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