QJEEM22Phf0b2: JEE Main 2022 Physics - Class 12 JEEM22Phf0b2 Solution: An AC source is connected to an inductance of 100 mH, a…

L = 100 $\times$ 10 ^{$-$ 3} H

C = 100 $\times$ 10 ^{$-$ 6} F

R = 120 $\Omega$

$\omega$ L = 10 $\Omega$

${1 \over {\omega C}} = {1 \over {{{10}^4} \times {{10}^{ - 6}}}} = 100\,\Omega$

$\Rightarrow$ X_C $-$ X_L = 90 $\Omega$

$\Rightarrow Z = \sqrt {{{90}^2} + {{120}^2}} = 150\,\Omega$

$\Rightarrow {I_{rms}} = {{20} \over {150}} = {2 \over {15}}A$

For heat resistance by 16 $^\circ$ C heat required = 32 J

$\Rightarrow {\left( {{2 \over {15}}} \right)^2} \times (120) \times t = 32$

$t = {{32 \times 15 \times 15} \over {4 \times 120}} = 15$