Question

A capacitor is discharging through a resistor $R$. Consider in time $t_1$, the energy stored in the capacitor reduces to half of its initial value and in time $t_2$, the charge stored reduces to one eighth of its initial value. The ratio $\frac{t_1}{t_2}$ will be

• A. $\frac{1}{6}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{6}$

For a discharging capacitor when energy reduces to half the charge would become $\frac{1}{\sqrt{2}}$ times the initial value.
$\Rightarrow {\left(\frac{1}{2}\right)}^{1/2}=e^{-t_1/\pi }$
$\text{similarly},{\left(\frac{1}{2}\right)}^3=e^{-t_2/\pi }$
$\Rightarrow \frac{t_1}{t_2}=\frac{1}{6}$