In the series L-C-R circuit shown,the impedan | vedclass

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(D) Given: Inductance $L = 1 	ext{ H}$,Capacitance $C = 20 \mu	ext{F} = 20 	imes 10^{-6} 	ext{ F}$,Resistance $R = 300 \Omega$,Frequency $f = \fra

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$500$

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(D) Given: Inductance $L = 1 	ext{ H}$,Capacitance $C = 20 \mu	ext{F} = 20 	imes 10^{-6} 	ext{ F}$,Resistance $R = 300 \Omega$,Frequency $f = \frac{50}{\pi} 	ext{ Hz}$.First,calculate the inductive reactance $X_{L}$:$X_{L} = 2 \pi f L = 2 \pi \left(\frac{50}{\pi}ight) 	imes 1 = 100 \Omega$.Next,calculate the capacitive reactance $X_{C}$:$X_{C} = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \left(\frac{50}{\pi}ight) 	imes 20 	imes 10^{-6}} = \frac{1}{100 	imes 20 	imes 10^{-6}} = \frac{1}{2 	imes 10^{-3}} = 500 \Omega$.The impedance $Z$ of a series $L-C-R$ circuit is given by:$Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}$.Substituting the values:$Z = \sqrt{(300)^{2} + (500 - 100)^{2}} = \sqrt{300^{2} + 400^{2}} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \Omega$.

In the series $L-C-R$ circuit shown,the impedance is (in $\Omega$)

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