Question

An uncharged capacitor is connected to a $15\text{V}$ battery through a resistance of $10\Omega$. It is found that in a time of $2\mu \text{s},$ potential difference across the capacitor becomes $5\text{V}$. Find the capacitance of the capacitor. Take $(\ln 1.5=0.4)$

• A. $0.5\mu \text{F}$
• B. $1.5\mu \text{F}$
• C. $2.5\mu \text{F}$
• D. $3.5\mu \text{F}$

Answer 1

The correct option is A $0.5\mu \text{F}$

Given,
Potential difference across the capacitor when $t=0$ is, $V_i=15\text{V}$,
Resistance, $R=10\Omega$,

We know that charge on the capacitors at any time is given by $$q=Q(1-e^{-t/\tau })$$Where, $Q=V_{i}C=15C.$

Here, charge $q$ at any time is given by $q=V_{t}C$ where $V_t$ is potential difference across capacitor at that time.

Here, $V_t=5\text{V}$, so $q=5C$.

Putting the values:

$5C=15C(1-e^{-t/\tau })$

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

Taking $\ln$ both sides,

$\Rightarrow \frac{t}{\tau }=\ln \frac{3}{2}$

As, $\tau =RC$

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

$\Rightarrow C=\frac{t}{R\ln 1.5}=\frac{2\times {10}^{-6}}{10\times 0.4}$

$\Rightarrow C=0.5\mu \text{F}$

Hence, option (a) is the correct answer.