Physics - EMI/AC Question with Solution | TestHub

Impedance across $AB,RC$ part of the circuit $\begin{array}{rl}{\mathrm{Z}}_1=& \sqrt{X_c^2+R_1^2}=\sqrt{{\left(\frac{1}{\omega C}\right)}^2+R_1^2}\\ =& \sqrt{(100{)}^2+(100{)}^2}=100\sqrt{2}\end{array}$ $\therefore I_1=\frac{V}{Z_1}=\frac{20}{100\sqrt{2}}$ $[$ leads emf by ${\varphi }_1]$ where $\cos {\varphi }_1=\frac{R}{Z_1}=\frac{100}{100\sqrt{2}}=\frac{1}{\sqrt{2}}\Rightarrow \theta =45^{\circ }$
Impedance across $CD,LR$ part of the circuit. ${\mathrm{Z}}_2=\sqrt{X_L^2+R_2^2}=\sqrt{(\omega L{)}^2+R_2^2}$ $=\sqrt{(0.5\times 100{)}^2+(50{)}^2}=50\sqrt{2}\Omega$ $\therefore I_2=\frac{V}{Z_2}=\frac{20}{50\sqrt{2}}$
where $\cos {\varphi }_2=\frac{R}{Z_2}=\frac{50}{50\sqrt{2}}=\frac{1}{\sqrt{2}}\Rightarrow {\varphi }_2=45^{\circ }$ $\therefore$ Current $I$ from the circuit $I=\frac{20}{100\sqrt{2}}+\frac{20}{50\sqrt{2}}={\mathrm{I}}_1+{\mathrm{I}}_2\simeq 0.3\mathrm{A}$