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)

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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