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 resistor. The time in which the potential will fall half of 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$

Charge on capacitor during discharging is given by ,

$q=q_0{e}^{-t/RC}$

$⟹\frac{q_0}{2}=q_0{e}^{-t/RC}$

$⟹t=RCln2$

$=(10\times {10}^6)\times (0.1\times {10}^{-6})\times 0.693$

$=0.693s$