Question

In the circuit shown, the symbols have their usual meanings. The cell has emf E. X is initially joined to Y for a long time. Then, X is joined to Z. The maximum charge on C at any later time will be: 

• A. \( \dfrac{E}{R\sqrt{LC}} \)
• B. \( \dfrac{ER}{2\sqrt{LC}} \)
• C. \( \dfrac{E\sqrt{LC}}{2R} \)
• D. dfracE√(LC)R

Hint:

In ideal LC oscillations, maximum charge equals the initial charge stored in the capacitor.

Answer 1

The Correct Option is D

Step 1:
When the switch is at Y for a long time, the capacitor charges fully:
\( Q_0 = CE \)
Step 2:
After switching to Z, the circuit becomes an LC circuit.
Step 3:
Maximum charge in LC oscillations remains equal to the initial charge:
\( Q_{\text{max}} = CE \)
Step 4:
Using \( \omega = \dfrac{1}{\sqrt{LC}} \):
\( Q_{\text{max}} = \dfrac{ER}{\sqrt{LC}} \)