In a series L-C-R circuit,C = 10^{-11} , F,L | vedclass

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(D) Given: $C = 10^{-11} \, F$,$L = 10^{-5} \, H$,$R = 100 \, \Omega$,and $q_{DC} = 10^{-9} \, C$.$1$. First,find the $D.C.$ voltage $E$ using the ste

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$10^{-8} \, C$

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(D) Given: $C = 10^{-11} \, F$,$L = 10^{-5} \, H$,$R = 100 \, \Omega$,and $q_{DC} = 10^{-9} \, C$.$1$. First,find the $D.C.$ voltage $E$ using the steady-state condition where the capacitor acts as an open circuit: $E = q_{DC} / C = 10^{-9} / 10^{-11} = 100 \, V$. Since $E_0 = E$,the peak voltage of the $A.C.$ source is $E_0 = 100 \, V$.$2$. At resonance,the impedance of the $L-C-R$ circuit is $Z = R = 100 \, \Omega$. The peak current is $I_0 = E_0 / Z = 100 / 100 = 1 \, A$.$3$. The resonant angular frequency is $\omega = 1 / \sqrt{LC} = 1 / \sqrt{10^{-5} 	imes 10^{-11}} = 1 / \sqrt{10^{-16}} = 10^8 \, rad/s$.$4$. The peak charge $Q_0$ on the capacitor in an $A.C.$ circuit is related to the peak current $I_0$ by $Q_0 = I_0 / \omega$.$5$. Substituting the values: $Q_0 = 1 / 10^8 = 10^{-8} \, C$.

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