In a series LCR circuit,R=300 Omega, L=0.9 te | vedclass

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(B) Given: $R = 300 \Omega, L = 0.9 	ext{ H}, C = 2.0 \mu	ext{F} = 2.0 	imes 10^{-6} 	ext{ F}, \omega = 1000 	ext{ rad/s}$.The impedance $Z$ of a

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$500$

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(B) Given: $R = 300 \Omega, L = 0.9 	ext{ H}, C = 2.0 \mu	ext{F} = 2.0 	imes 10^{-6} 	ext{ F}, \omega = 1000 	ext{ rad/s}$.The impedance $Z$ of a series $LCR$ circuit is given by the formula:$Z = \sqrt{R^2 + (X_L - X_C)^2}$First,calculate the inductive reactance $X_L$:$X_L = \omega L = 1000 	imes 0.9 = 900 \Omega$Next,calculate the capacitive reactance $X_C$:$X_C = \frac{1}{\omega C} = \frac{1}{1000 	imes 2.0 	imes 10^{-6}} = \frac{1}{2 	imes 10^{-3}} = 500 \Omega$Now,substitute these values into the impedance formula:$Z = \sqrt{300^2 + (900 - 500)^2}$$Z = \sqrt{90000 + (400)^2}$$Z = \sqrt{90000 + 160000}$$Z = \sqrt{250000} = 500 \Omega$Therefore,the impedance of the circuit is $500 \Omega$.

In a series $LCR$ circuit,$R=300 \Omega, L=0.9 \text{ H}, C=2.0 \mu\text{F}$ and $\omega=1000 \text{ rad/s}$,then the impedance of the circuit is: (in $\Omega$)

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