A series LCR circuit containing a resistance | vedclass

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Question

(A) At resonance, the current $I_{rms}$ in the circuit is given by $I_{rms} = \frac{V_R}{R} = \frac{60 \, V}{120 \, \Omega} = 0.5 \, A.$The voltage ac

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$20 \, mH, 25/8 \, \mu F$

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(A) At resonance, the current $I_{rms}$ in the circuit is given by $I_{rms} = \frac{V_R}{R} = \frac{60 \, V}{120 \, \Omega} = 0.5 \, A.$The voltage across the inductor is $V_L = I_{rms} 	imes X_L = I_{rms} 	imes (\omega L).$Substituting the values: $40 = 0.5 	imes (4 	imes 10^3) 	imes L.$$40 = 2000 	imes L \implies L = \frac{40}{2000} = 0.02 \, H = 20 \, mH.$At resonance, the voltage across the capacitor $V_C$ is equal to the voltage across the inductor $V_L$, so $V_C = 40 \, V.$$V_C = I_{rms} 	imes X_C = I_{rms} 	imes \left(\frac{1}{\omega C}ight).$$40 = 0.5 	imes \left(\frac{1}{4 	imes 10^3 	imes C}ight).$$40 = \frac{0.5}{4000 	imes C} \implies C = \frac{0.5}{4000 	imes 40} = \frac{0.5}{160000} = \frac{5}{1600000} = \frac{1}{320000} \, F.$$C = \frac{1}{320000} 	imes 10^6 \, \mu F = \frac{100}{32} \, \mu F = 3.125 \, \mu F = \frac{25}{8} \, \mu F.$

$A$ series $LCR$ circuit containing a resistance of $120 \, \Omega$ has an angular resonance frequency of $4 \times 10^3 \, rad \, s^{-1}.$ At resonance, the voltages across the resistance and the inductance are $60 \, V$ and $40 \, V$ respectively. The values of $L$ and $C$ are respectively:

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