A capacitor of capacity $0.1 \mu \mathrm{F}$ connected in series to a resistor of $10 \mathrm{M} \Omega$ is…
A capacitor of capacity $0.1 \mu \mathrm{F}$ connected in series to a resistor of $10 \mathrm{M} \Omega$ is charged to a certain potential and then made to discharge through the resistor. The time in which the potential will take to fall to half its original value is ______ $\times 10^{-3}$ (Given, $\log _{10} 2=0.3010$)
Solution
By equation of charging
$q=q_0\left(1-e^{-t / C R}\right)$
According to question
$\frac{q}{q_0}=\frac{1}{2}=0.50$
$\therefore \quad 0.50=1-e^{-t / C R}$
$e^{-t / C R}=1-0.50=0.50$
$e^{t / C R}=2$
or $\quad \frac{t}{C R}=\log _e 2$
or $\quad \frac{t}{C R}=2.3026 \log _{10} 2$
or $\quad t=C R \times 2.3026 \log _{10} 2$
or $t=0.1 \times 10^{-6} \times 10 \times 10^6 \times 2.3026 \log _{10} 2$
or $\quad t=2.3026 \times 0.3010$
or $\quad t=0.693 \mathrm{~s}$