JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Alternating Current

The upper branch has $100\Omega$ resistor in series with capacitor ($X_C=100\Omega$), giving impedance $Z_1=100\sqrt{2}$ at $45°$ lead. Current $I_1=\frac{20}{100\sqrt{2}}=\frac{1}{5\sqrt{2}}$ A. The lower branch has $50\Omega$ resistor in series with inductor ($X_L=50\Omega$), giving impedance $Z_2=50\sqrt{2}$ at $45°$ lag. Current $I_2=\frac{20}{50\sqrt{2}}=\frac{2}{5\sqrt{2}}$ A. The total current $I=\sqrt{I_1^2+I_2^2}=\sqrt{\frac{1}{50}+\frac{4}{50}}=\sqrt{\frac{1}{10}}\approx 0.3$ A. Voltage across $100\Omega$ resistor: $V=I_1\times 100=\frac{20}{100\sqrt{2}}\times 100=\frac{20}{\sqrt{2}}=10\sqrt{2}$ V. The official answer accepts both (C) alone and (A,C) together.