An R-L-C circuit consists of a 150 Omega resi | vedclass

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(D) In an $R-L-C$ series circuit,the given values are:$R = 150 \Omega$$C = 20 \mu F = 20 	imes 10^{-6} F = 2 	imes 10^{-5} F$$L = 500 mH = 0.5 H$$\o

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$	an^{-1}(0.5)$

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(D) In an $R-L-C$ series circuit,the given values are:$R = 150 \Omega$$C = 20 \mu F = 20 	imes 10^{-6} F = 2 	imes 10^{-5} F$$L = 500 mH = 0.5 H$$\omega = 400 rad s^{-1}$First,calculate the inductive reactance $X_L$:$X_L = \omega L = 400 	imes 0.5 = 200 \Omega$Next,calculate the capacitive reactance $X_C$:$X_C = \frac{1}{\omega C} = \frac{1}{400 	imes 2 	imes 10^{-5}} = \frac{1}{800 	imes 10^{-5}} = \frac{10^5}{800} = \frac{1000}{8} = 125 \Omega$The phase angle $\phi$ is given by the formula:$	an \phi = \frac{X_L - X_C}{R}$Substituting the values:$	an \phi = \frac{200 - 125}{150} = \frac{75}{150} = 0.5$Therefore,the phase angle is $\phi = 	an^{-1}(0.5)$.

An $R-L-C$ circuit consists of a $150 \Omega$ resistor,$20 \mu F$ capacitor,and a $500 mH$ inductor connected in series with a $100 V$ $AC$ supply. The angular frequency of the supply voltage is $400 rad s^{-1}$. The phase angle between the current and the applied voltage is

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