Verified Solution • 2025–26 Session moderate

QJEEM22Ph0a46: JEE Main 2022 Physics - Class 12 JEEM22Ph0a46 Solution: An alternating emf is applied to a circuit containing an…

An alternating emf E=440sin⁡100πt\mathrm{E}=440 \sin 100 \pi \mathrm{t} is applied to a circuit containing an inductance of 2πH\frac{\sqrt{2}}{\pi} \mathrm{H} . If an a.c. ammeter is connected in the circuit, its reading will be :

Multiple Choice Options

  • A 4.4 A
  • B 1.55 A
  • C 2.2 A
  • D 3.11 A

Step-by-Step Verified Solution

I=VωLI = {V \over {\omega L}}

=440100π×2π=44102= {{440} \over {100\pi \times {{\sqrt 2 } \over \pi }}} = {{44} \over {10\sqrt 2 }}

⇒Irms=I2=4420=2.2 A\Rightarrow {I_{rms}} = {I \over {\sqrt 2 }} = {{44} \over {20}} = 2.2\,A

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