Question

A charged capacitor is connected with a resistor. After how many time constants, does the energy stored in the capacitor becomes ${\left(\frac{1}{10}\right)}^{\text{th}}$ of its initial value?
$[\text{Take}\log {{}_e}^{10}=2.303\log {{}_{10}}^{10}]$

• A. $2.3$
• B. $1.15$
• C. $0.69$
• D. None of the above

Answer 1

The correct option is B $1.15$


While discharging $q=q_0{e}^{-t/\tau }$
So, $U\left(t\right)=\frac{q^2}{2C}=\frac{q{{}_0}^2}{2C}\cdot e^{-2t/\tau }$

And, $U_{t=0}=\frac{q{{}_0}^2}{2C}$

Let, after $n$ time constants, the energy stored in the capacotor becomes ${\left(\frac{1}{10}\right)}^{\text{th}}$ of initial value.

So, at $t=n\tau ,$ $U_f=\frac{U_i}{10}$

i.e. $\frac{q{{}_0}^2}{2C}\cdot e^{-2t/\tau }=\left(\frac{q{{}_0}^2}{2C}\right)\left(\frac{1}{10}\right)$

$\Rightarrow e^{-2t/\tau }=\frac{1}{10}$

$\Rightarrow e^{-2n\tau /\tau }=\frac{1}{10}$

$e^{-2n}=\frac{1}{10}$

$2n=\log {{}_e}^{10}$

$2n=2.303\log {{}_{10}}^{10}$

$2n=2.303$

$\therefore n=1.15$

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