Question

A capacitor loses a certain fraction of its charge in $30\text{s}$ because of humidity in the air giving rise to leakage between its terminals. When a $4M\Omega$ resistance is connected between its terminals, in the presence of humidity, the same fraction of charge is lost in $7.5\text{s}$. Find the leakage resistance due to humidity $\left(\text{in M}\Omega \right)$

Answer 1

Leakage resistance $R_l$

External resistance $=R=4M\Omega$

Charge on the discharging capacitor after time $t$ is,

$q=q_0{e}^{\left(-\frac{t}{RC}\right)}$

Given, final charge on the capacitor is same

$q_0{e}^{-\left(\frac{t_1}{R_{l}C}\right)}=q_0{e}^{-\left(\frac{t_2(R_l+R)}{R_{l}RC}\right)}$

Hence, $\frac{t_1}{R_{l}C}=\frac{t_2(R_l+R)}{R_{l}RC}$

Substituting the known values we get,

$\frac{30}{R_{l}C}=\frac{7.5(R_l+4)}{R_{l}4C}$

$\therefore R_l=\frac{90}{7.5}=12M\Omega$