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

Answer: A::CSolution: Impedance across AB
$\left(Z_1\right)=\sqrt{\left(X_c^2\right)+\left(R_1^2\right)}={\sqrt{\frac{1}{\omega C}}}^2+\left(R_1^2\right))$
$=\sqrt{\left({100}^2\right)+\left({100}^2\right)}=100\sqrt{2}$
$\therefore \left(I_1\right)=\frac{V}{Z_1}=\frac{20}{100\sqrt{2}}\left[\le adsemfby\left({\varphi }_1\right)\right]$
where $\cos \left({\varphi }_1\right)=\frac{R}{100\sqrt{2}}=\frac{1}{\sqrt{2}}\Rightarrow \theta ={45}^{\circ }$

Impendance across CD is
$\left(Z_2\right)=\sqrt{X_L^2+R_2^2}$
$=\sqrt{\left(0.5\times {100}^2\right)+\left({50}^2\right)}=50\sqrt{2}\left(\omega \right)$
$\therefore \left(I_2\right)=\frac{V}{Z_2}=\frac{20}{50\sqrt{2}}$ [lead emf by $\left({\varphi }_1\right)$]

where cos$\left({\varphi }_2\right)=\frac{R}{Z_2}=\frac{50}{50\sqrt{2}}=\frac{1}{\sqrt{2}}\Rightarrow \left({\varphi }_2\right)={45}^{\circ }$
The current I from the circuit is
$I=\left(I_1\right)+\left(I_2\right)\stackrel{\sim }{=}0.3A$.