A single-loop circuit consists of a $\mathbf{7}\mathbf{.}\mathbf{20}\mathit{\Omega }$resistor, an $\mathbf{12}\mathbf{.}\mathbf{0}\mathit{H}$inductor, and acapacitor. Initially, the capacitor has a charge of $\mathbf{6}\mathbf{.}\mathbf{20}\mathbf{}\mathit{\mu }\mathit{C}$and the current is zero. (a) Calculate the charge on the capacitor Ncomplete cycles later for N=5. (b) Calculate the charge on the capacitor Ncomplete cycles later for N=10. (c) Calculate the charge on the capacitor Ncomplete cycles later for N=100.

A single loop circuits consisting of:

(a) Resistance,$R=7.2\Omega$

(b) Inductance,$L=12H$

(c) Capacitance,$C=3.20\mu F$

(d) Capacitor charge,$q_0=6.2\mu C$

The damped oscillations in RLC (resistance, inductor, and capacitor) circuit contain electromagnetic energy. Damped oscillations mean oscillations, which fade away with time. In this problem, the charge on the capacitor is given. After N completes cycles of oscillations, the charge on the capacitor will vary. To calculate the change in the charge on the capacitor, we have the given equation of damped oscillations. As given in the problem, t changes with the number of cycles. Therefore, the time value will increase per cycle, t = NT.

Formulae:

Theequation for damped oscillation charge capacitor,$q=Q{e}^{-Rt/2L}\text{cos}\left(\omega \text{'}t+\phi \right)$ (i)

The period of oscillation, $T=2\pi /\omega$ (ii)

For N cycles, the time taken for the stored charge by the capacitor is given using equation (ii) by:

$$\begin{gathered} t=NT \\ =2\pi N/\omega \text{'} \end{gathered}$$

The charge q after N cycles is obtained by substitutingthe above value of time in the damped equation (i) as follows:

$$\begin{gathered} q=Q{e}^{-R2\pi N/\omega 2L}\text{cos}\left(\omega \text{'}\left(2\pi N/\omega \text{'}\right)+\phi \right) \\ =Q{e}^{-RN(2\pi \sqrt{LC}/2L)}\text{cos}\left(2\pi N+\phi \right)\text{(}\because \omega =\sqrt{LC}\text{)} \\ =Q{e}^{-RN\left(2\pi \sqrt{LC}/2\sqrt{L}\sqrt{L}\right)}\text{cos}\left(2\pi N+\phi \right) \\ =Q{e}^{-N\pi R\sqrt{C/L}}\cos \left(\phi \right)\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(1\right) \end{gathered}$$

We note that the initial charge (for) is given using equation (1) as:

$q_0=Qcos\phi \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(2\right)$

where, localid="1663161415633" $q_0=6.2\mu \Omega$is given.

Thus, fromequation (1) and (2), we can write the final equation for charge capacitor for N cycles as follows:

$q_N=q_{0}exp\left(-N\pi R\sqrt{\frac{C}{L}}\right)........................................\left(3\right)$

Putting the given values of N=5 in equation (3), we get the charge on the capacitor as follows:

$$\begin{gathered} q_5=\left(6.2\times {10}^{-6}\right)\exp \left[-5\pi \left(7.2\right)\left(\sqrt{3.2\times {10}^{-6}/12}\right)\right] \\ =5.85\times {10}^{-6}\text{C} \end{gathered}$$

Hence, the value of the charge is$5.85\times {10}^{-6}\text{C}$.'1

Similarly, putting the given values of in equation (3), we get the charge on the capacitor as follows:

$$\begin{gathered} q_{10}=\left(6.2\times {10}^{-6}\right)\exp \left[-10\pi \left(7.2\right)\left(\sqrt{3.2\times {10}^{-6}/12}\right)\right] \\ =5.52\times {10}^{-6}\text{C} \end{gathered}$$

Hence, the value of the charge is $5.52\times {10}^{-6}\text{C}$.

Similarly, putting the given values of in equation (3), we get the charge on the capacitor as follows:

$$\begin{gathered} q_{100}=\left(6.2\times {10}^{-6}\right)\exp \left[-100\pi \left(7.2\right)\left(\sqrt{3.2\times {10}^{-6}/12}\right)\right] \\ =1.93\times {10}^{-6}\text{C} \end{gathered}$$

Hence, the value of the charge is $1.93\times {10}^{-6}\text{C}$.