What resistance Rshould be connected in series with an inductance L=220mHand capacitance C=12.0Ffor the maximum charge on the capacitor to decay to 99.0% of its initial value in$\mathbf{50}\mathbf{.}\mathbf{0}$ cycles?

(Assume${\mathit{\omega }}_{\mathbf{1}}\mathbf{\approx }\mathit{\omega }$ .)

(a) Series inductance$L=220\text{mH}=220\times {10}^{-3}\text{H}$

(b) Series capacitance role="math" localid="1663164194202" $C=12.0\mu F=12.0\times {10}^{-6}\text{F}$

(c) 99% of initial charge decays in 50 cycles

Formulae:

The charge equation for decay within the capacitor or equation of a damped oscillation, $q\left(t\right)=Q{e}^{\frac{-Rt}{2L}}\text{cos}\left(\omega t+\varphi \right)$ .........................(i)

The angular frequency of oscillation, $\omega =\frac{2\pi }{T}$........................(ii)

The frequency of LC oscillations, $\omega =\frac{1}{\sqrt{LC}}$...........................(iii)

Here L is the inductance of the coil, C is the capacitance of the capacitor and T is the time period.

Using equations (ii) and (iii), the period of one cycle can be given as follows:

$T=2\pi \sqrt{LC}$

Then, the time taken for 50 cycles can be given as:

$$\begin{gathered} t=50\times 2\pi \sqrt{LC} \\ =50\times 2\pi \sqrt{\left(220\times {10}^{-3}H\right)\left(12.0\times {10}^{-6}F\right)} \\ =0.51s \end{gathered}$$

Charge has decayed 99% in 50 cycles.

Thus, the charge equation from equation (i) becomes:

$$\begin{gathered} 0.99Q=Q{e}^{\frac{-Rt}{2L}}\text{cos}\left(\omega t+\varphi \right) \\ 0.99=e^{\frac{-Rt}{2L}}\text{cos}\left(\omega t+\varphi \right) \\ 0.99=e^{\frac{-Rt}{2L}} \end{gathered}$$

Taking natural log on both sides, we can solve for the value of series resistance as follows:

$$\begin{gathered} \ln \left(0.99\right)=\ln \left(e^{\frac{-Rt}{2L}}\right) \\ -0.01=\frac{-Rt}{2L} \\ R=\frac{0.01\Omega \times 2\text{L}}{t} \\ =\frac{0.01\Omega \times 2\times \left(220\times {10}^{-3}\right)\text{m}}{0.51\text{s}} \\ =8.66\times {10}^{-3}\Omega \end{gathered}$$

Hence, the value of resistance is $8.66\times {10}^{-3}\Omega$.