QJEEM22Ph6d62: JEE Main 2022 Physics - Class 12 JEEM22Ph6d62 Solution: To light, a lamp is connected, in series with a capacitor of…

${X_C} = {1 \over {wc}} = {{\pi \sqrt x } \over {2\pi \times 50 \times 50}} \times {10^6}$

$v_R^2 + v_C^2 = {(200)^2}$

$v_C^2 = {200^2} - {100^2}$

${v_C} = 100\sqrt 3 \,V$

${v_R} = 100\,V$

$P = {{{V^2}} \over R}$

$R = {{100 \times 100} \over {50}} = 200\,\Omega$

${i_{rm}} = {1 \over 2}\,A$

${1 \over 2} \times {x_C} = 100\sqrt 3 \Rightarrow {10^{ - 6}} \times {{\sqrt x } \over {5000}} \times {1 \over 2} = 100\sqrt 3$

${{{{10}^{ - 6}}\sqrt x } \over {10000 \times 100}} = \sqrt 3$

$\sqrt x = \sqrt 3$

$x = 3$