In the given L-C circuit, charge on the capacitor is maximum at 𝑡 = 0, find time at which charge becomes 25% of it's initial value first time.

Correct Option: $\sqrt{LC}{\cos }^{-1}\left(\frac{1}{4}\right)$

Step-by-Step Explanation:

In an ideal LC circuit consisting of an inductor of inductance L and a capacitor of capacitance C, charge on the capacitor oscillates simple harmonically with time according to the equation:

$q\left(t\right)=q_0\cos \left(\omega t\right)$

where:

• $q_0$ is the maximum initial charge at $t=0$.
• $\omega$ is the angular frequency of LC oscillations, given by $\omega =\frac{1}{\sqrt{LC}}$.

We are required to find the time t when the charge becomes 25% of its initial maximum value for the first time.

25% of the initial charge $q_0$ can be written as:

$q\left(t\right)=25\%\text{of}{q}_0=\frac{25}{100}{q}_0=\frac{1}{4}{q}_0$

Substituting $q\left(t\right)=\frac{1}{4}{q}_0$ into the charge oscillation equation:

$\frac{1}{4}{q}_0=q_0\cos \left(\omega t\right)$

Dividing both sides by $q_0$:$\cos \left(\omega t\right)=\frac{1}{4}$

Taking the inverse cosine on both sides:

$\omega t={\cos }^{-1}\left(\frac{1}{4}\right)$

Rearranging for time t:

$t=\frac{1}{\omega }$ ${\cos }^{-1}\left(\frac{1}{4}\right)$

Since $\omega =\frac{1}{\sqrt{LC}}$, we have $\frac{1}{\omega }$ = $\sqrt{LC}$. Substituting this back gives:

$t=\sqrt{LC}{\cos }^{-1}\left(\frac{1}{4}\right)$

Thus, the time at which the charge on the capacitor becomes 25% of its initial value for the first time is $\sqrt{LC}{\cos }^{-1}\left(\frac{1}{4}\right)$.