In the given circuit the AC source has ω = 100   rad   s - 1 . Considering the inductor and…

$Z_C=\sqrt{{\left(\frac{1}{\omega C}\right)}^2+R^2}$

$=\sqrt{{\left(\frac{1}{100\times 100\times {10}^{-6}}\right)}^2+{100}^2}$

$=\sqrt{{100}^2+{100}^2}$

$=100\sqrt{2} \Omega$

$Z_L=\sqrt{(\omega L{)}^2+R^2}=\sqrt{(100\times 0.5{)}^2+{50}^2}=50\sqrt{2} \Omega$

$i_L=\frac{200}{{\mathrm{Z}}_{\mathrm{L}}}=\frac{200}{50\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2} \text{A}$

${\mathrm{i}}_{\mathrm{C}}=\frac{200}{{\mathrm{Z}}_{\mathrm{C}}}=\frac{200}{100\sqrt{2}}=\sqrt{2} \text{A}$

$\cos {\varphi }_1=\frac{100}{100\sqrt{2}}=\frac{1}{\sqrt{2}}, {\varphi }_1=45°$

$\cos {\varphi }_2=\frac{50}{50\sqrt{2}}=\frac{1}{\sqrt{2}}, {\varphi }_2=45°$

$I=\sqrt{I_C^2+I_L^2}; I=\sqrt{8+2}; I=\sqrt{10}=3.16 \mathrm{A}$

This question is bonus as the correct answer was not available for students to mark as option in original paper by NTA.