Obtain the resonant frequency omega_{r} of a | vedclass

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(A) Given values are $L = 2.0 \; H$,$C = 32 	imes 10^{-6} \; F$,and $R = 10 \; \Omega$.The resonant frequency $\omega_{r}$ is given by the formula $\

Vedclass · Accepted Answer

$25$

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(A) Given values are $L = 2.0 \; H$,$C = 32 	imes 10^{-6} \; F$,and $R = 10 \; \Omega$.The resonant frequency $\omega_{r}$ is given by the formula $\omega_{r} = \frac{1}{\sqrt{LC}}$.Substituting the values: $\omega_{r} = \frac{1}{\sqrt{2.0 	imes 32 	imes 10^{-6}}} = \frac{1}{\sqrt{64 	imes 10^{-6}}} = \frac{1}{8 	imes 10^{-3}} = 125 \; rad/s$.The $Q$-value (Quality factor) of a series $LCR$ circuit is given by $Q = \frac{\omega_{r} L}{R}$.Substituting the values: $Q = \frac{125 	imes 2.0}{10} = \frac{250}{10} = 25$.Alternatively,$Q = \frac{1}{R} \sqrt{\frac{L}{C}} = \frac{1}{10} \sqrt{\frac{2.0}{32 	imes 10^{-6}}} = \frac{1}{10} \sqrt{\frac{1}{16 	imes 10^{-6}}} = \frac{1}{10} 	imes \frac{1}{4 	imes 10^{-3}} = \frac{1000}{40} = 25$.Thus,the $Q$-value is $25$.

Obtain the resonant frequency $\omega_{r}$ of a series $LCR$ circuit with $L=2.0 \;H, C=32\; \mu F$ and $R=10\; \Omega$. What is the $Q$-value of this circuit?

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