Verified Solution • 2025–26 Session moderate

QJEEM22Phaa82: JEE Main 2022 Physics - Class 12 JEEM22Phaa82 Solution: A series LCR circuit has and and it is connected to ac voltage…

A series LCR circuit has L=0.01 H,R=10 Ω\mathrm{L}=0.01\, \mathrm{H}, \mathrm{R}=10\, \Omega and C=1μF\mathrm{C}=1 \mu \mathrm{F} and it is connected to ac voltage of amplitude (Vm)50 V\left(\mathrm{V}_{\mathrm{m}}\right) 50 \mathrm{~V} . At frequency 60%60 \% lower than resonant frequency, the amplitude of current will be approximately :

Multiple Choice Options

  • A 466 mA
  • B 312 mA
  • C 238 mA
  • D 196 mA

Step-by-Step Verified Solution

ω=0.4ω0\omega = 0.4{\omega _0} ...... (i)

⇒I=VZ=50R2+(ωL−1ωC)2\Rightarrow I = {V \over Z} = {{50} \over {\sqrt {{R^2} + {{\left( {\omega L - {1 \over {\omega C}}} \right)}^2}} }} ..... (ii)

⇒I=238\Rightarrow I = 238 mA

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