In the given circuit,the AC source has omega | vedclass

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(A, C, D) Given: $V_{rms} = 20 \ V$,$\omega = 100 \ rad/s$,$C = 100 \ \mu F$,$L = 0.5 \ H$.$1$. Impedance of the upper branch ($RC$ series): $X_C = \f

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$(A, C)$

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(A, C, D) Given: $V_{rms} = 20 \ V$,$\omega = 100 \ rad/s$,$C = 100 \ \mu F$,$L = 0.5 \ H$.$1$. Impedance of the upper branch ($RC$ series): $X_C = \frac{1}{\omega C} = \frac{1}{100 	imes 100 	imes 10^{-6}} = 100 \ \Omega$.$Z_1 = \sqrt{R_1^2 + X_C^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2} \ \Omega$.$I_{1,rms} = \frac{V_{rms}}{Z_1} = \frac{20}{100\sqrt{2}} = \frac{1}{5\sqrt{2}} \ A$.Voltage across $100 \ \Omega$ resistor: $V_{R1} = I_{1,rms} 	imes R_1 = \frac{1}{5\sqrt{2}} 	imes 100 = \frac{20}{\sqrt{2}} = 10\sqrt{2} \ V$. (Option $C$ is correct).$2$. Impedance of the lower branch ($RL$ series):$X_L = \omega L = 100 	imes 0.5 = 50 \ \Omega$.$Z_2 = \sqrt{R_2^2 + X_L^2} = \sqrt{50^2 + 50^2} = 50\sqrt{2} \ \Omega$.$I_{2,rms} = \frac{V_{rms}}{Z_2} = \frac{20}{50\sqrt{2}} = \frac{2}{5\sqrt{2}} \ A$.Voltage across $50 \ \Omega$ resistor: $V_{R2} = I_{2,rms} 	imes R_2 = \frac{2}{5\sqrt{2}} 	imes 50 = \frac{20}{\sqrt{2}} = 10\sqrt{2} \ V$. (Option $D$ is correct).$3$. Total current $I_{rms}$:The phase angle $\phi_1 = 	an^{-1}(\frac{-X_C}{R_1}) = -45^\circ$ and $\phi_2 = 	an^{-1}(\frac{X_L}{R_2}) = 45^\circ$.The phase difference between $I_1$ and $I_2$ is $90^\circ$. $I_{rms} = \sqrt{I_{1,rms}^2 + I_{2,rms}^2} = \sqrt{(\frac{1}{5\sqrt{2}})^2 + (\frac{2}{5\sqrt{2}})^2} = \sqrt{\frac{1}{50} + \frac{4}{50}} = \sqrt{\frac{5}{50}} = \frac{1}{\sqrt{10}} \approx 0.316 \ A \approx 0.3 \ A$. (Option $A$ is correct).

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