A capacitor (C$=40\mu$F) is connected through a resistor ($R=2.5$M$\Omega$) across a battery of negligible internal resistance of voltage $12$ volts. The time after which the potential difference across the capacitor becomes three times to that of resistor is (ln $2=0.693$).

Given: Capacitor $C$ =$40\mu F$

Resistor $R = 2.5 M\Omega$

Voltage V = 12 V

Solution: The charge q is given by,

$q = C\varepsilon (1-e^{\frac{-t}{RC}})$

$i=\frac{\varepsilon }{R}e^{\frac{-t}{RC}}$

According to the question,

$3V_{R}=V_{C}$

$\varepsilon\left ( 1-e^{\frac{-t}{RC}} \right )=3\varepsilon e^{\frac{-t}{RC}}\Rightarrow e^{\frac{-t}{RC}}=\frac{1}{4}$

$\frac{t}{RC}= 2 \ln 2$

$\therefore t=20\times 0.693=13.86 sec$