AC Circuits II: AC Power Generator

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SUMMARY

The discussion focuses on solving an AC circuit problem involving an AC generator with an rms voltage of 110 V at 60.0 Hz, connected to a 0.300 H inductor, a 5.80 μF capacitor, and a 236 Ω resistor. Key calculations include determining the impedance (Z), rms current, average power, peak current, and peak voltages across the inductor and capacitor. The participant specifically seeks assistance with calculating the peak voltages and understanding the relationship between peak and rms voltage.

PREREQUISITES
  • Understanding of AC circuit components: inductors, capacitors, and resistors
  • Familiarity with impedance calculations in AC circuits
  • Knowledge of rms and peak voltage relationships
  • Ability to apply relevant equations: Z=[R^2+(XL-Xc)^2]^(1/2), XL=2πfL, Xc=1/(2πfc)
NEXT STEPS
  • Calculate the average power dissipated in the circuit using the formula P=I_rms^2 * R
  • Learn about resonance in RLC circuits and how to calculate the resonance frequency
  • Explore the relationship between peak voltage and rms voltage in AC circuits
  • Investigate the effects of changing frequency on circuit impedance and current
USEFUL FOR

Students studying electrical engineering, educators teaching AC circuit theory, and anyone involved in practical applications of AC power systems.

OsDaJu
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Homework Statement



A An AC generator supplies an rms voltage of 110 V at 60.0 Hz. It is connected in series with a 0.300 H inductor, a 5.80 μF capacitor and a 236 Ω resistor.
What is the impedance of the circuit?

B What is the rms current through the resistor?

C What is the average power dissipated in the circuit?

D What is the peak current through the resistor?

E What is the peak voltage across the inductor?

F What is the peak voltage across the capacitor?

G The generator frequency is now changed so that the circuit is in resonance. What is that new (resonance) frequency?

Homework Equations



Z=[R^2+(XL-Xc)^2]^(1/2)
XL=2pifL
Xc=1/(2pifc)

Part E: VL=[(2^.5)*Vrms]/Z
Part F: VL=[(2^.5)*Vrms]/Xc

The Attempt at a Solution



I need help for part E and F

I substituted the numbers in the equation and I got the wrong answer. Can someone point me in the right direction?

Xc=457.34
Vrms=110
Z=4.17×10^2
 
Last edited:
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Vrms across C = Irms* Xc
Now what is the relation between peak voltage and rms voltage?
 
Thank you!
 

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