You connect a 100-$\Omega$ resistor,a 800-mH inductor,and a 10.0-$\mu$F capacitor in series across a 60.0-Hz,120-V (peak)source.The impedance of your circuit is approximately
A)377 $\Omega$
B)302 $\Omega$
C)265 $\Omega$
D)100 $\Omega$
E)107 $\Omega$

Question

Quizplus · Accepted Answer

The impedance $Z$ of a series RLC circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$, where $R$ is the resistance, $X_L = 2\pi fL$ is the inductive reactance, and $X_C = \frac{1}{2\pi fC}$ is the capacitive reactance. Given $R = 100\ \Omega$, $L = 800\ \text{mH} = 0.8\ \text{H}$, $C = 10.0\ \mu\text{F} = 10^{-5}\ \text{F}$, and $f = 60.0\ \text{Hz}$, we calculate $X_L$ and $X_C$ and then $Z$. The calculation yields an impedance close to $107\ \Omega$, making option E the correct choice.

You Connect a 100- Ω\OmegaΩ Resistor,a 800-MH Inductor,and a 10 μ\muμ F Capacitor in Series Across a 60

You Connect a 100- $\Omega$ Resistor,a 800-MH Inductor,and a 10 $\mu$ F Capacitor in Series Across a 60