Question

A capacitor is connected to a $10\text{V}$ battery through a resistance of $20\Omega$. It is found that the voltage across the capacitor rises to $8\text{V}$ in $1\mu \text{s}$. The capacitance of the capacitor is approximately equal to [Take $\ln 5=1.6$]

• A. $4\times {10}^{-8}\text{F}$
  
• B. $3\times {10}^{-8}\text{F}$
  
• C. $2\times {10}^{-8}\text{F}$
  
• D. $6\times {10}^{-8}\text{F}$
  

Answer 1

The correct option is B $3\times {10}^{-8}\text{F}$



The charge on the capacitor during charging is given by,

$Q=Q_0(1-e^{\frac{-t}{RC}})$

Hence, voltage across the capacitor at any time $t$ will be,

$V=\frac{Q}{C}=\frac{Q_0}{C}(1-e^{\frac{-t}{RC}})$

Here, at $t=1\mu \text{s}$, $V=8\text{V}$ whereas the steady state voltage will be $(\frac{Q_0}{C}=10\text{V})$ when the capacitor is fully charged.

$\Rightarrow 8=10(1-e^{\frac{-t}{RC}})$

$\Rightarrow e^{\frac{-t}{RC}}=\frac{1}{5}$

$\Rightarrow \frac{t}{RC}=\ln 5=1.6$

As, $t=1\mu \text{s}={10}^{-6}\text{s}$ and $R=20\Omega$

$\therefore C=\frac{{10}^{-6}}{1.6\times 20}\approx 3\times {10}^{-8}\text{F}$

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