Question

A capacitor of capacitance $1\mu \text{F}$ is discharging through a resistance. Initially, the potential difference across the capacitor is $12\text{V}$. At $t=\frac{1}{10}\text{sec}$, potential difference across capacitor becomes $6\text{V}$. Find the value of resistance $R$.
[Take $\ln 2=0.693$]

• A. $1.443\times {10}^4\Omega$
• B. $2.886\times {10}^4\Omega$
• C. $1.443\times {10}^5\Omega$
• D. $2.886\times {10}^5\Omega$

Answer 1

The correct option is C $1.443\times {10}^5\Omega$


While discharging, let the initial charge is $q_0$

Now, $q=q_o{e}^{-t/\tau }$
So, potential across the capacitor is,
$V=\left(\frac{q}{C}\right)=\left(\frac{q_0}{C}\right)e^{-t/\tau }$

It is given that $V_0=\frac{q_0}{C}=12$

So, $V=12e^{-t/RC}$

Also, at $t=\frac{1}{10}\text{sec}$, $V=6$

$\Rightarrow 6=12e^{\left(\frac{-1}{10RC}\right)}$

$\Rightarrow \frac{1}{2}=e^{\left(\frac{-1}{10RC}\right)}$

$\Rightarrow ln2=\frac{1}{10RC}$

$\Rightarrow 0.693=\frac{1}{10R\times {10}^{-6}}$

$\therefore R=\frac{{10}^5}{0.693}=1.443\times {10}^5\Omega$

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