1:14 AM loncapa1.fsu.edu AC Power Generator .\'ll AT&T;? An AC generator supplie

Question

1:14 AM loncapa1.fsu.edu AC Power Generator .'ll AT&T;? An AC generator supplies an rms voltage of 230 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 5.20 ?F capacitor and a 266 ? resistor. Part A What is the impedance of the circuit? Submit Answr Tries 0/1s Part B What is the rms current through the resistor? Submit AnswerTries 0/15 Part C What is the average power dissipated in the circuit? Submit AnswrTries 0/15 Part D What is the peak current through the resistor? Submit AnswerTries 0/15 Part E What is the peak voltage across the inductor? Submit Answer Tries o/15 Part F What is the peak voltage across the capacitor? Submit Arswer Tries 0/15 Part G The generator frequency is now changed so that the circuit is in resonance. What is that new Submit AnswrTries 0/15 1998-2018 by Florida State University. All rights

Explanation / Answer

Given,

Vrms = 230 V ; f = 50 Hz ; L = 0.650 H ; C = 5.2 uF ; R = 266 Ohm

a)We know that, impedence is:

Z = sqrt (R^2 + (Xl - Xc)^2)

Xl = 2 pi f L

Xl = 2 x 3.14 x 50 x 0.65 = 204.1 Ohm

Xc = 1/(2 pi f C)

Xc = 1/(2 x 3.14 x 50 x 5.2 x 10^-6) = 612.44 ohm

Z = sqrt [266^2 + (204.1 - 612.44)^2] = 487.34 Ohm

Hence, Z = 487.34 Ohm

B)Irms = Vrms/z

Irms = 230/487.34 = 0.472 A

Hence, Irms = 0.472 A

C)Pav = VI cos(phi)

Pav = 230 x 0.472 x 0.55 = 59.71 W

Hence, P = 59.71 w

D)Ip = Irms x 1.414

Ip = 0.472 x 1.414 = 0.667 A

Hence, Ip = 0.667 A

E)VL = I Wl

VL = 0.667 x 204.1 = 136.13 V

VL = 136.13 V

F)VC = I Xc

Vc = 0.667 x 612.44 = 408.5

Vc = 408.5 V

G)at resonance

XL = XC

f = 1/2 pi sqrt (LC)

f = 1/(4 s 3.14^2 sqrt (0.65 x 5.2 x 10^-6)) = 86.5

f = 86.5 Hz

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