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 :
After the capacitor gets fully charged, \(S_1\) is opened and \(S_2\) is closed so that the inductor is connected in series with the capacitor. Then,
- A at \(t=0\), energy stored in the circuit is purely in the form of magnetic energy
- B at any time \(t>0\), current in the circuit is in the same direction
- C at \(t>0\), there is no exchange of energy between the inductor and capacitor
- D at any time \(t>0\), maximum instantaneous current in the circuit may \(V \sqrt{\frac{C}{L}}\)
Answer & Solution
Correct Answer
(D) at any time \(t>0\), maximum instantaneous current in the circuit may \(V \sqrt{\frac{C}{L}}\)
Step-by-step Solution
Detailed explanation
From conservation of energy, \(\frac{1}{2} L I_{\max }^2=\frac{1}{2} C V^2\)
\(\therefore \quad I_{\max }=V \sqrt{\frac{C}{L}}\)
\(\therefore \quad I_{\max }=V \sqrt{\frac{C}{L}}\)
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