In an oscillating LCcircuit with$\mathit{C}\mathbf{=}\mathbf{}\mathbf{64}\mathbf{.}\mathbf{0}\mathit{\mu }\mathit{F}$, the current is given by$\mathit{i}\mathbf{=}\mathbf{}\left(1.60\right)\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{2500}\mathit{t}\mathbf{+}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{680}\mathbf{)}$, where tis in seconds, Iin amperes, and the phase constant in radians.(a) How soon after$\mathit{t}\mathbf{=}\mathbf{}\mathbf{0}$will the current reach its maximum value? (b) What is the inductance L? (c) What is the total energy?

1. The value of the capacitor, $C=0.64\mathrm{\mu F}\mathrm{or}0.64\times {10}^{-6}\mathrm{F}$
2. The current equation,$i=\left(1.60\right)sin\left(2500t+0.680\right)$

As current is a function of time, we can compare the current equation with the given current equation to find the time for the current to reach its maximum value. As we know the relation between the inductance and the angular frequency, we can calculate the inductance. After that, we calculate the energy of the LC circuit.

Formulae:

The basic current equation of the LC circuit, $i=Isin\left(\omega t+\varphi \right)$ (i)

The angular frequency of the LC oscillation, $\omega =\frac{1}{\sqrt{LC}}$ (ii)

The electric energy stored in the capacitor, $U=\frac{1}{2}L{I}^2$ (iii)

For maximum current value, the term from equation (i) should be

$\begin{array}{c}\sin (\omega t+\varphi )=1\\ \omega t+\varphi =\frac{\pi }{2}\end{array}$

Now, as current is a function of time,so by comparing equation (i) with the given current equation we get

$\begin{array}{c}\frac{\pi }{2}=2500t+0.680\\ 2500t=\frac{3.14}{2}-0.680\\ 2500t=0.89\\ t=\frac{0.89}{2500}\\ =3.56\times {10}^{-4}\mathrm{s}\end{array}$

Hence, the value of the time is $3.56\times {10}^{-4}\mathrm{s}$.

The angular frequency is given by comparing equation (i) and the given current equation as: $\omega =2500\mathrm{rad}/\sec$

Now, substituting this value in equation (ii), we can get the value of the inductance as follows:

$\begin{array}{c}L=\frac{1}{{\omega }^{2}C}\\ =\frac{1}{\left(2500\right)\left(64.0\times {10}^{-6}\right)}\\ =2.50\times {10}^{-3}\mathrm{H}\end{array}$

Hence, the value of inductance is $2.50\times {10}^{-3}\mathrm{H}$.

The total energy of the LC oscillator is given using the given data in equation (iii) as follows:

$\begin{array}{c}U=\frac{1}{2}\times 2.50\times {10}^{-3}\times {\left(1.60\right)}^2\\ =3.20\times {10}^{-3}\mathrm{J}\end{array}$

Hence, the value of the energy is $3.20\times {10}^{-3}\mathrm{J}$.