In the given circuit, the \(\mathrm{AC}\) source has \(\omega=100 \mathrm{rad} / \mathrm{s}\). Considering the inductor and capacitor to be ideal, the correct choice(s) is (are)

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 \(\cos \phi_{1}=\frac{R}{Z_{1}}=\frac{100}{100 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}\)



Impedance across \(C D, L R\) part of the circuit.

\(\mathrm{Z}_{2}=\sqrt{X_{L}^{2}+R_{2}^{2}}=\sqrt{(\omega L)^{2}+R_{2}^{2}}\)

\(=\sqrt{(0.5 \times 100)^{2}+(50)^{2}}=50 \sqrt{2} \Omega\)

\(\therefore \quad I_{2}=\frac{V}{Z_{2}}=\frac{20}{50 \sqrt{2}}\)


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

\(I=\frac{20}{100 \sqrt{2}}+\frac{20}{50 \sqrt{2}}=\mathrm{I}_{1}+\mathrm{I}_{2} \simeq 0.3 \mathrm{~A}\)