Question

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.

Short answer

(a) Charge on capacitor for N=5 is$5.85\mu C$ .

(b) Charge on capacitor for N=10 is $5.52\mu C$.

(c) Charge on capacitor for N=100 is$1.93\mu C$.

Step-by-step solution

The given data

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,$q_0=6.2\mu C$

Understanding the concept of damped oscillations

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)

a) Calculation of the charge on the capacitor for N=5 cycles

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

b) Calculation of the charge on the capacitor for N=10 cycles

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 $5.52\times {10}^{-6}\text{C}$.

c) Calculation of the charge on the capacitor for N=100 cycles

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 $1.93\times {10}^{-6}\text{C}$.