In the given circuit, the AC source has $\mathrm{\omega }$ = 100 rad/s . Considering the inductor and capacitor to be ideal, the correct choice(s) is (are)
 

${\mathrm{X}}_{\mathrm{L}}=\mathrm{\omega L}=100\times 0.5=50\mathrm{\Omega }$
 

$$\begin{gathered} {\text{X}}_{\mathrm{C}}=\frac{1}{\mathrm{\omega C}}=\frac{1}{100\times 100\times {10}^{-6}}=100\mathrm{\Omega } \\ \begin{array}{l}{\mathrm{Z}}_1=100\sqrt{2},{\mathrm{I}}_1=\frac{20}{100\sqrt{2}}=\frac{1}{5\sqrt{2}}\\ {\mathrm{V}}_{\mathrm{across} }100\mathrm{\Omega }=\frac{1}{5\sqrt{2}}\times 100=\frac{20}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=10\sqrt{2}\end{array} \end{gathered}$$
Phase difference between I₁ and V$\Rightarrow \cos {\mathrm{\varphi }}_1=\frac{{\mathrm{R}}_1}{{\mathrm{Z}}_1}=\frac{100}{100\sqrt{2}}$
${\mathrm{\varphi }}_1=\mathrm{\pi }/4$
I₁ lead V

$$\begin{gathered} {\text{Z}}_2=50\sqrt{2};{\mathrm{I}}_2=\frac{20}{50\sqrt{2}}=\frac{2}{5\sqrt{2}} \\ {\mathrm{V}}_{\mathrm{rms}} \mathrm{across} 50\mathrm{\Omega }=\frac{2}{5\sqrt{2}}\times 50=10\sqrt{2} \end{gathered}$$
Phase difference between $\frac{\mathrm{\pi }}{2}$ and
$\begin{array}{l}\mathrm{V}=\cos {\mathrm{\varphi }}_2=\frac{{\mathrm{R}}_2}{{\mathrm{Z}}_2}=\frac{50}{50\sqrt{2}}=\frac{1}{\sqrt{2}}\\ {\mathrm{\varphi }}_2=\mathrm{\pi }/4\\ {\mathrm{I}}_2\mathrm{lag}\mathrm{Vby\pi }/4\\ \mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2\\ {\mathrm{I}}_{\text{net }}=\sqrt{{\mathrm{I}}_1^2+{\mathrm{I}}_2^2}\\ \mathrm{I}=\sqrt{\frac{4}{25\times 2}+\frac{1}{25\times 2}}=\sqrt{\frac{5}{50}}=\frac{1}{\sqrt{10}}\end{array}$
 

= 0.316