Question

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

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

Answer 1

Answer: (B)

By equation of charging
$q=q_0\left(1-e^{-t/CR}\right)$
According to question
$\frac{q}{q_0}=\frac{1}{2}=0.50$
$\therefore 0.50=1-e^{-t/CR}$
$e^{-t/CR}=1-0.50=0.50$
$e^{t/CR}=2$
or $\frac{t}{CR}={\log }_{e}2$
or $\frac{t}{CR}=2.3026{\log }_{10}2$
or $t=CR\times 2.3026{\log }_{10}2$
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 $t=0.693s$