In the given circuit the AC source has ω = 100rad s ^ - 1 ....

${Z}_{C} = \sqrt{{\left(\dfrac{1}{\omegaC}\right)}^{2} + {R}^{2}}$

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

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

$= 100\sqrt{2}Ω$

${Z}_{L} = \sqrt{( \omegaL{)}^{2} + {R}^{2}} = \sqrt{( 100 \times 0 . 5{)}^{2} + {50}^{2}} = 50\sqrt{2}Ω$

${i}_{L} = \dfrac{200}{{Z}_{L}} = \dfrac{200}{50\sqrt{2}} = \dfrac{4}{\sqrt{2}} = 2\sqrt{2}\text{A}$

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

$cos{ϕ}_{1} = \dfrac{100}{100\sqrt{2}} = \dfrac{1}{\sqrt{2}}, {ϕ}_{1} = 45 °$

$cos{ϕ}_{2} = \dfrac{50}{50\sqrt{2}} = \dfrac{1}{\sqrt{2}}, {ϕ}_{2} = 45 °$

$I = \sqrt{{I}_{C}^{2} + {I}_{L}^{2}}; I = \sqrt{8 + 2}; I = \sqrt{10} = 3 . 16A$

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