A series LCR circuit with L=0.5 text{ H} and | vedclass

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(A) Key Idea: The current amplitude in a series $LCR$ circuit is maximum at resonance, where the inductive reactance equals the capacitive reactance,

Vedclass · Accepted Answer

$3000$

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(A) Key Idea: The current amplitude in a series $LCR$ circuit is maximum at resonance, where the inductive reactance equals the capacitive reactance, i.e., $\omega L = \frac{1}{\omega C}$.Given: $L = 0.5 	ext{ H}$, $R = 10 \Omega$, $V_{rms} = 200 	ext{ V}$, and $f = \frac{150}{\pi} 	ext{ Hz}$.Angular frequency $\omega = 2 \pi f = 2 \pi 	imes \frac{150}{\pi} = 300 	ext{ rad/s}$.At resonance, the impedance $Z = R = 10 \Omega$.The maximum current $I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{10} = 20 	ext{ A}$.The $rms$ voltage across the inductor is $V_L = I_{rms} 	imes X_L$, where $X_L = \omega L$.$V_L = 20 	imes (300 	imes 0.5) = 20 	imes 150 = 3000 	ext{ V}$.

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