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

${\mathrm{X}}_{\mathrm{C}}=\frac{1}{\mathrm{\omega C}}=\frac{1}{100\times \left(100\times {10}^{-6}\right)}=100 \mathrm{\Omega }$

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

For the upper branch, ${\mathrm{I}}_{\text{upper }}=\frac{\mathrm{V}}{\sqrt{{{\mathrm{X}}_{\mathrm{C}}}^2+{{\mathrm{R}}_1}^2}}=\frac{20}{\sqrt{{100}^2+{100}^2}}=\frac{1}{5\sqrt{2}} \mathrm{A}$.

Current in the upper branch leads voltage by  ${\mathrm{\varphi }}_1={\tan }^{-1}\left(\frac{{\mathrm{X}}_{\mathrm{C}}}{{\mathrm{R}}_1}\right)={\tan }^{-1}\left(\frac{100}{100}\right)=\frac{\mathrm{\pi }}{4}$.

For the lower branch, ${\mathrm{I}}_{\text{lower }}=\frac{\mathrm{V}}{\sqrt{{{\mathrm{X}}_{\mathrm{L}}}^2+{{\mathrm{R}}_2}^2}}=\frac{20}{\sqrt{{50}^2+{50}^2}}=\frac{2}{5\sqrt{2}} \mathrm{A}$.

Current in the lower branch lags voltage by  ${\mathrm{\varphi }}_2={\tan }^{-1}\left(\frac{{\mathrm{X}}_{\mathrm{L}}}{{\mathrm{R}}_2}\right)={\tan }^{-1}\left(\frac{50}{50}\right)=\frac{\mathrm{\pi }}{4}$.

As the angle between the currents in upper and lower branches is  $\mathrm{\varphi }={\mathrm{\varphi }}_1+{\mathrm{\varphi }}_2=\frac{\mathrm{\pi }}{2}$, 

total current  $\mathrm{I}=\sqrt{{{\mathrm{I}}_{\text{upper}}}^2+{{\mathrm{I}}_{\text{lower}}}^2}=\sqrt{{\left(\frac{1}{5\sqrt{2}}\right)}^2+{\left(\frac{2}{5\sqrt{2}}\right)}^2}=\frac{1}{\sqrt{10}}\approx 0.3 \mathrm{A}$.

Voltage across 100 Ω resistor, ${\mathrm{V}}_1={\mathrm{I}}_{\text{upper}}{\mathrm{R}}_1=\frac{1}{5\sqrt{2}}\times 100=10\sqrt{2} \mathrm{V}$.

Voltage across 50 Ω resistor, ${\mathrm{V}}_2={\mathrm{I}}_{\text{lower}}{\mathrm{R}}_2=\frac{2}{5\sqrt{2}}\times 50=10\sqrt{2} \mathrm{V}$.