Physics - EMI/AC Question with Solution | TestHub

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

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

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

$=100\sqrt{2}\Omega$

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

$i_L=\frac{200}{Z_L}=\frac{200}{50\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\mathrm{A}$

$i_C=\frac{200}{Z_C}=\frac{200}{100\sqrt{2}}=\sqrt{2}\mathrm{A}$

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

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

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

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