(8c31p61) In the figure, R = 26.0 , C = 7.60 F, and L = 50.0 mH. The generator p

ID: 1647958 • Letter: #

Question

(8c31p61) In the figure, R = 26.0 , C = 7.60 F, and L = 50.0 mH. The generator provides a sinusoidal voltage of 50 V (rms) and frequency f = 690 Hz .

A) Calculate the rms current.

B) Find the rms voltage, Vab.

C) Find the rms voltage, Vbc.

D) Find the rms voltage, Vcd.

E) Find the rms voltage, Vbd.

F) Find the rms voltage, Vad.

G) At what average rate is energy dissipated by the capacitor?

H) At what average rate is energy dissipated by the inductor?

I) At what average rate is energy dissipated by the resistor?

capacitive impedance=Xc=1/(2*pi*f*C)=1/(2*pi*690*7.6*10^(-6))=30.35 ohms

inductive reactance=Xl=2*pi*f*L=2*pi*690*50*0.001=216.77 ohms

total impedance=Z=R+j(Xl-Xc)=26+j186.42 ohms

current=50/(26+j186.42)=0.036694 - j 0.263094 A

part A:

rms current=rms voltage/impedance magnitude=50/sqrt(26^2+186.42^2)=0.26564 A

part B:

rms voltage Vab=magnitude of voltage across R=26*0.26564=6.9067 volts

part C:

rms voltage Vbc=magnitude of voltage across C=30.35*0.26564=8.0622 volts

part d:

rms voltage Vcd=magnitude of voltage across L=216.77*0.26564=57.583 volts

part e:

rms voltage Vbd=magnitude of voltage across C and L=magnitude of (0.036694 - j 0.263094)*j(Xl-Xc)

=49.521 volts

part f:

rms voltage Vad=magnitude of voltage across R,C and L=magnitude of (0.036694 - j 0.263094)*(26+j186.42)=50 volts

part G:

power dissipated by caapcitor=0

part H:

power dissipated by inductor=0

part I:

average rate of energy dissipated by resistor=rms current^2*resistance

=0.26564^2*26

=1.8347 W

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