91Ó°ÊÓ

1. The inductance is $L=1.10\text{mH or}1.10×{10}^{-3}\text{H}$.
2. The capacitance is $C=4.00\mathrm{μ}F$or $4.00×{10}^{-6}\text{F}$.
3. The maximum charge on the capacitor is $Q=3.00\mathrm{μ}C$or $3.00×{10}^{-6}\text{C}$.

According to energy conservation, we can say that electrical energy is equal to magnetic energy, and by substituting the energy values in the energy conservation equation, we can find the equation for the maximum current. Using this equation of current, we can find the required value of the maximum current in the circuit.

Formulae:

The energy stored in the electric field of the capacitor, ${\mathit{U}}_{\mathbf{E}}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{2}\mathbf{C}}$ (i)

The energy stored in the magnetic field of the inductor at any time, ${\mathit{U}}_{\mathbf{B}}\mathbf{=}\frac{\mathbf{L}{\mathbf{i}}^{\mathbf{2}}}{\mathbf{2}}$ (ii)

The energy in an oscillating LC circuit, $\mathbf{U}\mathbf{=}{\mathbf{U}}_{\mathbf{E}}\mathbf{\left(}\mathbf{o}\mathbf{r}{\mathbf{U}}_{\mathbf{B}}\mathbf{\right)}$ (iii)

If $Q$is the maximum charge on the capacitor and role="math" localid="1662802970777" $I$is the maximum current, then we know that the energy in an oscillating LC circuit is given by substituting equations (i) and (ii) in equation (iii) as follows: (according to energy conservation)

$\frac{Q^2}{2C}=\frac{L{i}^2}{2}$

Therefore, the maximum current is given using above equation as follows:

$$\begin{gathered} dI=\frac{Q}{\sqrt{LC}} \\ =\frac{\left(3.00×{10}^{-6}\text{C}\right)}{\sqrt{\left(1.10×{10}^{-3}\text{H}\right)\left(4.00×{10}^{-6}\text{F}\right)}} \\ =4.52×{10}^{-2}\text{A} \end{gathered}$$

Hence, the value of the maximum current is $4.52×{10}^{-2}\text{A}$.