Question

In an oscillating LCcircuit, $\mathit{L}\mathbf{=}\mathbf{}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{H}$and $\mathit{C}\mathbf{=}\mathbf{}\mathbf{7}\mathbf{.}\mathbf{80}\mathbf{}\mathit{\mu }\mathit{F}$. At time $\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$ the current is $\mathbf{9}\mathbf{.}\mathbf{20}\mathbf{}\mathit{m}\mathit{A}$, the charge on the capacitor is $\mathbf{3}\mathbf{.}\mathbf{80}\mathbf{}\mathit{\mu }\mathit{C}$, and the capacitor is charging. (a) What is the total energy in the circuit? (b) What is the maximum charge on the capacitor? (c) What is the maximum current? (d) If the charge on the capacitor is given by $\mathit{q}\mathbf{=}\mathbf{}\mathit{Q}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{\omega t}\mathbf{+}\mathbf{}\mathbf{\phi }\mathbf{)}\mathbf{,}$what is the phase angle $\mathit{\phi }$? (e) Suppose the data are the same, except that the capacitor is discharging at $\mathit{t}\mathbf{=}\mathbf{}\mathbf{0}$. What then is $\mathit{\phi }$?

Short answer

1. Total energy in the circuit is$1.98\times {10}^{-6}\mathrm{J}$.
2. Maximum charge on the capacitor is$5.56\times {10}^{-6}\mathrm{C}$.
3. Maximum charge current is$1.26\times {10}^{-2}\mathrm{A}$.
4. Phase angle $\phi$is$\pm 46.9°$.
5. When capacitor is discharged, phase angle $\phi$ will be $+46.9°$.

Step-by-step solution

The given data

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

Understanding the concept of oscillations of LC circuit

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)

a) Calculation of the total energy in the circuit

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:

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

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

b) Calculation of the maximum charge

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

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

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

c) Calculation of the maximum current

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

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

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

d) Calculation of the phase angle

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

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

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

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

e) Calculation of the phase angle when capacitor is discharged

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

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