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 capacitor 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 capacitor is:
A)2.5 V
B)3.4 V
C)6.7 V
D)10.0 V
E)10.8 V

Quizplus · Accepted Answer

Using the impedance formula $Z = \sqrt{R^2+(\omega L - \frac{1}{\omega C})^2}$, we can find the impedance of the series circuit:

$$Z = \sqrt{(50\ \Omega)^2 + (200\ 	ext{mH} 	imes 200\ 	ext{rad/s} - \frac{1}{200\ 	ext{rad/s} 	imes 200\ 	ext{μF}})^2} \approx 22.2\ \Omega$$

Using Ohm's Law, we can find the current in the circuit:

$$I = \frac{V}{Z} = \frac{10\ 	ext{V}}{22.2\ \Omega} \approx 0.450\ 	ext{A}$$

The voltage across the capacitor can be found using the equation $V_C = IXC$, where $XC$ is the capacitive reactance:

$$V_C = IX_C = 0.450\ 	ext{A} 	imes \frac{1}{200\ 	ext{μF}} = 2.25\ 	ext{V}$$

Since this is the voltage across the capacitor at a single instant, and the problem asks for the rms voltage, we need to divide this value by $\sqrt{2}$:

$$V_{C, 	ext{rms}} = \frac{2.25\ 	ext{V}}{\sqrt{2}} \approx 1.59\ 	ext{V}$$

Therefore, the closest answer is B) 3.4 V.

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