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 t ~s \ 25 \% of the initial energy

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Quantrex Academy · Accepted Answer

Correct answer: B. π L C 6

In an LC circuit, energy oscillates sinusoidally.
Initial energy in capacitor is

E_0 = \frac{1}{2}CV_0^2

.
When 25% is transferred to the inductor, 75% remains in the capacitor.
Since energy in capacitor varies as

E_C(t) = E_0\cos^2(\omega t)

, we need

\cos^2(\omega t) = 0.75

.
This gives

\sin^2(\omega t) = 0.25

, so

\sin(\omega t) = 0.5

, yielding

\omega t = \frac{\pi}{6}

.
With

\omega = \frac{1}{\sqrt{LC}}

, we get

t = \frac{\pi\sqrt{LC}}{6}

seconds.

A capacitor C is first charged fully with potential difference of...

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