QJEEM22Phdb42: JEE Main 2022 Physics - Class 12 JEEM22Phdb42 Solution: The frequencies at which the current amplitude in an LCR series …

${i \over {{i_{\max }}}} = {1 \over {\sqrt 2 }}$

$= {{{{{V_0}} \over Z}} \over {{{{V_0}} \over R}}}$

$\Rightarrow {R \over Z} = {1 \over {\sqrt 2 }}$

and ${1 \over {212C}} - 212L = 232L - {1 \over {232C}}$

so $212L = {1 \over {232C}}$

so ${R \over {\sqrt {{R^2} + {{\left( {232L + {1 \over {232C}}} \right)}^2}} }} = {1 \over {\sqrt 2 }}$

${{{R^2}} \over {{R^2} + {{(20L)}^2}}} = {1 \over 2}$

$400{L^2} = {R^2}$

$L = {5 \over {20}}$

$H = {5 \over {20}} \times 1000$ mH

$= 250$ mH