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7.Alternating Current
hard
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 $.......A$
A
$4.4$
B
$1.55$
C
$2.2$
D
$3.1$
(JEE MAIN-2022)
Solution
$E =440 \sin 100 \pi t , L =-\frac{\sqrt{2}}{\pi} H$
$X _{ L }=\omega L =100 \pi \frac{\sqrt{2}}{\pi}=100 \sqrt{2} \Omega$
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 $I_{\text {ms }}$
$I_{\text {mus }}=\frac{I_{0}}{\sqrt{2}}=2.2\,A$
Standard 12
Physics
