Question

In a charging circuit, after how many time constant will the energy stored in the capacitor reach one-half of its equilibrium value?$[\text{Take},\ln \frac{\sqrt{2}}{\sqrt{2}-1}=1.23]$

• A. $1.23\tau$
• B. $1\tau$
• C. $2.23\tau$
• D. $3\tau$

Answer 1

The correct option is A $1.23\tau$

At steady state, let charge on the capcitor be $q_0$, then the energy stored on the capacitor is,$$U_i=\frac{q_0^2}{2C}$$
And after time $t$, let charge on the capacitor be $q$, then the energy be,

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

According to problem after time $t$,

$U_f=\frac{U_i}{2}$

$\Rightarrow \frac{q^2}{2C}=\frac{1}{2}\times \frac{q_0^2}{2C}$

$\Rightarrow q=\frac{q_0}{\sqrt{2}}$

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

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

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

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

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

Taking $\ln$ on both sides,

$\Rightarrow \frac{t}{\tau }=\ln \frac{\sqrt{2}}{\sqrt{2}-1}$

$\therefore t=\tau \ln \frac{\sqrt{2}}{\sqrt{2}-1}=1.23\tau$

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