# A fully charge capacitor C either initial charge q is connected to a coil of self inductance. L at $t=0$ the time at which the energy is stored equally between electric and magnetic field is

$(a)\;\pi \sqrt {LL} \\ (b)\;\frac{\pi}{4} \sqrt {Ll} \\(c)\;2 \pi \sqrt {Ll} \\(d)\;\sqrt {LL}$

In Lc oscillator
maximum energy in $C=\large\frac{V^2_{max}}{2C}$
Maximum energy in $L= \large\frac{4}{2} $$L I_{max ^2} We have I= I_{max} \sin wt ----(i) Also equal energy will be when \large\frac{1}{2}$$LI^2$ is half of maximum value.
$\large\frac{1}{2} $$LI^2=\large\frac{1}{2} \bigg[\frac{1}{2}$$L I^2_{max}\bigg]$
$I^2=\large\frac{1}{2} $$I_{max}^2 I= \large\frac{1}{\sqrt 2}$$I_{max}$-------(ii)
From (i) and (ii)
$I_{max} \sin wt =\large\frac{1}{\sqrt 2} $$I_{max} \sin wt =\large\frac{1}{\sqrt 2} wt =\large\frac{\pi}{4} \large\frac{2 \pi}{T}$$t=\large\frac{\pi}{4}$
$t= \large\frac{T}{8}$
or $t= \large\frac{2 \pi \sqrt {Ll}}{8}$
$\qquad= \large\frac{\pi}{4}$$\sqrt {Ll}$
Hence b is the correct answer.

edited Sep 24, 2014