In Figure, $\mathit{R}\mathbf{=}\mathbf{}\mathbf{14}\mathbf{.}\mathbf{0}\mathbf{\Omega }\mathbf{,}\mathbf{}\mathbf{C}\mathbf{=}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{\mu F}$,and $\mathit{L}\mathbf{=}\mathbf{}\mathbf{54}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mH}$, and the ideal battery has emf $\mathit{\epsilon }\mathbf{=}\mathbf{34}\mathbf{.}\mathbf{0}\mathbf{V}$. The switch is kept at a for a long time and then thrown to position b.

(a)What is the frequency? (b)What is the current amplitude of the resulting oscillations?

1. Resistance attached in the circuit $R=14\mathrm{\Omega }$,
2. Inductance of the inductor, role="math" localid="1663075735599" $L=54.0\mathrm{mH}\mathrm{or}54\times {10}^{-3}\mathrm{H}$
3. Capacitance of a capacitor,$C=6.20\mathrm{\mu F}\mathrm{or}6.20\times {10}^{-6}\mathrm{F}$
4. The emf voltage, $\epsilon =34.0\mathrm{V}$

When a switch is kept at a point the capacitor is charged. After that, the switch is thrown to position b; then the circuit is an LC circuit. From that, we can calculate the angular frequency and frequency. Now capacitor is charged up to maximum voltage, and we can calculate the maximum charge on the capacitor. From that, we can calculate the current amplitude by taking the relation between the current and charge and angular frequency.

Formulae:

The angular frequency of an LC oscillation, $\omega =\frac{1}{\sqrt{LC}}\left(\mathrm{i}\right)$

The charge of the capacitor, $Q=CV\left(\mathrm{ii}\right)$

The current amplitude of the LC circuit, $i=\omega Q\left(\mathrm{iii}\right)$

The frequency of an oscillation,$f=\frac{\omega }{2\pi }\left(\mathrm{iv}\right)$

When the switch is thrown to position b, the circuit is an LC circuit. The frequency of the oscillation can be given using equation (i) in equation (iv) as follows:

$\begin{array}{c}f=\frac{1}{2\pi \sqrt{LC}}\\ =\frac{1}{2\pi \sqrt{\left(54\times {10}^{-3}\right)\left(6.20\times {10}^{-6}\right)}}\\ =275\mathrm{Hz}\end{array}$

Hence, the value of the frequency is $275\mathrm{Hz}$.

After that, the switch is thrown to the point b so that the capacitor is charged$V=34.0\mathrm{V}.$Now, the maximum charge on the capacitor can be calculated using equation (ii) as follows:

$\begin{array}{c}Q=6.20\times {10}^{-6}\times 34.0\\ =2.11\times {10}^{-4}\mathrm{C}\end{array}$

So now, the current amplitude can be calculated using equation (iv) in equation (iii) as follows:

$\begin{array}{c}i=2\pi fQ\\ =2\pi \times 275\times 2.11\times {10}^{-4}\\ =0.365\mathrm{A}\end{array}$

Hence, the value of the current amplitude is $0.365\mathrm{A}$.