In a series R-L-C circuit,R = 300 , Omega,L = | vedclass

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(C) The impedance $Z$ of a series $R-L-C$ circuit is given by the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2}$.First,calculate the inductive reactance $X

Vedclass · Accepted Answer

$500$

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(C) The impedance $Z$ of a series $R-L-C$ circuit is given by the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2}$.First,calculate the inductive reactance $X_L = \omega L = 1000 	imes 0.9 = 900 \, \Omega$.Next,calculate the capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{1000 	imes 2.0 	imes 10^{-6}} = \frac{10^6}{2000} = 500 \, \Omega$.Now,substitute these values into the impedance formula:$Z = \sqrt{300^2 + (900 - 500)^2}$$Z = \sqrt{300^2 + 400^2}$$Z = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, \Omega$.

In a series $R-L-C$ circuit,$R = 300 \, \Omega$,$L = 0.9 \, H$,$C = 2.0 \, \mu F$,and $\omega = 1000 \, rad/s$. The impedance of the circuit is ........ $\Omega$.

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