Question

The time when the voltage across the resistor drops to nearly $37\%$ of the value just after the switch $S_w$ is closed $(R=100k\Omega ,C=1\mu \text{F})$ is:

• A. $0.15\text{s}$
• B. $0.30\text{s}$
• C. $0.45\text{s}$
• D. $0.60\text{s}$

Answer 1

The correct option is A $0.15\text{s}$

Equivalnet capacitance,

$C_{\text{eq}}=\left(\frac{C\times C}{C+C}\right)+C=\frac{C}{2}+C=\frac{3C}{2}$

Circuit can be redrawn as:


Time constant, $\tau =\frac{3RC}{2}$

After switch is closed,

$i=i_0{e}^{\frac{-t}{\tau }}$

$\Rightarrow iR=i_{0}R{e}^{\frac{-t}{\tau }}$

$\Rightarrow v=\epsilon e^{-\frac{t}{\tau }}$

$\Rightarrow 0.37\epsilon =\epsilon e^{-\frac{t}{\tau }}$

$\Rightarrow \frac{t}{\tau }=-\ln \left(0.37\right)$

$\Rightarrow t=\frac{3RC}{2}\left(0.99\right)$

$\Rightarrow t=\frac{3\times 100\times {10}^3\times {10}^{-6}}{2}\times \left(0.99\right)$

$\therefore t\approx 0.15\text{s}$