In the given L-C circuit, charge on the capacitor is maximum at \( t = 0 \), find the time at which charge becomes 25% of its initial value for the first time.
Show Hint
In L-C circuits, the charge on the capacitor oscillates with time, and the time at which the charge becomes a fraction of its initial value can be calculated using the inverse cosine function.
Step 1: General formula for charge in L-C circuit.
The charge on the capacitor in an L-C circuit varies with time \( t \) as:
\[
q(t) = q_0 \cos \left( \omega t \right)
\]
where \( q_0 \) is the initial charge on the capacitor, and \( \omega \) is the angular frequency, given by:
\[
\omega = \frac{1}{\sqrt{LC}}
\]
Step 2: Find the time when the charge is 25% of its initial value.
We are given that the charge becomes 25% of its initial value. Therefore, at that time:
\[
q(t) = 0.25 q_0
\]
Substituting into the formula for \( q(t) \):
\[
0.25 q_0 = q_0 \cos \left( \frac{t}{\sqrt{LC}} \right)
\]
Dividing both sides by \( q_0 \):
\[
0.25 = \cos \left( \frac{t}{\sqrt{LC}} \right)
\]
Step 3: Solve for \( t \).
Taking the inverse cosine of both sides:
\[
\frac{t}{\sqrt{LC}} = \cos^{-1} \left( \frac{1}{4} \right)
\]
Therefore, the time \( t \) is:
\[
t = \sqrt{LC} \cos^{-1} \left( \frac{1}{4} \right)
\]
Thus, the time when the charge becomes 25% of its initial value is \( \sqrt{LC} \cos^{-1} \left( \frac{1}{4} \right) \).
