A series RLC circuit is driven by a 1.0-kHz oscillator.The circuit parameters are V_rms = 12 V,L = 5.0 mH,C = 4.0 $\mu$F,and R = 10 $\Omega$.Under steady-state conditions,the rms potential difference across the resistor will be A)5.13 V B)9.19 V C)12.3 V D)14.7 V E)18.8 V

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The rms potential difference across the resistor in a series RLC circuit can be found using Ohm's law, $V_R = I \times R$, where $I$ is the current through the resistor. The current in the circuit can be calculated using the formula $I = \frac{V_{S}}{Z}$, where $V_{S}$ is the source voltage and $Z$ is the impedance of the circuit. The impedance of a series RLC circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$, where $X_L = 2\pi fL$ is the inductive reactance and $X_C = \frac{1}{2\pi fC}$ is the capacitive reactance. Given $f = 1.0 \, kHz$, $L = 5.0 \, mH$, $C = 4.0 \, \mu F$, and $R = 10 \, \Omega$, we can calculate $X_L$ and $X_C$, and subsequently $Z$, to find the current $I$. Finally, using $I$ and $R$, we can find $V_R$.Given values:- $f = 1.0 \, kHz = 1000 \, Hz$- $L = 5.0 \, mH = 5.0 \times 10^{-3} \, H$- $C = 4.0 \, \mu F = 4.0 \times 10^{-6} \, F$- $R = 10 \, \Omega$- $V_{S} = 12 \, V$Calculating $X_L$ and $X_C$:- $X_L = 2\pi fL = 2\pi (1000)(5.0 \times 10^{-3}) = 31.4 \, \Omega$- $X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi (1000)(4.0 \times 10^{-6})} = 39.8 \, \Omega$Calculating $Z$:- $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{10^2 + (31.4 - 39.8)^2} = \sqrt{100 + (-8.4)^2} = \sqrt{100 + 70.56} = \sqrt{170.56} = 13.06 \, \Omega$Calculating $I$:- $I = \frac{V_{S}}{Z} = \frac{12}{13.06} = 0.918 \, A$Calculating $V_R$:- $V_R = I \times R = 0.918 \times 10 = 9.18 \, V$Therefore, the rms potential difference across the resistor is approximately 9.19 V, making option B the correct answer.

A Series RLC Circuit Is Driven by a 1 μ\muμ F,and R = 10

A Series RLC Circuit Is Driven by a 1 $\mu$ F,and R = 10