Question

A capacitor of capacity $0.1\, \mu F$ connected in series to a resistor of $10\, M \Omega$ 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 $\left(\right.$ Given, $\left.\log _{10} 2=0.3010\right)$

• A. 2 s
• B. 0.693 s
• C. 0.5 s
• D. 1.0 s

Answer 1

Answer: (B)

By equation of charging
$q=q_{0}\left(1-e^{-t / C R}\right)$
According to question
$\frac{q}{q_{0}}=\frac{1}{2}=0.50$
$\therefore 0.50=1-e^{-t / C R}$
$e^{-t / C R} =1-0.50=0.50$
$e^{t / C R} =2$
or $\frac{t}{C R}=\log _{e} 2$
or $\frac{t}{C R}=2.3026 \log _{10} 2$
or $t=C R \times 2.3026 \log _{10} 2$
or $t=0.1 \times 10^{-6} \times 10 \times 10^{6} \times 2.3026 \log _{10} 2$
or $t=2.3026 \times 0.3010$
or $t=0.693\, s$