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JEE Advanced 2006
Exam
JEE Advanced
Year
2006
Subject
Physics
Chapter
Alternating Current

JEE Advanced 2006 · Physics · Alternating Current

Paragraph: The capacitor of capacitance C can be charged (with the...

Paragraph: The capacitor of capacitance CC can be charged (with the help of a resistance RR ) by a voltage source VV, by closing switch S1S_1 while keeping switch S2S_2 open. The capacitor can be connected in series with an inductor LL by closing switch S2S_2 and opening S1S_1.Paragraph: The capacitor of capacitance C can be charged (with the help of a resistance R ) by a voltage source V , by
Question: If the total charge stored in the LCL C circuit is Q0Q_0, then for t≥0t \geq 0

Options

  1. Athe charge on the capacitor is $Q=Q_0 \cos \left(\frac{\pi}{2}+\frac{t}{\sqrt{L C}}\right)
  2. Bthe charge on the capacitor is $Q=Q_0 \cos \left(\frac{\pi}{2}-\frac{t}{\sqrt{L C}}\right)
  3. Cthe charge on the capacitor is Q=−LCd2Qdt2Q=-L C \frac{d^2 Q}{d t^2}
  4. Dthe charge on the capacitor is Q=−1LCd2Qdt2Q=-\frac{1}{\sqrt{L C}} \frac{d^2 Q}{d t^2}

Correct answer

C. the charge on the capacitor is Q=−LCd2Qdt2Q=-L C \frac{d^2 Q}{d t^2}

Step-by-step solution

Comparing the LC oscillations with normal SHM, we get \frac{d^2 Q}{d t^2}=-\omega^2 Q Here, ω2=1LCQ=−LCd2Qdt2\begin{aligned}\omega^2 & =\frac{1}{L C}\\ Q & =-L C \frac{d^2 Q}{d t^2} \end{aligned}\therefore \quadQ=−LCQ=-L C\frac{d^2 Q}{d t^2}$
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