QJEEM23Phc89a: JEE Main 2023 Physics - Class 12 JEEM23Phc89a Solution: For the given figures, choose the correct options :

For (a), $i = {V \over R} = {{220} \over {40}} = 5.5\,A$

for (b), ${X_C} = {1 \over {2\pi fC}} = {1 \over {2\pi \times 50 \times 0.5 \times {{10}^{ - 6}}}} = {{{{10}^6}} \over {50\pi }}\Omega$

${X_L} = 2\pi fL = 2\pi \times 50 \times 50 \times {10^{ - 3}} = 50\pi \,\Omega$

{X_C} > {X_L} , hence impedance is greater than 40 $\Omega$ .

${i_{rms}} = {{220} \over Z}$

$\therefore$ {\left. {{i_{rms}}} \right|_b} < {\left. {{i_{rms}}} \right|_a}