Question

An oscillating LCcircuit has current amplitude of $\mathbf{7}\mathbf{.}\mathbf{50}\mathbf{}\mathit{m}\mathit{A}$, potential amplitude of$\mathbf{250}\mathbf{}\mathit{m}\mathit{V}$, and a capacitance of$\mathbf{220}\mathbf{}\mathit{n}\mathit{F}$. (a) What is the period of oscillation? (b) What is the maximum energy stored in the capacitor? (c) What is the maximum energy stored in the inductor? (d) What is the maximum rate at which the current changes? (e) What is the maximum rate at which the inductor gains energy?

Short answer

1. The period of oscillationis $46.1\mathrm{\mu s}$.
2. The maximum energy stored in the capacitor is $6.88\times {10}^{-9}\mathrm{J}$.
3. The maximum energy stored in inductor is $6.88\times {10}^{-9}\mathrm{J}$.
4. The maximum rate of change of current is $1.02\times {10}^3\mathrm{A}/\mathrm{s}$.
5. The maximum rate of gain of energy by inductor is $1.02\times {10}^3\mathrm{A}/\mathrm{s}$.

Step-by-step solution

The given data

1. Current amplitude $I=7.50\mathrm{mA}\mathrm{or}7.50\times {10}^{-3}\mathrm{A}$,
2. Potential amplitude across capacitor $V=250\mathrm{mA}\mathrm{or}250\times {10}^{-3}\mathrm{A}$,
3. Capacitance $C=220\mathrm{nF}\mathrm{or}220\times {10}^{-9}\mathrm{F}$,

Understanding the concept of current and energy of LC oscillations

By using the relation between the potential difference across capacitors and the period, we can find the period of oscillation. After that, by using the concept of energy stored in the capacitor, we find the maximum energy stored in the capacitor. Similarly, we can find the maximum energy stored in the inductor, and from Faraday’s law, we can find the maximum rate of change of current. As the current is a function of time, we can find the maximum rate of gain of energy by the inductor.

Formulae:

Voltage drop across the capacitor, $V=I·X_C\left(\mathrm{i}\right)$

The period of oscillation, $T=\frac{2\pi }{\omega }\left(\mathrm{ii}\right)$

The capacitive reactance of the capacitor, $X_C=\frac{1}{\omega C}\left(\mathrm{iii}\right)$

Maximum energy stored in the capacitor,$U_C=\frac{1}{2}·C{V}^2\left(\mathrm{iv}\right)$

Maximum energy stored in the inductor,$U_L=\frac{L{i}^2}{2}\left(\mathrm{v}\right)$

Magnitude of emf due to the current in the inductor,$V\left(\mathrm{or}\epsilon \right)=L\frac{di}{dt}\left(\mathrm{vi}\right)$

The charge equation on the plate of capacitor as function of time,$q=q_{0}sin\omega t\left(\mathrm{vii}\right)$

The current equation due to the rate of charge,role="math" localid="1663136121465" $i=\frac{dq}{dt}\left(\mathrm{viii}\right)$

a) Calculation of the period of oscillation

Substituting the reactance vale from equation (iii) in equation (i), we can get that the angular frequency of the oscillation can be given as follows:

$\omega =\frac{I}{VC}$

Now, substituting the above value in equation (ii), we can get the period of oscillation as follows:

$\begin{array}{c}T=\frac{2\pi VC}{I}\\ =\left(\frac{6.28\times 250\times {10}^{-3}\times 220\times {10}^{-9}}{7.50\times {10}^{-3}}\right)\\ =46.1\times {10}^{-6}\mathrm{sor}46.1\mathrm{\mu s}\end{array}$

Hence, the value of the period is $46.1\mathrm{\mu s}$.

b) Calculation of the maximum energy stored in the capacitor

Maximum energy stored in the capacitor can be calculated using equation (iv) as follows:

$\begin{array}{c}U=0.5\times 220\times {10}^{-9}\times {\left(250\right)}^2\times {10}^{-6}\\ =6.88\times {10}^{-9}\mathrm{J}\end{array}$

Hence, the value of the maximum energy in the capacitor is $6.88\times {10}^{-9}\mathrm{J}$.

c) Calculation of the maximum energy stored in inductor

Maximum energy stored in inductor can be given by equation (v).

Thus, equating equations (iv) and (v), we can get the inductance value as:

$L=\frac{C{V}^2}{i^2}...............................\left(a\right)$

Thus, the value of the maximum energy using equation (v) and the above value can be given as:

$\begin{array}{c}{U}_L=\frac{C{V}^2}{2i^2}·i^2\\ =\frac{C{V}^2}{2}\\ =6.88\times {10}^{-9}\mathrm{J}\end{array}$

Hence, the value of the maximum energy stored in inductor is$6.88\times {10}^{-9}\mathrm{J}$.

d) Calculation of the maximum rate of current change

From equation (vi), we get the maximum rate of change in current value as follows:

$\begin{array}{c}\frac{di}{dt}=\frac{V}{L}\\ =\left(\frac{V}{\frac{C{V}^2}{i^2}}\right)(\mathrm{from}\mathrm{equation}\left(\mathrm{a}\right))\\ =\frac{I^2}{CV}\\ =\left(\frac{{\left(7.50\times {10}^{-3}\right)}^2}{220\times {10}^{-9}\times 250\times {10}^{-3}}\right)\\ =\left(\frac{56.25\times {10}^{-6}}{220\times 250\times {10}^{-12}}\right)\\ =\left(\frac{56.25\times {10}^6}{220\times 250}\right)\\ =1.02\times {10}^3\mathrm{A}/\mathrm{s}\end{array}$

Hence, the maximum rate of change of current is $1.02\times {10}^3\mathrm{A}/\mathrm{s}$.

e) Calculation of the maximum rate of energy gain by the inductor

The rate of charge that is current equation can be given using equation (vii) in equation (viii) as follows:

$\begin{array}{c}i=q_0\omega cos\omega t\\ =Icos\omega t(I=q_0\omega ,\mathrm{maximum}\mathrm{current}\mathrm{amplitude})\\ i^2=I^{2}cos\omega t\end{array}$

Now, substituting the above value in equation (v), we get the maximum rate of energy gain by the inductor as follows:

$\begin{array}{c}\frac{d{U}_L}{dt}=\frac{L}{2}·\left(\frac{d}{dt}\left\{I^{2}cos\omega t\right\}\right)\\ =\frac{L{I}^2}{2}·\omega sin2\omega t\\ =\frac{L{I}^2}{2}·\omega \\ =0.5\times IV\\ =0.5\times 7.50\times {10}^{-3}\times 250\times {10}^{-3}\\ =0.938\mathrm{mW}\end{array}$

Hence, the value of the maximum rate of energy gain is $0.938\mathrm{mW}$.