A parallel plate capacitor in series with a r | vedclass

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Question

(C) In a series $R-L-C$ circuit, the resonance frequency is given by $f_0 = \frac{1}{2\pi\sqrt{LC}}$.Given: $R = 100 \Omega$, $L = 20 	ext{ mH} = 20

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$0.2 	ext{ mC}$

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(C) In a series $R-L-C$ circuit, the resonance frequency is given by $f_0 = \frac{1}{2\pi\sqrt{LC}}$.Given: $R = 100 \Omega$, $L = 20 	ext{ mH} = 20 	imes 10^{-3} 	ext{ H}$, $f_0 = \frac{1250}{\pi} 	ext{ Hz}$.Substituting the values into the resonance formula:$\frac{1250}{\pi} = \frac{1}{2\pi\sqrt{20 	imes 10^{-3} 	imes C}}$$1250 = \frac{1}{2\sqrt{0.02 	imes C}}$$2500 = \frac{1}{\sqrt{0.02 	imes C}}$Squaring both sides:$6.25 	imes 10^6 = \frac{1}{0.02 	imes C}$$C = \frac{1}{0.02 	imes 6.25 	imes 10^6} = \frac{1}{0.125 	imes 10^6} = 8 	imes 10^{-6} 	ext{ F} = 8 \mu	ext{F}$.When charged by a $DC$ source of $V = 25 	ext{ V}$, the charge $Q$ stored is:$Q = C 	imes V = 8 	imes 10^{-6} 	ext{ F} 	imes 25 	ext{ V} = 200 	imes 10^{-6} 	ext{ C} = 0.2 	imes 10^{-3} 	ext{ C} = 0.2 	ext{ mC}$.

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