A 400 Omega resistor, a 250 text{ mH} inducto | vedclass

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(D) Given: $R = 400 \Omega$, $L = 250 	ext{ mH} = 0.25 	ext{ H}$, $C = 2.5 \mu 	ext{F} = 2.5 	imes 10^{-6} 	ext{ F}$, $V_0 = 5 	ext{ V}$, $\omeg

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$5$

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(D) Given: $R = 400 \Omega$, $L = 250 	ext{ mH} = 0.25 	ext{ H}$, $C = 2.5 \mu 	ext{F} = 2.5 	imes 10^{-6} 	ext{ F}$, $V_0 = 5 	ext{ V}$, $\omega = 2000 	ext{ rad/s}$.First, calculate the inductive reactance: $X_L = \omega L = 2000 	imes 0.25 = 500 \Omega$.Next, calculate the capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{2000 	imes 2.5 	imes 10^{-6}} = \frac{1}{0.005} = 200 \Omega$.The impedance of the circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{400^2 + (500 - 200)^2} = \sqrt{400^2 + 300^2} = 500 \Omega$.The peak current in the circuit is $I_0 = \frac{V_0}{Z} = \frac{5}{500} = 0.01 	ext{ A}$.The peak voltage across the capacitor is $(V_C)_0 = I_0 X_C = 0.01 	imes 200 = 2 	ext{ V}$.The peak electrostatic energy stored in the capacitor is $(U_C)_{\max} = \frac{1}{2} C (V_C)_0^2 = \frac{1}{2} 	imes 2.5 	imes 10^{-6} 	imes (2)^2 = 5 	imes 10^{-6} 	ext{ J} = 5 \mu 	ext{J}$.

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