Question

In the given circuit, switch is closed at $t=0$. The time after which the current becomes $2.5\mu \text{A}$ is given by $(\text{Take}\ln 2=0.69)$

• A. $10\text{s}$
• B. $5\text{s}$
• C. $6.93\text{s}$
• D. $0.693\text{s}$

Answer 1

The correct option is C $6.93\text{s}$

According to problem,

$RC=(2\times {10}^6)\times (5\times {10}^{-6})=10\text{s}$

In the case of discharging, current at time $t$,$$I=I_0{e}^{-t/RC}$$Substituting, $I_0=\frac{q_0}{RC}$ and the values of different parameters provided in the problem,

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

$\Rightarrow 2.5\times {10}^{-6}=\frac{50\times {10}^{-6}}{10}{e}^{-t/RC}$

$\Rightarrow 2.5\times {10}^{-6}=5\times {10}^{-6}{e}^{-t/10}$

$\Rightarrow e^{-t/10}=\frac{1}{2}$

$\Rightarrow e^{t/10}=2$

Taking $\ln$ on the both the sides,

$\Rightarrow \frac{t}{10}=\ln 2$

$\Rightarrow t=10\ln 2=6.93\text{s}$

Hence, option $\left(c\right)$ is the correct answer.