Explanation:
C Given,
rms voltage, $\mathrm{V}_{\mathrm{rms}}=220 \mathrm{~V}$
Frequency, $\mathrm{f}=50 \mathrm{~Hz}$
Inductance, $\mathrm{L}=1 \mathrm{H}$
We know that,
Inductance Reactance $\mathrm{X}_{\mathrm{C}}=\omega \mathrm{L}$
$=2 \pi \mathrm{f} . \mathrm{L} \quad(\omega=2 \pi \mathrm{f})$
$=2 \pi 50 \times 1=314 \Omega$
So, the rms value of current, $I_{r m s}=\frac{V_{\text {rms }}}{X_{L}}$
$I_{\mathrm{rms}}=\frac{220}{314}$
$I_{\mathrm{rms}}=0.7 \mathrm{~A}$
Peak value of current, $\mathrm{I}_{0}=\sqrt{2} \mathrm{I}_{\mathrm{rms}}$
$\mathrm{I}_{0}=1.414 \times 0.7=1 \mathrm{~A}$
Hence, the peak value of current, $\mathrm{I}_{0}=1 \mathrm{~A}$.