A sinusoidal voltage with a frequency of 50 | vedclass

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(D) The impedance $Z$ for a series $LCR$ circuit is given by the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Here,$R = 5 \ \Omega$,$L = 20 \ mH = 20 	i

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$5$

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(D) The impedance $Z$ for a series $LCR$ circuit is given by the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Here,$R = 5 \ \Omega$,$L = 20 \ mH = 20 	imes 10^{-3} \ H$,and $C = 500 \ \mu F = 500 	imes 10^{-6} \ F$.The angular frequency is $\omega = 2 \pi f = 2 \pi 	imes 50 = 100 \pi \ rad/s$.Inductive reactance $X_L = \omega L = 100 \pi 	imes 20 	imes 10^{-3} = 2 \pi \approx 6.28 \ \Omega$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \pi 	imes 500 	imes 10^{-6}} = \frac{1}{0.05 \pi} = \frac{20}{\pi} \approx 6.37 \ \Omega$.Now,$Z = \sqrt{5^2 + (6.28 - 6.37)^2} = \sqrt{25 + (-0.09)^2} = \sqrt{25 + 0.0081} = \sqrt{25.0081} \approx 5 \ \Omega$.

$A$ sinusoidal voltage with a frequency of $50 \ Hz$ is applied to a series $LCR$ circuit with a resistance of $5 \ \Omega$,inductance of $20 \ mH$ and a capacitance of $500 \ \mu F$. The magnitude of impedance of the circuit is close to (in $\Omega$)

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