In the given circuit, the AC source has ω=100 rad / s . Considering...

Impedance across $A B, R C$ part of the circuit $$\begin{aligned} \mathrm{Z}_{1}=& \sqrt{X_{c}^{2}+R_{1}^{2}}=\sqrt{\left(\frac{1}{\omega C}\right)^{2}+R_{1}^{2}} \\=& \sqrt{(100)^{2}+(100)^{2}}=100 \sqrt{2} \end{aligned}$$\therefore \quad I_{1}=\frac{V}{Z_{1}}=\frac{20}{100 \sqrt{2}} \quad$\left[\right.$ leads emf by \left.\phi_{1}\right]$where$\cos \phi_{1}=\frac{R}{Z_{1}}=\frac{100}{100 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}$<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/ed64m2hdoG2W80ioJw89XZcyqG4WBGJtaNWRyTdBbrs.original.fullsize.png"><br> Impedance across$C D, L R$part of the circuit.$\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 \quad I_{2}=\frac{V}{Z_{2}}=\frac{20}{50 \sqrt{2}}$<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/srQHHHyf8YGl8W5QJp-lTf3vffyvgV-w4nhnlGsmfq0.original.fullsize.png"><br> where$\cos \phi_{2}=\frac{R}{Z_{2}}=\frac{50}{50 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow \phi_{2}=45^{\circ} \therefore \quad$Current$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}$