An alternating emf E = 440 100 is applied to a circuit containing...

Given that $E = 440\sin 100\pit, L = \dfrac{\sqrt{2}}{\pi}H$

Angular frequency of the source is $\omega = 100\pi rad{s}^{- 1}$.

Now the reactance of the inductor will be,

${X}_{L} = \omegaL = 100\pi\dfrac{\sqrt{2}}{\pi} = 100\sqrt{2}Ω$

Therefore, the peak current ${I}_{0} = \dfrac{{E}_{0}}{{X}_{L}} = \dfrac{440}{100\sqrt{2}} = 2 . 2\sqrt{2}A$

$AC$ ammeter reads RMS value therefore reading will be ${I}_{rms}$

${I}_{rms} = \dfrac{{I}_{0}}{\sqrt{2}} = 2 . 2A$