Question

An L-C circuit consists of an inductor with $ L = 0.0900H $ and a capacitor of $ C = 4 \times {10^{ - 4}}F $ . The initial charge on the capacitor is $ 5.00\mu C $ , and the initial current in the inductor is zero. When the current in the inductor has half its maximum value, what is the charge on the capacitor?
 $ (A)1.33 \times {10^6}C $
 $ (B)9.33 \times {10^{ - 6}}C $
 $ (C)4.33 \times {10^{ - 6}}C $
 $ (D)7.33 \times {10^6}C $

Answer 1

Hint: This question will be solved by using the expression of charge which is stored in a capacitor in terms of time. On putting the given value of current, we will find the value of the sine trigonometric function. Then we will find the value of the cosine trigonometric function and with the help of it we will find the required value.

Complete Step By Step Answer:

Given,
Inductor which has inductance $ L = 0.0900H $
Capacitor of capacitance $ C = 4 \times {10^{ - 4}}F $ .
The initial charge which is present on the capacitor is $ 5.00\mu C $
On converting the initial charge in $ C $ , we get,
The initial charge which is present on the capacitor is $ 5 \times {10^{ - 6}}C $
We also know that the charge which is stored in a capacitor in terms of time is,
 $ Q = {Q_i}\cos \omega t $
Similarly, the current as a function of time is,
 $ I = {I_o}\sin \omega t $
Now, we have to find the charge on a capacitor, when the current in the inductor has half its maximum value. So,
 $ I = \dfrac{{{I_o}}}{2} $
On putting the above value in the equation for the current, we get,
 $ \dfrac{{{I_o}}}{2} = {I_o}\sin \omega t $
On cancelling the same terms on both the sides, we get,
 $ \sin \omega t = \dfrac{1}{2} $
 $ \sin \omega t = 0.5 $
According to the trigonometric functions,
 $ \cos \omega t = \sqrt {1 - {{\sin }^2}\omega t} $
On putting the value of $ \sin \omega t = 0.5 $ in the above equation,
 $ \cos \omega t = \sqrt {1 - {{0.5}^2}} $
On further solving, we get,
 $ \cos \omega t = \sqrt {1 - 0.25} $
 $ \cos \omega t = \sqrt {0.75} $
On solving the square root,
 $ \cos \omega t = 0.866 $
On putting the above value in the charge which is stored in a capacitor in terms of time,
 $ Q = 5 \times {10^{ - 6}} \times 0.866 $
 $ Q = 4.33 \times {10^{ - 6}}C $
So, the charge on the capacitor is $ Q = 4.33 \times {10^{ - 6}}C $ .

Note:
LC circuit is defined as a closed loop which consists of just two elements: a capacitor and an inductor. It also has a special resonance property like mechanical systems like a pendulum or even a mass on a spring. There is also a particular frequency at which it likes to oscillate at, and therefore it responds strongly to it.