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 reduced 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. 2
• B. 1
• C. 1/2
• D. 1/4

Answer 1

The correct option is D 1/4

$\begin{array}{c}q=q_0{e}^{\frac{-t}{t}}\\ U=\frac{q^2}{2C}\\ \Rightarrow U=U_i{e}^{\frac{-2t}{t}}\end{array}$

On substituting, $q=\frac{q_0}{4}$ and $=\frac{U_t}{2}$,

hence, Option $D$ is correct.