In the given circuit, the AC source has ω=100rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are): The current through the circuit, I is 0.3 A. The current through the circuit, I is 0.3√2 A. The voltage across 100Ω resistor = 10√2 V. The voltage across 50Ω resistor = 10V.

In the upper branch the net impedance,(C=100μF,R=100Ω)Therefore, the net impedance will be:Z1=√(1/ωC)²+R²=√[100² + 100²=100√2ΩThe current flowing the upper branch:I1=V/Z1=20/100√2 Awhich leads by(45°)w.r.t. voltage.In the lower branch the net impedance,(L=0.5 H ,R=50Ω)Therefore, the net impedance will be:Z2=√(ωL)²+R²=√(0.5×100)²+100²=50√2ΩThe current flowing the lower branch:I2=V/Z2=20/50√2 Awhich lags by(45°)w.r.t. voltage.Thus total current I is given by summation of phasors(I1)and(I2)which differ by(90°)in phase and henceI=√I₁²+I₂²=1/√10 A≈0.3 AAlso voltage across 100Ω resistor=I1R1=(20/100√2)×100=10√2V)Similarly across 50Ω resistor=I2R2=(20/50√2)×50=10√2VSo, options A and C is correct.