Question

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 (in $2=0.693$).

• A. $13.86$ sec.
• B. $6.48$ sec.
• C. $3.24$ sec.
• D. $20.52$ sec.

Answer 1

The correct option is A $13.86$ sec.

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

Resistor $R=2.5M\Omega$

Voltage V = 12 V

Solution: The charge q is given by,

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

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

According to the question,

$3V_R=V_C$

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

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

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

Hence, the correct option is (A).