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)

$U = \frac{1}{2} \frac{q^{2}}{C} = \frac{1}{2C}\left(q_{0}e^{-t / T}\right)^{2} = \frac{q^{2}_{0}}{2C} = e^{-2t / T}$ (where $\tau = CR$)
$U = U_{i}e^{-2t /\tau}$
$\frac{1}{2}U = U_{i}e^{-2t /\tau }$
$\frac{1}{2} = e^{-2t /\tau } \Rightarrow t_{1} = \frac{T}{2} ln\,2$
Now $q = q_{0}e^{-t /T}$
$\frac{1}{4} q = q_{0}e^{-t /2T}$
$t_{2} = T\,ln\,4 = 2T\,ln\,2$
$\therefore \quad \frac{t_{1}}{t_{2}} = \frac{1}{4}$