You connect a 250-$\Omega$ resistor,a 1.20-mH inductor,and a 1.80-$\mu$F capacitor in series across a 60.0-Hz,120-V (peak)source.When you connect a voltmeter across the resistor in this circuit,the meter reads approximately
A)25.7 mV
B)83.6 V
C)120 V
D)14.2 V
E)84.9 V

Question

Quizplus · Accepted Answer

We can first find the total impedance of the circuit using the formula:

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

where $R$ is the resistance, $X_L$ is the inductive reactance, and $X_C$ is the capacitive reactance.

$R = 250 \,\Omega$

$X_L = 2\pi f L = 2\pi (60.0 \,	ext{Hz})(1.20 \,	ext{mH}) = 452 \,\Omega$

$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi (60.0 \,	ext{Hz})(1.80 \,	ext{mF})} = 147 \,\Omega$

$Z = \sqrt{(250 \,\Omega)^2 + (452 \,\Omega - 147 \,\Omega)^2} = 506 \,\Omega$

Using Ohm's law, we can then find the current through the circuit:

$I = \frac{V_{	ext{peak}}}{Z} = \frac{120 \,	ext{V}}{506 \,\Omega} = 0.237 \,	ext{A (peak)}$

Finally, we can find the voltage drop across the resistor using Ohm's law:

$V_R = IR = (0.237 \,	ext{A}) (250 \,\Omega) = 59.3 \,	ext{V (peak)}$

Since the voltmeter measures the RMS voltage, which is equal to the peak voltage divided by the square root of two:

$V_R = \frac{59.3 \,	ext{V (peak)}}{\sqrt{2}} \approx \boxed{	extbf{(D) }14.2 \,	ext{V (RMS)}}$

You Connect a 250- Ω\OmegaΩ Resistor,a 1 μ\muμ F Capacitor in Series Across a 60

You Connect a 250- $\Omega$ Resistor,a 1 $\mu$ F Capacitor in Series Across a 60