Question

In a charging circuit having resistance $R=0.5\text{M}\Omega$ and capacitance, $C=2\mu \text{F}$, after what time will the energy stored in the capacitor reach to $75\%$ of its maximum value?

• A. $\ln \frac{4}{4-\sqrt{3}}\text{s}$
• B. $\ln \frac{2}{2-\sqrt{3}}\text{s}$
• C. $2\ln \frac{2}{2-\sqrt{3}}\text{s}$
• D. $\ln \frac{2}{2+\sqrt{3}}\text{s}$

Answer 1

The correct option is B $\ln \frac{2}{2-\sqrt{3}}\text{s}$

At steady state, let charge on the capacitor be $q_o$, then the energy stored in the capacitor is, $$U_i=\frac{q_o^2}{2C}$$
And after time $t$, let charge on the capacitor be $q$, then the energy will be,

$U_f=\frac{q^2}{2C}$

According to problem, after time $t$,

$U_f=\frac{75U_i}{100}=\frac{3U_i}{4}$

$\Rightarrow \frac{q^2}{2C}=\frac{3}{4}\times \frac{q_0^2}{2C}$

$\Rightarrow q=\frac{\sqrt{3}{q}_0}{2}$

In the case of charging capacitor, charge at any time $t$,

$$q=q_0[1-e^{-t/\tau }]$$

$\Rightarrow \frac{\sqrt{3}{q}_0}{2}=q_0[1-e^{-t/\tau }]$

$\Rightarrow \frac{\sqrt{3}}{2}=1-e^{-\frac{t}{\tau }}$

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

Taking $\ln$ on both sides,

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

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

As, $\tau =RC$

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

$\Rightarrow t=0.5\times {10}^6\times 2\times {10}^{-6}\ln \frac{2}{2-\sqrt{3}}$

$\therefore t=\ln \frac{2}{2-\sqrt{3}}\text{s}$

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