Question

An inductor of inductance $ 2.0mH $ is connected across a charge capacitor of capacitance $ 5.0\mu F $ , and the resulting LC circuit is set oscillating at natural frequency. Let $ Q $ denote the instantaneous charge on the capacitor, and $ I $ the current in the circuit. It is found the maximum value of $ Q $ is $ 200\mu C $ .When $ Q = 100\mu C $ the value of $ \dfrac{{dI}}{{dt}} $ ​ is found to be $ {10^x}A/s $ . Find the value of $ x $ is then
(A) $ x = - 4 $
(B) $ x = - 5 $
(C) $ x = 4 $
(D) $ x = 5 $

Answer 1

Hint :For solving this question, we will first find the natural frequency of the given LC circuit. After that using its relation with charge, and further the relation of charge with current, we will determine the value of $ \dfrac{{dI}}{{dt}} $ . Finally we will compare the determined value with the given value to find the value of $ x $ .
 $ \omega = \dfrac{1}{{\sqrt {LC} }} $ , where, $ \omega $ is the natural frequency of oscillation, $ L $ is the inductance and $ C $ is the capacitance.
 $ Q = {Q_0}\cos \omega t $ , where, $ Q $ is the instantaneous charge on the capacitor, $ {Q_0} $ is the maximum charge, $ \omega $ is the natural frequency of oscillation and $ t $ is the time.
 $ I = - \dfrac{{dQ}}{{dt}} $ , where, $ I $ the current in the circuit, $ Q $ is the instantaneous charge on the capacitor and $ t $ is the time.

Complete Step By Step Answer:
We are given $ L = 2.0mH = 2 \times {10^{ - 3}}H $ , $ C = 5.0\mu F = 5 \times {10^{ - 6}}F $ , $ {Q_0} = 200\mu C = 200 \times {10^{ - 6}}C $ and $ Q = 100\mu C = 100 \times {10^{ - 6}}C = {10^{ - 4}}C $ .
First, we will find the natural frequency of oscillation.
 $ \omega = \dfrac{1}{{\sqrt {LC} }} = \dfrac{1}{{\sqrt {2 \times {{10}^{ - 3}} \times 5 \times {{10}^{ - 6}}} }} = {10^4}rad $
Now, we know that the instantaneous charge of the capacitor is given by
 $ Q = {Q_0}\cos \omega t $
And as per the definition of the current of the capacitor,
 $
  I = - \dfrac{{dQ}}{{dt}} \\
   \Rightarrow I = {Q_0}\omega \sin \omega t \\
  $
Now, let us determine the value of $ \dfrac{{dI}}{{dt}} $ .
 $ \dfrac{{dI}}{{dt}} = {Q_0}{\omega ^2}\cos \omega t = {\omega ^2}\left( {{Q_0}{\omega ^2}\cos \omega t} \right) = {\omega ^2}Q $
 $ \dfrac{{dI}}{{dt}} = {\left( {{{10}^4}} \right)^2} \times {10^{ - 4}} = {10^4}A/s $
We are given that the value of $ \dfrac{{dI}}{{dt}} $ ​ is found to be $ {10^x}A/s $ .
Thus the value of $ x $ is 4.
Hence, option C is the right answer.

Note :
While solving this question, we need to be careful for the sign at the time of differentiation. For example, we have determined $ I = - \dfrac{{dQ}}{{dt}} \Rightarrow I = {Q_0}\omega \sin \omega t $ . This is because the differentiation of $ $
 $ \cos x $ is $ - \sin x $ . Also in the next step, we have used the rule that the differentiation of $ \sin x $ is $ \cos x $ .