JEE Mains · Physics · STD 12 -7. Alternating current
A capacitor C is first charged fully with potential difference of \( V_{0} \) and disconnected from the battery. The charged capacitor is connected across an inductor having inductance L. In time t, 25% of the initial energy in the capacitor is transferred to the inductor. The value of t is
- A \(\frac{\pi \sqrt{ LC }}{3}\)
- B \(\frac{\pi \sqrt{ LC }}{6}\)
- C \(\frac{\pi \sqrt{ LC }}{2}\)
- D \(\pi \sqrt{\frac{ LC }{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi \sqrt{ LC }}{6}\)
Step-by-step Solution
Detailed explanation
\( U_{c_{f}}=75\%U_{c_{i}} \) \( Q_{F}^{2}=\frac{3}{4}Q_{1}^{2} \) \( Q_{i}cos~\omega t=\frac{\sqrt{3}}{2}Q_{i}\Rightarrow t=\frac{T}{12} \) \( t=\frac{\pi}{6}\sqrt{LC} \)
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