91Ó°ÊÓ

Consider that energy falls to half of its maximum value, and energy is maximum when$q=Qatt=0s$.

A circuit formed by connecting a resistor, an inductor, and a capacitor in series with each other, the circuit is known as an LCR circuit. The electrical energy and charge decay can be determined for series RLC circuits in terms of the initial charge, resistance, inductance, and time. To find the energy that falls to half of its maximum value, energy is maximum when t=0s.

The electric energy stored by the capacitor, $U_E=\frac{q^2}{2C}$................. (i)

The charge equation of damped oscillations, localid="1664186035739" $q\left(t\right)=Q{e}^{\frac{-Rt}{2L}}\text{cos}\left(\mathrm{Ó\neg }\text{'}t+\mathrm{Ï•}\right)$ ...................(ii)

Here L is the inductance of the coil, C is the capacitance of the capacitor, R is the resistance of the resistor and ${\mathrm{Ó\neg }}^1$ is the angular frequency.

Substituting the charge value from equation (ii) in equation (i), we can get the energy equation as follows:

$$\begin{gathered} U_E=\frac{1}{2C}{Q}^2{e}^{\frac{-2Rt}{2L}}co{s}^2\left(\mathrm{Ó\neg }\text{'}t+\mathrm{Ï•}\right) \\ =\frac{Q^2}{2C}{e}^{\frac{-Rt}{L}}co{s}^2\left(\mathrm{Ó\neg }\text{'}t+\mathrm{Ï•}\right)....................\left(a\right) \end{gathered}$$

When the maximum energy falls to half of its initial value, initial electrical energy is given using equation (i) by:

$$\begin{gathered} U_E=0.5U_i \\ =0.5\frac{Q^2}{2C} \end{gathered}$$

Now, substituting this above value in equation (a), we can get

$$\begin{gathered} 0.5\frac{Q^2}{2C}=\frac{Q^2}{2C}{e}^{\frac{-Rt}{L}}co{s}^2\left(\mathrm{Ó\neg }\text{'}t+\mathrm{Ï•}\right) \\ 0.5=e^{\frac{-Rt}{L}}co{s}^2\left(\mathrm{Ó\neg }\text{'}t+\mathrm{Ï•}\right) \\ 0.5=e^{\frac{-Rt}{L}} \end{gathered}$$

Taking the natural log on both sides, we can get the time requiredfor the energy fall to half of its maximum value in series RLC network as follows:

$$\begin{gathered} \text{ln}\left(0.5\right)=ln\left(e^{\frac{-Rt}{L}}\right) \\ -0.693=\frac{-Rt}{L} \\ t=\frac{0.693L}{R}\text{s} \end{gathered}$$

Hence, the value of required time is $t=\frac{0.693L}{R}\text{s}$.