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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} $

1 Answer

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.


answered Mar 25, 2014 by meena.p
edited Sep 24, 2014 by thagee.vedartham

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