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A fully charged capacitor $C$ with initial charge $q_{0}$ is connected to a coil of self inductance $L$ at $t=0$ The time at which the energy is stored equally between the electric and the magnetic fields is

Option 1)
$\pi \sqrt{LC}$

Option 2)
$\frac{\pi}{4} \sqrt{LC}$

Option 3)
$2\pi \sqrt{LC}$

Option 4)
$\sqrt{LC}$

Answers (1)

best_answer

As we learnt in

$q=q_{o}sin(wt+\phi )$

LC Circuit voltage -

$V_{L}=\: {I}\! X_{L}$

$V_{C}={I}\! X_{C}$

- wherein

at t=0, $q=q_{o}$

$\phi =90^{\circ}$

$q=q_{o}cos\:wt$

maximum energy= $\frac{q_{o}^{2}}{2c}$

When energy is equally distributed among capacitor and inductor then

$\frac{q^{2}}{2c}=(\frac{q_0^{2}}{2c}).\frac{1}{2}$

$\therefore q=\frac{qo}{\sqrt{2}}=q_{o}cos\:\omega t$

$\omega t=\frac{\pi}{4}$ or $t=\frac{\pi}{4\omega}$

$\therefore \omega=\frac{1}{\sqrt{Lc}}$

$\therefore t=\frac{\pi}{4}.\sqrt{LC}$

Option 1)

$\pi \sqrt{LC}$

This option is incorrect

Option 2)

$\frac{\pi}{4} \sqrt{LC}$

This option is correct

Option 3)

$2\pi \sqrt{LC}$

This option is incorrect

Option 4)

$\sqrt{LC}$

This option is incorrect

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prateek

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