An inductor of left(frac{100}{pi}right) mH,a | vedclass

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(D) Given: Inductance $L = \frac{100}{\pi} 	ext{ mH} = \frac{0.1}{\pi} 	ext{ H}$,Capacitance $C = \frac{10^{-3}}{2\pi} 	ext{ F}$,Resistance $R = 10

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(D) Given: Inductance $L = \frac{100}{\pi} 	ext{ mH} = \frac{0.1}{\pi} 	ext{ H}$,Capacitance $C = \frac{10^{-3}}{2\pi} 	ext{ F}$,Resistance $R = 10 \Omega$,Frequency $f = 50 	ext{ Hz}$.First,calculate the inductive reactance $X_L = 2\pi f L = 2\pi 	imes 50 	imes \frac{0.1}{\pi} = 100 	imes 0.1 = 10 \Omega$.Next,calculate the capacitive reactance $X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi 	imes 50 	imes \frac{10^{-3}}{2\pi}} = \frac{1}{50 	imes 10^{-3}} = \frac{1000}{50} = 20 \Omega$.The tangent of the phase angle $\phi$ is given by $	an \phi = \frac{X_L - X_C}{R}$.Substituting the values: $	an \phi = \frac{10 - 20}{10} = \frac{-10}{10} = -1$.The magnitude of the tangent of the phase angle is $|	an \phi| = 1$.

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