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(C) Given: $\omega = 100 \, rad/s$,$L = \frac{\sqrt{3}}{10} \, H$,$R_1 = 10 \, \Omega$,$C = \frac{\sqrt{3}}{2} 	imes 10^{-3} \, F$,$R_2 = 20 \, \Omeg

Vedclass · Accepted Answer

$90^\circ$

Vedclass · Answer

(C) Given: $\omega = 100 \, rad/s$,$L = \frac{\sqrt{3}}{10} \, H$,$R_1 = 10 \, \Omega$,$C = \frac{\sqrt{3}}{2} 	imes 10^{-3} \, F$,$R_2 = 20 \, \Omega$.For the $L-R_1$ branch:Inductive reactance $X_L = \omega L = 100 	imes \frac{\sqrt{3}}{10} = 10\sqrt{3} \, \Omega$.The phase angle $\phi_1$ of current $I_1$ with respect to voltage $V$ is given by $	an \phi_1 = -\frac{X_L}{R_1} = -\frac{10\sqrt{3}}{10} = -\sqrt{3}$.Thus,$\phi_1 = -60^\circ$.For the $C-R_2$ branch:Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 	imes \frac{\sqrt{3}}{2} 	imes 10^{-3}} = \frac{1}{\frac{\sqrt{3}}{20}} = \frac{20}{\sqrt{3}} \, \Omega$.The phase angle $\phi_2$ of current $I_2$ with respect to voltage $V$ is given by $	an \phi_2 = \frac{X_C}{R_2} = \frac{20/\sqrt{3}}{20} = \frac{1}{\sqrt{3}}$.Thus,$\phi_2 = 30^\circ$.The phase difference between $I_1$ and $I_2$ is $|\phi_2 - \phi_1| = |30^\circ - (-60^\circ)| = 90^\circ$.

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