QJEEM22Phf7dc: JEE Main 2022 Physics - Class 12 JEEM22Phf7dc Solution: A capacitor of capacitance 500 F is charged completely using a…

At steady state charge stored on the capacitor,

${q_{\max }} = CV$

$= 500 \times {10^{ - 6}} \times 100$

$= 5 \times {10^{ - 2}}\,C$

Energy stored in the capacitor,

${U_{\max }} = {{q_{\max }^2} \over {2C}}$

Now, when electrostatic energy of capacitor converted to magnetic field energy then all energy of capacitor is transferrd to the inductor.

$\therefore$ Maximum energy stored in the inductor

${U_{L\,\max }} = {1 \over 2}L\,I_{\max }^2$

$\therefore$ ${1 \over 2}L\,I_{\max }^2 = {{q_{\max }^2} \over {2C}}$

$\Rightarrow {I_{\max }} = {{{q_{\max }}} \over {\sqrt {LC} }}$

$= {{5 \times {{10}^{ - 2}}} \over {\sqrt {50 \times {{10}^{ - 3}} \times 500 \times {{10}^{ - 6}}} }}$

$= {{5 \times {{10}^{ - 2}}} \over {5 \times {{10}^{ - 3}}}}$

$= 10\,A$