JEE Advanced · Physics · 22. AC Circuits
Paragraph :
The capacitor of capacitance \(C\) can be charged (with the help of a resistance \(R\) ) by a voltage source \(V\), by closing switch \(S_1\) while keeping switch \(S_2\) open. The capacitor can be connected in series with an inductor \(L\) by closing switch \(S_2\) and opening \(S_1\).

Question :
If the total charge stored in the \(L C\) circuit is \(Q_0\), then for \(t \geq 0\)
- A the charge on the capacitor is \(Q=Q_0 \cos \left(\frac{\pi}{2}+\frac{t}{\sqrt{L C}}\right)\)
- B the charge on the capacitor is \(Q=Q_0 \cos \left(\frac{\pi}{2}-\frac{t}{\sqrt{L C}}\right)\)
- C the charge on the capacitor is \(Q=-L C \frac{d^2 Q}{d t^2}\)
- D the charge on the capacitor is \(Q=-\frac{1}{\sqrt{L C}} \frac{d^2 Q}{d t^2}\)
Answer & Solution
Correct Answer
(C) the charge on the capacitor is \(Q=-L C \frac{d^2 Q}{d t^2}\)
Step-by-step Solution
Detailed explanation
Comparing the LC oscillations with normal SHM, we get
\(
\frac{d^2 Q}{d t^2}=-\omega^2 Q
\)
Here,
\(
\begin{aligned}
\omega^2 & =\frac{1}{L C} \\
Q & =-L C \frac{d^2 Q}{d t^2}
\end{aligned}
\)
\(
\therefore \quad Q=-L C \frac{d^2 Q}{d t^2}
\)
\(
\frac{d^2 Q}{d t^2}=-\omega^2 Q
\)
Here,
\(
\begin{aligned}
\omega^2 & =\frac{1}{L C} \\
Q & =-L C \frac{d^2 Q}{d t^2}
\end{aligned}
\)
\(
\therefore \quad Q=-L C \frac{d^2 Q}{d t^2}
\)
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