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  4. An alternating emf E =440 sin 100 π t is applited to a circuit containing an inductance of (√2/π) H. If an a.c. ammeter is connected in the circuit, its reading will be :

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

JEE MainJEE Main 2022Alternating Current

Solution:

$ E =440 \operatorname{Sin} 100 \pi t, L=-\frac{\sqrt{2}}{\pi} H $
$ X _{ L }=\omega L =100 \pi \frac{\sqrt{2}}{\pi}=100 \sqrt{2} \Omega$
$ \text { Peak current } I _0=\frac{ E _0}{ X _{ L }}=\frac{440}{100 \sqrt{2}}=2.2 \sqrt{2} A$
$AC$ ammeter reads RMS value therefore reading will be $I _{ rms }$
$I _{ rms }=\frac{ I _0}{\sqrt{2}}=2.2 A$