Question

A capacitor is connected to a $12\text{V}$ battery through a resistance of $10\Omega$. It is found that the potential difference across the capacitor rises to $4.0\text{V}$ in $1\mu \text{s}$. Find the capacitance of the capacitor. $(\ln \frac{3}{2}=0.405)$

• A. $0.25\mu \text{F}$
• B. $0.5\mu \text{F}$
• C. $2.5\mu \text{F}$
• D. $0.05\mu \text{F}$

Answer 1

The correct option is A $0.25\mu \text{F}$

The charge on the capacitor during charging is given by $$Q=Q_0(1-e^{\frac{-t}{RC}})$$Hence, the potential difference across the capacitor is, $$V=\frac{Q}{C}=\frac{Q_0}{C}(1-e^{\frac{-t}{RC}})$$
Here, at $t=1\mu \text{s}$, the potential difference is $4\text{V}$, whereas the steady potential difference is $\frac{Q_0}{C}=12\text{V}$.

So,
$4=12(1-e^{\frac{-t}{RC}})$

$\Rightarrow 1-e^{\frac{-t}{RC}}=\frac{1}{3}$

$\Rightarrow e^{\frac{-t}{RC}}=\frac{2}{3}$

$\Rightarrow \frac{t}{RC}=\ln \frac{3}{2}=0.405$

$\Rightarrow RC=\frac{t}{0.405}=\frac{1\mu \text{s}}{0.405}=2.469\mu \text{s}$

$\therefore C=\frac{2.469\mu \text{s}}{10\Omega }=0.25\mu \text{F}$

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