91Ó°ÊÓ

1. The inductance and capacitance in the LC circuit,$L=25.0\mathrm{mH}\mathrm{or}25×{10}^{-3}\mathrm{H}$,$C=7.80\mathrm{μ¹ó}\mathrm{or}7.80×{10}^{-6}\mathrm{F}$
2. The current value at$t=0$,$i=9.20\mathrm{mA}\mathrm{or}9.20×{10}^{-3}\mathrm{A}$
3. The charge value on the capacitor, $q=3.80\mathrm{μ°ä}\mathrm{or}3.80×{10}^{-6}\mathrm{C}$
4. The equation of charge on the capacitor, $q=Qcos\left(\mathrm{Ó\neg }t+\mathrm{φ}\right)$

In an oscillating circuit, oscillations of the capacitor in an electric field and oscillations of an inductor in the magnetic field cause electromagnetic oscillations. The energy stored in the capacitor's electric field (E) at any time is due to the charge stored by the capacitor. The energy stored in the magnetic field (B) at any time is due to the current flow in the inductor coil. And total energy is a combination of both.

Formulae:

Maximum energy stored in electric field by the capacitor, $U_E=\frac{q^2}{2C}$ (i)

Maximum energy stored in magnetic field by the inductor, $U_B=\frac{L{i}^2}{2}$ (ii)

The total energy in the circuit is due to the charge storing by the capacitor and the current flow through the inductor. Thus, it is given using given data, equations (i) and (ii) as follows:

$\begin{array}{c}U=\frac{q^2}{2C}+\frac{L{i}^2}{2}\\ =\frac{{(3.80×{10}^{-6})}^2}{2×\left(7.80×{10}^{-6}\right)}+\frac{\left(25×{10}^{-3}\right){(9.20×{10}^{-3})}^2}{2}\\ =1.98×{10}^{-6}\mathrm{J}\end{array}$

Hence, the total energy is $1.98×{10}^{-6}\mathrm{J}$.

Using the above energy value in equation (i), we can get the maximum charge stored by the capacitor as follows:

$\begin{array}{c}Q=\sqrt{2UC}\\ =\sqrt{2×7.80×{10}^{-6}×1.98×{10}^{-6}}\\ =5.66×{10}^{-6}\mathrm{C}\end{array}$

Hence, the value of the charge is $5.66×{10}^{-6}\mathrm{C}$.

Using the energy value from part (a) in equation (ii), we can get the maximum current stored by the capacitor as follows:

$\begin{array}{c}i=\sqrt{\frac{2U}{L}}\\ =\frac{2×1.98×{10}^{-6}}{25×{10}^{-3}}\\ =1.26×{10}^{-2}\mathrm{A}\end{array}$

Hence, the value of the current is $1.26×{10}^{-2}\mathrm{A}$.

At$t=0$, the phase angle can be found by using given charge value and maximum charge value in the given charge equation$q=Qcos\left(\mathrm{Ó\neg }t+\mathrm{φ}\right)$as follows:

$\begin{array}{c}\mathrm{φ}=co{s}^{-1}\left(\frac{q}{Q}\right)\\ =co{s}^{-1}\left(\frac{3.80×{10}^{-6}}{5.56×{10}^{-6}}\right)\\ =Â\pm 46.9°\end{array}$

By taking the derivative of the equation at t = 0, we obtainrole="math" localid="1663160844647" $-\mathrm{Ó\neg }Qsin\mathrm{φ}$, which is negative. Therefore,the sign indicates that for $\mathrm{φ}=+46.9°,$the charge on the capacitor is decreasing, and for$\mathrm{φ}=-46.9°$, the charge on the capacitor is increasing.

Hence, the value of the phase angle is $Â\pm 46.9°$.

For the given data to be the same, and only the capacitor is discharging at $t=0$; then the derivative will be negative, and $sin\mathrm{φ}$ will be positive. Thus, $\mathrm{φ}=+46.9°$.

Hence, the value of the phase angle for case of discharging is $+46.9°$.