Answer:
The times is 0.688 ms.
Explanation:
Given that,
Inductance L= 0.048 H
Capacitor C= 4\mu F
The current is initially maximum
[tex]I=I_{max}[/tex]
We know that,
[tex]I=I_{max}\cos\omega t[/tex]
Where, [tex]\omega=\dfrac{1}{\sqrt{LC}}[/tex]
We need to calculate the time
[tex]\dfrac{dq}{dt}=I[/tex]
[tex]dq=Idt[/tex]....(I)
On integrating equation (I)
[tex]\int_{0}^{q}{dq}=\int_{0}^{t}{I_{max}\cos\omega t}dt[/tex]
[tex]q=I_{max}(\dfrac{\sin\omega t}{\Omega})_{0}^{t}[/tex]
[tex]q=\dfrac{I_{max}}{\omega}\sin\omega t[/tex]
If [tex]q = q_{max}[/tex]
[tex]\sin\omega t=1[/tex]
[tex]\omega t=\dfrac{\pi}{2}[/tex]
[tex]t =\dfrac{\pi}{2\omega}[/tex]
Put the value of [tex]\omega[/tex]
[tex]t=\dfrac{\pi}{2\times\dfrac{1}{\sqrt{LC}}}[/tex]
[tex]t=\dfrac{\pi\sqrt{LC}}{2}[/tex]
[tex]t=\dfrac{\pi}{2}\sqrt{LC}[/tex]
[tex]t=\dfrac{\pi}{2}\sqrt{0.048\times4\times10^{-6}}[/tex]
[tex]t=0.688\ msec[/tex]
Hence, The times is 0.688 ms.
