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Question: An L-C circuit consists of an inductor with \( L = 0.0900H \) and a capacitor of \( C = 4 \times {10...

An L-C circuit consists of an inductor with L=0.0900HL = 0.0900H and a capacitor of C=4×10−4FC = 4 \times {10^{ - 4}}F . The initial charge on the capacitor is 5.00μC5.00\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)1.33×106C(A)1.33 \times {10^6}C
(B)9.33×10−6C(B)9.33 \times {10^{ - 6}}C
(C)4.33×10−6C(C)4.33 \times {10^{ - 6}}C
(D)7.33×106C(D)7.33 \times {10^6}C

Explanation

Solution

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=0.0900HL = 0.0900H
Capacitor of capacitance C=4×10−4FC = 4 \times {10^{ - 4}}F .
The initial charge which is present on the capacitor is 5.00μC5.00\mu C
On converting the initial charge in CC , we get,
The initial charge which is present on the capacitor is 5×10−6C5 \times {10^{ - 6}}C
We also know that the charge which is stored in a capacitor in terms of time is,
Q=Qicos⁡ωtQ = {Q_i}\cos \omega t
Similarly, the current as a function of time is,
I=Iosin⁡ωtI = {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=Io2I = \dfrac{{{I_o}}}{2}
On putting the above value in the equation for the current, we get,
Io2=Iosin⁡ωt\dfrac{{{I_o}}}{2} = {I_o}\sin \omega t
On cancelling the same terms on both the sides, we get,
sin⁡ωt=12\sin \omega t = \dfrac{1}{2}
sin⁡ωt=0.5\sin \omega t = 0.5
According to the trigonometric functions,
cos⁡ωt=1−sin⁡2ωt\cos \omega t = \sqrt {1 - {{\sin }^2}\omega t}
On putting the value of sin⁡ωt=0.5\sin \omega t = 0.5 in the above equation,
cos⁡ωt=1−0.52\cos \omega t = \sqrt {1 - {{0.5}^2}}
On further solving, we get,
cos⁡ωt=1−0.25\cos \omega t = \sqrt {1 - 0.25}
cos⁡ωt=0.75\cos \omega t = \sqrt {0.75}
On solving the square root,
cos⁡ωt=0.866\cos \omega t = 0.866
On putting the above value in the charge which is stored in a capacitor in terms of time,
Q=5×10−6×0.866Q = 5 \times {10^{ - 6}} \times 0.866
Q=4.33×10−6CQ = 4.33 \times {10^{ - 6}}C
So, the charge on the capacitor is Q=4.33×10−6CQ = 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.