A capacitor C is first charged fully with potential difference of V_0$V_{0}$ and disconnected from the battery. The charged capacitor is connected across an inductor having inductance L. In t text s$t \text{ s}$ 25% of the initial energy in the capacitor is transferred to the inductor. The value of t is ________ s.

### Related Formula For LC oscillations, charge varies as: Q(t) = Q_0 cos(omega t)$$Q(t) = Q_0 \cos(\omega t)$$ Where omega = frac1sqrtLC$\omega = \frac{1}{\sqrt{LC}}$ Energy in capacitor: U_C = fracQ^22C$$U_C = \frac{Q^2}{2C}$$ ### Core Logic Since 25% of the initial energy is transferred to the inductor, the remaining energy in the capacitor is 75% of its initial value. U_C_f = 75\% text of U_C_i$$U_{C_f} = 75\% \text{ of } U_{C_i}$$ fracQ_f^22C = frac34 times fracQ_0^22C$$\frac{Q_f^2}{2C} = \frac{3}{4} \times \frac{Q_0^2}{2C}$$ Q_f^2 = frac34 Q_0^2$$Q_f^2 = \frac{3}{4} Q_0^2$$ Q_f = fracsqrt32 Q_0$$Q_f = \frac{\sqrt{3}}{2} Q_0$$ ### Step 1: Finding Time Using the equation for charge variation: Q_0 cos(omega t) = fracsqrt32 Q_0$$Q_0 \cos(\omega t) = \frac{\sqrt{3}}{2} Q_0$$ cos(omega t) = fracsqrt32$$\cos(\omega t) = \frac{\sqrt{3}}{2}$$ omega t = fracpi6$$\omega t = \frac{\pi}{6}$$ ### Step 2: Final Conclusion Substitute omega = frac1sqrtLC$\omega = \frac{1}{\sqrt{LC}}$: frac1sqrtLC t = fracpi6$$\frac{1}{\sqrt{LC}} t = \frac{\pi}{6}$$ t = fracpi sqrtLC6$$t = \frac{\pi \sqrt{LC}}{6}$$ ### Pattern Recognition Energy is proportional to charge squared. 75\%$75\%$ energy remaining means charge is sqrt0.75 = fracsqrt32$\sqrt{0.75} = \frac{\sqrt{3}}{2}$ of the original. Cosine of pi/6$\pi/6$ yields this exact ratio. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction Class 12 Physics: Alternating Current