Question

In an LC oscillator circuit $L = 10mH$, $C = 40\mu F$. If initially at $t = 0$ the capacitor is fully charged with $4\mu C$ then find the current in the circuit when the capacitor and inductor share equal energies.
A. 0.2mA
B. 4.4mA
C. 0.3mA
D. 2mA

Answer 1

Hint: In this question we have to apply the concept of current electricity. We have to use the condition of the L.C oscillation circuit. We have to also use the concepts of charging and discharging of a capacitor. The current in this case is AC and we have to use the formulas of AC where every quantity varies with time.

Complete step by step answer:For Alternating current electricity if we consider the charge in a capacitor, it is given by and varies as;
$Q = {Q_0}\cos \omega t$
If we put the value of the mean charge form the question we get:
$Q = (4\mu C)\cos \omega t$
Therefore, we understand that $4\mu C$ is the mean value of charge.
Now we have been told that the energy of the conductor and the inductor is same.
The energy stored in capacitor is given by:
${E_C} = \dfrac{{{Q_0}^2}}{{2C}}{\cos ^2}\omega t$
And the energy stored in an inductor is given by:
${E_L} = \dfrac{{{Q_0}^2}}{{2C}}{\sin ^2}\omega t$
If we equate the two equations as said in the question we get:
\[\begin{array}{l}
\dfrac{{{Q_0}^2}}{{2C}}{\cos ^2}\omega t = \dfrac{{{Q_0}^2}}{{2C}}{\sin ^2}\omega t\\
\Rightarrow {\cos ^2}\omega t = {\sin ^2}\omega t\\
\Rightarrow \cos \omega t = \sin \omega t\\
\Rightarrow \tan \omega t = 1\\
\Rightarrow \omega t = \dfrac{\pi }{4},\dfrac{{3\pi }}{4}
\end{array}\]
now we have to find the current where \[\omega t = \dfrac{\pi }{4},\dfrac{{3\pi }}{4}\]
The current is given as:
$I = {I_0}\sin \omega t$
And the value of ${I_0}{\rm{ is }}{Q_0}\omega $ putting this value we get:
$I = {Q_0}\omega \sin \omega t$
Now we know the value of \[\omega t\], ${Q_0}$ and $\omega = \dfrac{1}{{\sqrt {LC} }}$, putting all these values in the above equation we get:
$\begin{array}{l}
I = {Q_0}\omega \sin \omega t\\
\Rightarrow I = \left( {4 \times {{10}^{ - 6}}} \right)\left( {\dfrac{1}{{\sqrt {LC} }}} \right)\left( {\sin \dfrac{\pi }{4}} \right)\\
\Rightarrow I = \left( {4 \times {{10}^{ - 6}}} \right)\left( {\dfrac{1}{{\sqrt {10 \times {{10}^{ - 3}} \times 40 \times {{10}^{ - 6}}} }}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right)\\
\Rightarrow I = 4.4 \times {10^{ - 3}}A\\
\Rightarrow I = 4.4mA
\end{array}$
Therefore, the current is 4.4mA and the correct option is (B).

Note:For questions like these the formulas of $\omega $, the mean current and the understanding of the mean charge is very important. Also, the factor of \[\omega t\]is very important and helps to find the required condition. For questions like this where we have to deal with large calculations students should be cautious about not making mistakes.