Impedance of circuit
$\begin{aligned}
& \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}{\mathrm{L}}\right)^2}=\sqrt{\mathrm{R}^2+\left(\omega{\mathrm{L}}\right)^2} \
& =\sqrt{(100 \pi)^2+(2 \pi \times 50 \times 1)^2} \
& =\sqrt{(100 \pi)^2+(100 \pi)^2} \
& =\sqrt{2} \times 100 \pi \
& \mathrm{I}{\mathrm{rms}}=\frac{V}{2}=\frac{100 \pi}{\sqrt{2} \times 100 \pi}=\frac{1}{\sqrt{2}} \
& \mathrm{I}{\max }=\sqrt{2} \mathrm{I}_{\mathrm{rms}}=\sqrt{2} \times \frac{1}{\sqrt{2}}=1 \text { Ampere }
\end{aligned}$
Correct Answer : 1