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 inductor 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

The inductive reactance is XL=2πfL=2π(60Hz)(1.20mH)=0.452ΩThe capacitive reactance is XC=1/(2πfC)=1/(2π(60Hz)(1.80μF))=142ΩThe impedance of the circuit is Z=√(R2+(XL−XC)2)=√(2502+(0.452−142)2)=250ΩThe current in the circuit is I=V/Z=120V/250Ω=0.480AThe voltage across the inductor is VL=IXL=0.480A×0.452Ω=0.216VThe voltmeter reads approximately 0.216 V, which is closest to choice A, 25.7 mV.

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