Question

A charged capacitor is connected with a resistor. After how many time constants, does the energy of the capacitor become $\frac{1}{10}th$ of its initial value? $(\ln 10=2.303)$

• A. $2.3$
• B. $1.15$
• C. $0.69$
• D. $1.38$

Answer 1

The correct option is B $1.15$


While discharging, $q=q_0{e}^{\frac{-t}{\tau }}$

So, energy stored,

$U=\frac{q^2}{2C}=\frac{q_0^2}{2C}{e}^{(-\frac{2t}{\tau })}$

Energy at $t=0$ is, $\Rightarrow U_i=\frac{q_0^2}{2C}$

Let say after $n$ time constant energy becomes $\frac{1}{10}th$ of its initial energy, thus $t=n\tau$

$\Rightarrow U_f=\frac{U_i}{10}$

$\Rightarrow \frac{q_0^2}{2C}{e}^{\frac{-2t}{\tau }}=\left(\frac{q_0^2}{2C}\right)\left\{\frac{1}{10}\right\}$

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

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

$\Rightarrow e^{2n}=10$

Taking $\ln$ on both sides we get,

$2n=\ln 10$

$\Rightarrow n=\frac{2.303}{2}=1.15$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.