Question

For the arrangement shown in the figure, the switch is closed at $t=0$. The time after which the current becomes $2.5\mu A$ is given by (take ln 2 = 0.69)

• A. 10 $s$
• B. 5 $s$
• C. 7 $s$
• D. 0.693 $s$

Answer 1

The correct option is C

7 $s$

In the case of discharging, $I=I_0{e}^{-\frac{t}{RC}}$ ...$\left(i\right)$
Given that
Initial charge on capacitor $q_0=50\times {10}^{-6}C$
Capacitance $C=5\times {10}^{-6}$
$R=2\times {10}^6\Omega$
Substituting these in eq$\left(i\right)$
Since $RC$ is the time constant for the $R$-$C$ circuit
$\therefore I_0=\frac{q_0}{RC}$

$2.5\times {10}^{-6}=\frac{q_0}{RC}{e}^{-\frac{t}{RC}}=5\times {10}^{-6}{e}^{-\frac{t}{10}}$

$e^{\frac{t}{10}}=2$

Taking log on both the sides, we get

$\frac{t}{10}=ln2$

$t=7s$