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. Then the ratio $t_{1}/t_{2}$ will be

Question

Dr. Abubacker · Accepted Answer

(d) $U = \frac{1}{2}CV^{2}$
$\frac{U_{0}}{2} = \frac{1}{2}CV_{0}^{2}e^{- 2t_{1}/RC}$
$\frac{1}{2} = e^{- 2t_{1}/RC}\ \ \ \ \ \lbrack	herefore U_{0} = \frac{1}{2}CV_{0}^{2}brack$
$- \frac{2t_{1}}{RC} = \ln 2$
$t_{1} = \frac{RCln\ 2}{2}$ …(i)
And $\frac{q_{0}}{4} = q_{0}e^{- t_{2}/RC}$
$- \frac{t_{2}}{RC} = 2\ln 2$
$t_{2} = 2RCln\ 2$ …(ii)
from equation (i) and (ii)
$\frac{t_{1}}{t_{2}} = \frac{1}{4}$

Explanation

Reviewed by Dr. Abubacker

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