Q. Let C be the capacitance of a capacitor discharging through a resister 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. Then the ratio $\frac{t_{1}\ }{t_{2}}$ will be
A) 1
B) $\frac{1}{2}$
C) $\frac{1}{4}$
D) 2
Explanation
(c) $U = \frac{1}{2}\frac{q^{2}}{C} = \frac{1}{2C}\left( q_{0}e^{- t/\tau} \right)^{2} = \frac{q_{0}^{2}}{2C}e^{- 2t/\tau}\ ($where$\ \tau = CR)$
$U = U_{i^{e^{- 2t/\tau}\ }}$
$\frac{1}{2}U_{i} = U_{i^{e^{- 2t_{1}/\tau}\ }}$
$\frac{1}{2} = e^{- 2t_{1}/\tau}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t_{1} = \frac{\tau}{2}\ln 2$
Now $q = q_{0^{e^{- t/\tau}}}$
$\frac{1}{2}q_{0} = q_{0^{e^{- t_{2}/\tau}}}$
$t_{2} = \tau\ln{4 = 2\tau\ln 2}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{t_{1}}{t_{2}} = \frac{1}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.