Alternating Current

JEE Main level, single correct, medium

20 of 190

Question 20 of 190

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

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Answer: (C) 2.2A2.2A

Given that E=440sin⁡100πt,L=2πHE=440\sin 100\pi t,L=\dfrac{\sqrt{2}}{\pi }H
Angular frequency of the source is ω=100πrads−1\omega =100\pi rad{s}^{-1}.
Now the reactance of the inductor will be,
XL=ωL=100π2π=1002Ω{X}_{L}=\omega L=100\pi \dfrac{\sqrt{2}}{\pi }=100\sqrt{2}\Omega
Therefore, the peak current I0=E0XL=4401002=2.22A{I}_{0}=\dfrac{{E}_{0}}{{X}_{L}}=\dfrac{440}{100\sqrt{2}}=2.2\sqrt{2}A
ACAC ammeter reads RMS value therefore reading will be Irms{I}_{rms}
Irms=I02=2.2A{I}_{rms}=\dfrac{{I}_{0}}{\sqrt{2}}=2.2A

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