A capacitance of left(frac{10^{-3}}{2 pi}righ | vedclass

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Question

(C) Given: Capacitance $C = \frac{10^{-3}}{2 \pi} F$,Inductance $L = \frac{100}{\pi} mH = \frac{100}{\pi} 	imes 10^{-3} H$,Resistance $R = 10 \Omega$

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) Given: Capacitance $C = \frac{10^{-3}}{2 \pi} F$,Inductance $L = \frac{100}{\pi} mH = \frac{100}{\pi} 	imes 10^{-3} H$,Resistance $R = 10 \Omega$,and Frequency $f = 50 Hz$.First,calculate the inductive reactance $X_L = \omega L = 2 \pi f L = 2 \pi 	imes 50 	imes \frac{100}{\pi} 	imes 10^{-3} = 10 \Omega$.Next,calculate the capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C} = \frac{1}{2 \pi 	imes 50 	imes \frac{10^{-3}}{2 \pi}} = \frac{1}{50 	imes 10^{-3}} = \frac{1000}{50} = 20 \Omega$.The phase angle $\phi$ is given by $	an \phi = \frac{X_C - X_L}{R}$.Substituting the values: $	an \phi = \frac{20 - 10}{10} = \frac{10}{10} = 1$.Therefore,$\phi = 	an^{-1}(1) = 45^{\circ}$.

$A$ capacitance of $\left(\frac{10^{-3}}{2 \pi}\right) F$,an inductance of $\left(\frac{100}{\pi}\right) mH$,and a resistance of $10 \Omega$ are connected in series with an $AC$ voltage source of $220 V, 50 Hz$. The phase angle of the circuit is (in $^{\circ}$)

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