Question

A capacitor charged to $50\text{V}$ is discharged by connecting the two plates. The connections are made at $t=0$. At $t=10\text{ms}$, the potential difference across the plates is found to be $1.0\text{V}$. Find the potential difference at $t=20\text{ms}$.

• A. $0.01\text{V}$
• B. $0.04\text{V}$
• C. $0.02\text{V}$
• D. $0.05\text{V}$

Answer 1

The correct option is C $0.02\text{V}$

We know that in a discharging circuit, charge on the capacitor at any time $t$ is given by $$q=q_o{e}^{-\frac{t}{\tau }}$$
Also, $V\propto q$, so similar relation can be written for the potential difference, $$V=V_0{e}^{\frac{-t}{\tau }}$$Where, $V_0=50\text{V}$

Given, at $t=10\text{ms}$, $V=1\text{V}$, putting

$1=50e^{-\frac{10\text{ms}}{\tau }}$

$\Rightarrow e^{-\frac{10}{\tau }}=\frac{1}{50}......\left(i\right)$

Let, at $t=20\text{ms},V=V_2$, then

$V_2=50e^{-\frac{20}{\tau }}$

$\Rightarrow V_2=50{\left[e^{-\frac{10}{\tau }}\right]}^2$

Using equation $\left(i\right)$,

$\Rightarrow V_2=50\times {\left[\frac{1}{50}\right]}^2$

$\Rightarrow V_2=0.02\text{V}$

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