$(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.

Ask Question

Tag:MathPhyChemBioOther

Take Test

...