The reactance of the given inductor is,
XL=ωL=2πfL=2π×50×0.2=20πΩ.
The maximum current in the given circuit will be,
i0=XL2Vrms=20π2×220=π242A
Therefore, a=242
JEE Main 2022 — Physics Electromagnetism
As shown in the figure an inductor of inductance 200mH is connected to an AC source of emf 220Vand frequency 50Hz. The instantaneous voltage of the source is 0V when the peak value of current is πaA. The value of a is _____ .

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