A resistor of 2 Omega,an inductor of 100 mu | vedclass

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(B) At resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$,and the impedance $Z$ of the circuit is equal to the resista

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$25$

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(B) At resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$,and the impedance $Z$ of the circuit is equal to the resistance $R$.Given: $R = 2 \ \Omega$,$L = 100 \ \mu H = 100 	imes 10^{-6} \ H$,$C = 400 \ pF = 400 	imes 10^{-12} \ F$,and $V_{rms} = 0.1 \ V$.The current in the circuit at resonance is $i = \frac{V_{rms}}{Z} = \frac{V_{rms}}{R} = \frac{0.1}{2} = 0.05 \ A$.The resonant angular frequency is $\omega = \frac{1}{\sqrt{LC}}$.The voltage drop across the inductor is $V_L = i X_L = i \omega L = i \left(\frac{1}{\sqrt{LC}}ight) L = i \sqrt{\frac{L}{C}}$.Substituting the values: $V_L = 0.05 	imes \sqrt{\frac{100 	imes 10^{-6}}{400 	imes 10^{-12}}} = 0.05 	imes \sqrt{\frac{100}{400} 	imes 10^6} = 0.05 	imes \sqrt{0.25 	imes 10^6} = 0.05 	imes 0.5 	imes 10^3 = 0.05 	imes 500 = 25 \ V$.

$A$ resistor of $2 \ \Omega$,an inductor of $100 \ \mu H$,and a capacitor of $400 \ pF$ are connected in series across an $A$.$C$. source of $e_{rms} = 0.1 \ V$. At resonance,the voltage drop across the inductor is: (in $V$)

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