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 resistor. The time in which the potential will take to fall to half its original value is
(Given, $lo{g}_{10}2=0.3010)$

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

Answer 1

The correct option is B 0.693 s

Capacitor discharging equation is
$V=V_0{e}^{-t/2}NowV=\frac{V_0}{2}{e}^{t/2}=2t=2ln2\tau =RC=1t=2.303lo{g}_{10}2=0.693s$