Obtain the voltage of the generator and the phase angle

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SUMMARY

The discussion focuses on calculating the voltage of an AC generator and the phase angle in a circuit containing a 231-Ω resistor and a 0.250-µF capacitor, with a frequency of 4.80 kHz. The voltage was correctly determined to be 10.7V using the formula V = IZ, where Z is the impedance calculated as 266.36Ω. The phase angle was calculated using the formula arccos(231/266.36), resulting in approximately 29.86 degrees. The user expressed uncertainty regarding the sign of the phase angle, indicating a potential area for further clarification.

PREREQUISITES
  • Understanding of AC circuit analysis
  • Familiarity with impedance and reactance calculations
  • Knowledge of trigonometric functions in electrical engineering
  • Basic concepts of capacitors and resistors in series circuits
NEXT STEPS
  • Study the concept of impedance in AC circuits
  • Learn about the relationship between phase angle and impedance
  • Explore the effects of frequency on capacitive reactance
  • Review the use of arccos and its implications in phase angle calculations
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in AC circuit analysis will benefit from this discussion, particularly those seeking to understand voltage and phase relationships in resistor-capacitor circuits.

cclement524
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Homework Statement




An ac generator has a frequency of 4.80 kHz and produces a current of 0.0400 A in a series circuit that contains only a 231-Ω resistor and a 0.250-µF capacitor. Obtain the voltage of the generator and the phase angle between the current and the voltage across the resistor/capacitor combination.

I have the voltage correct, I'm just getting an incorrect answer for the angle.


Homework Equations



Capaacitance of the capacitor is C = 0.250 *10^-6 F

resistor is R = 230Ω

phase angle between capacitor and resistor is tan^-1 XC / R

Capcitive reactance is X_C = 1/2 π f C


The Attempt at a Solution



?=2pf = 4.80kHz x 2p = 30.16 x 103rad/s

XC = 1/?C =1/(30.16*103*0.250*10-6) = 132.62 O
Z = 231 - 132.62j

Z = v(2312 + 132.622) =266.36O

V = IZ = 266.36O * 0.0400A = 10.7V (1)

arccos (231/266.36)= 0.5211 (in radian)

0.5211 * 180/p = 29.86 (degrees)
 
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Numerically your angle looks okay. It's just its sign that I have my suspicions about...
 
Thank you!
 

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