An ac generator producing 10 V (rms)at 200 rad/s is connected in series with a 50- $\Omega$ resistor, a 400-mH inductor, and a 200- $\mu$ F capacitor.The rms voltage across the inductor is:
A)2.5 V
B)3.4 V
C)6.7 V
D)10.0 V
E)10.8 V

Question

An ac generator producing 10 V (rms)at 200 rad/s is connected in series with a 50- $\Omega$  resistor, a 400-mH inductor, and a 200- $\mu$ F capacitor.The rms voltage across the inductor is:
A)2.5 V
B)3.4 V
C)6.7 V
D)10.0 V
E)10.8 V

Quizplus · Accepted Answer

The rms voltage across the inductor can be found using the formula for the inductive reactance $X_L = \omega L$, where $\omega = 200 \, \text{rad/s}$ and $L = 0.4 \, \text{H}$. The inductive reactance $X_L = 200 \times 0.4 = 80 \, \Omega$. The total impedance $Z$ of the circuit can be calculated using the formula $Z = \sqrt{R^2 + (X_L - X_C)^2}$, where $X_C = \frac{1}{\omega C} = \frac{1}{200 \times 200 \times 10^{-6}} = 25 \, \Omega$. Therefore, $Z = \sqrt{50^2 + (80 - 25)^2} = \sqrt{50^2 + 55^2} = \sqrt{2500 + 3025} = \sqrt{5525} \approx 74.3 \, \Omega$. The current $I = \frac{V}{Z} = \frac{10}{74.3} \approx 0.134 \, \text{A}$. The voltage across the inductor $V_L = I \times X_L = 0.134 \times 80 \approx 10.8 \, \text{V}$.

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