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

The correct option is C $\frac{1}{4}$

$U=\frac{q^2}{2C}=\frac{1}{2C}(q_0{e}^{-t/\tau }{)}^2=\frac{q_0^2}{2C}{e}^{-2t/\tau }$ (where $\tau =$ CR) (using the standard formula)
$U=U_i{e}^{2t/\tau }$
$\frac{1}{2}{U}_i=U_i{e}^{-2t_1/\tau }$
$\frac{1}{2}=e^{-2t_1/\tau }=t_1=\frac{\tau }{2}$ ln2
Now $q=q_0{e}^{-t/\tau }$
$\frac{1}{4}{q}_0=q_0{e}^{-t_2/\tau }$
$t_2=\tau$ $ln4$ $=2\tau$ $In2$
.$\cdot .\frac{t_1}{t_2}=\frac{1}{4}$