Question

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

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

Answer 1

Answer: (C)

$U=\frac{1}{2}\frac{q^2}{C}=\frac{1}{2C}{\left(q_0{e}^{-t/T}\right)}^2=\frac{q_0^2}{2C}=e^{-2t/T}$ (where $\tau =CR$)
$U=U_i{e}^{-2t/\tau }$
$\frac{1}{2}U=U_i{e}^{-2t/\tau }$
$\frac{1}{2}=e^{-2t/\tau }\Rightarrow t_1=\frac{T}{2}ln2$
Now $q=q_0{e}^{-t/T}$
$\frac{1}{4}q=q_0{e}^{-t/2T}$
$t_2=Tln4=2Tln2$
$\therefore \frac{t_1}{t_2}=\frac{1}{4}$