A series LCR circuit with L = frac{100}{pi} t | vedclass

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(B) Given: $L = \frac{100}{\pi} 	imes 10^{-3} 	ext{ H}$,$C = \frac{10^{-3}}{\pi} 	ext{ F}$,$R = 10 \ \Omega$,$f = 50 	ext{ Hz}$.First,calculate th

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(B) Given: $L = \frac{100}{\pi} 	imes 10^{-3} 	ext{ H}$,$C = \frac{10^{-3}}{\pi} 	ext{ F}$,$R = 10 \ \Omega$,$f = 50 	ext{ Hz}$.First,calculate the inductive reactance $X_L$:$X_L = \omega L = 2\pi f L = 2\pi 	imes 50 	imes \frac{100}{\pi} 	imes 10^{-3} = 100 	imes 100 	imes 10^{-3} = 10 \ \Omega$.Next,calculate the capacitive reactance $X_C$:$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} = \frac{1}{2\pi 	imes 50 	imes \frac{10^{-3}}{\pi}} = \frac{1}{100 	imes 10^{-3}} = \frac{1}{0.1} = 10 \ \Omega$.Since $X_L = X_C$,the circuit is in resonance.At resonance,the impedance $Z = R = 10 \ \Omega$.The power factor $\cos \phi = \frac{R}{Z} = \frac{10}{10} = 1$.

$A$ series $LCR$ circuit with $L = \frac{100}{\pi} \text{ mH}$,$C = \frac{10^{-3}}{\pi} \text{ F}$,and $R = 10 \ \Omega$ is connected across an $AC$ source of $220 \text{ V}, 50 \text{ Hz}$ supply. The power factor of the circuit would be . . . . . . .

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