An alternating emf E=440 sin100π t is applied to a circuit containing an inductance of √2 π H. If an a.c. ammeter is connected in the circuit, its…
JEE Main 2022 — Physics Electromagnetism
2022mcqeasy
An alternating emf E=440sin100πt is applied to a circuit containing an inductance of π2H. If an a.c. ammeter is connected in the circuit, its reading will be :
Official previous-year question
Held on 29 Jul 2022 · Verified 6 Jul 2026.
Options
A
4.4A
B
1.55A
C
2.2A
D
3.11A
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Solution
Given that E=440sin100πt,L=π2H
Angular frequency of the source is ω=100πrads−1.
Now the reactance of the inductor will be,
XL=ωL=100ππ2=1002Ω
Therefore, the peak current I0=XLE0=1002440=2.22A
AC ammeter reads RMS value therefore reading will be Irms
Irms=2I0=2.2A
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