Physics - EMI/AC Question with Solution | TestHub

Given that $E=440\sin 100\pi t,L=\frac{\sqrt{2}}{\pi }\mathrm{H}$

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

Now the reactance of the inductor will be,

$X_L=\omega L=100\pi \frac{\sqrt{2}}{\pi }=100\sqrt{2}\Omega$

Therefore, the peak current $I_0=\frac{E_0}{X_L}=\frac{440}{100\sqrt{2}}=2.2\sqrt{2}\mathrm{A}$

$A\mathrm{C}$ ammeter reads RMS value therefore reading will be $I_{rms}$

$I_{rms}=\frac{I_0}{\sqrt{2}}=2.2\mathrm{A}$