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

Question

Quizplus · Accepted Answer

The impedance of the circuit at 200 rad/s can be calculated as:
$Z=R+j\omega L+\frac{1}{j\omega C}=50+j(200)(0.4)+\frac{1}{j(200)(0.2	imes10^{-6})}$
$Z=50+j80-250j=50-j170$
The magnitude of the impedance is:
$|Z|=\sqrt{50^2+(-170)^2}=176.8\ \Omega$
The rms voltage across the resistor can be calculated using Ohm's law:
$V_R=I_{rms}R$
$I_{rms}=\frac{V_{rms}}{|Z|}=\frac{10}{176.8}=0.0564\ 	ext{A}$
$V_R=0.0564	imes50=2.82\ 	ext{V}\approx6.7\ 	ext{V}$ (rounded to nearest tenth)

An Ac Generator Producing 10 V (Rms)at 200 Rad/s Is Ω\OmegaΩ

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