Question

: In LC Oscillator , at time ,energy of inductor and energy of capacitor are equal ?
A. T/8
B. T/4
C. T/2
D. T

Answer 1

Hint:-
For solving the above problem we will use :
Energy of capacitor as: $\dfrac{1}{2}\dfrac{{{Q_0}^2}}{C}$
Energy of inductor as:$\dfrac{1}{2}L{I^2}$
Frequency of Oscillator :$\omega = \dfrac{1}{{\sqrt {LC} }}$

Complete step-by-step solution:Voltage in inductor given as : $L\dfrac{{di}}{{dt}}$ ..............1
Voltage in capacitor is given as: $\dfrac{q}{C}$ ..............2
Therefore we can write $L\dfrac{{di}}{{dt}}$$ = $ $\dfrac{q}{C}$...............3
Also we have $i$ = -$\dfrac{{dq}}{{dt}}$
In equation 3 we substitute the value of i
$
   \Rightarrow L\dfrac{{ddq}}{{dtdt}} = - \dfrac{q}{C} \\
   \Rightarrow \dfrac{{{d^2}q}}{{d{t^2}}} = - \dfrac{q}{{LC}} \\
 $ ( equation becomes the differential equation)
On solving this differential equation we will get :
Q=Qcos, q is maximum when cos$\omega t$ =1, $\omega = \dfrac{1}{{\sqrt {LC} }}$
When energy of capacitor and inductor are equal :
$\dfrac{1}{2}\dfrac{{{Q_0}^2}}{C} = $ $\dfrac{1}{2}L{I^2}$
We will substitute the value of we have calculated above but before that calculate .
         i = -$\dfrac{{dq}}{{dt}}$
$
   \Rightarrow i = - \dfrac{{dQ\cos \omega t}}{{dt}} \\
   \Rightarrow i = Q\omega \sin \omega t \\
 $ (we have differentiated q equation with respect to t)
Now we will substitute the equation of q we have calculated above and the value of i we got after differentiation in the equalized equation of energy.
         $\dfrac{1}{2}\dfrac{{{Q_0}^2}}{C} = $ $\dfrac{1}{2}L{I^2}$
      $
   \Rightarrow \dfrac{1}{2}\dfrac{{{{(Q\cos \omega t)}^2}}}{C} = \dfrac{1}{2}L{(Q\omega \sin wt)^2} \\
   \Rightarrow \dfrac{{{Q^2}{{\cos }^2}\omega t}}{C} = L{Q^2}{\sin ^2}wt \\
 $ (cancelling ½ term from LHS and RHS)
     $
   \Rightarrow \dfrac{1}{{LC}} = \dfrac{{\omega {Q^2}{{\sin }^2}\omega t}}{{{Q^2}{{\cos }^2}\omega t}} \\
   \Rightarrow \dfrac{1}{{LC}} = \omega {\tan ^2}\omega t \\
 $ (cancel Q2 and keep LC on LHS) (sin(wt) / cos(wt) = tan(wt))
     $ \Rightarrow \dfrac{1}{{\sqrt {LC} }} = \omega \tan \omega t$ (taking square roots on both LHS and RHS )
      $\omega = \dfrac{1}{{\sqrt {LC} }}$,$\omega = \dfrac{{2\pi }}{T}$
So we can write :
$
   \Rightarrow \omega = \omega \tan \omega t \\
   \Rightarrow 1 = \tan \omega t \\
   \Rightarrow {\tan ^{ - 1}}(1) = \omega t \\
   \Rightarrow {45^0} = \dfrac{{2\pi t}}{T} \\
 $ (Tan inverse 1 is equal to $45^0$ and substituting value of w from above )
We can write $45^0$ as $\dfrac{\pi}{4}$.
$
   \Rightarrow \dfrac{\pi }{4} = \dfrac{{2\pi t}}{T} \\
   \Rightarrow \dfrac{T}{4} = 2t \\
 $ (cancelling $\pi$ from both sides )
$ \Rightarrow t = \dfrac{T}{8}$
Option 1 is correct.

Note:- Oscillators are generally used to convert current flow form a direct current source to an alternating waveform which is of the frequency which is desired by us. Oscillators produce a continuous repeated, and alternating waveform without any input given to it.