/RC}}$
$ - \frac{{{t_2}}}{{RC}}\,\, = \,\,2\ln 2\,\,\,\, \Rightarrow {t_2}\,\, = \,\,2RC\ln 2\,\,\,\,\,\,\,\,\,\,\,...........\,\,(ii)$
$[ \frac{{{t_1}}}{{{t_2}}}\,\, = \,\,\frac{1}{4}]$
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