Question

In an oscillating series RLC circuit, find the time required for the maximum energy present in the capacitor during an oscillation to fall to half its initial value. Assume $\mathit{q}\mathbf{=}\mathit{Q}\mathbf{}\mathit{a}\mathit{t}\mathbf{}\mathit{t}\mathbf{=}\mathbf{0}\mathit{s}$

Short answer

The time required for the energy fall to half of its maximum value in series RLC network is $\left(\frac{0.693L}{R}\right)\text{s}$.

Step-by-step solution

The given data

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

Understanding the concept of electrical energy and charge decay

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" $q\left(t\right)=Q{e}^{\frac{-Rt}{2L}}\text{cos}\left(\omega \text{'}t+\varphi \right)$ ...................(ii)

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

Calculation of the required time for the energy fall

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

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

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

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

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 $t=\frac{0.693L}{R}\text{s}$.