Capacitor Peak Current and Voltage in an AC Circuit

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SUMMARY

The discussion focuses on calculating the capacitance of a capacitor in an AC circuit with a peak current of 240 μA and a peak voltage of 2.7 V at a frequency of 250 kHz. The correct capacitance is determined by converting the frequency from kHz to radians per second before applying the formula C = 1/(ωXc). The user initially calculated the capacitance incorrectly due to a failure to convert the frequency, but upon correction, the capacitance was found to be 3.56 x 10-10 F.

PREREQUISITES
  • Understanding of AC circuit principles
  • Familiarity with capacitive reactance (Xc)
  • Knowledge of angular frequency (ω) conversion from kHz to radians/sec
  • Proficiency in using Ohm's Law in AC circuits
NEXT STEPS
  • Learn about angular frequency conversion in AC circuits
  • Study capacitive reactance calculations in detail
  • Explore the relationship between frequency and capacitance
  • Investigate the effects of frequency changes on peak current in capacitors
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in AC circuit analysis and capacitor applications.

ReidMerrill
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Homework Statement


A capacitor has a peak current of 240 μA when the peak voltage at 250 kHz is 2.7 V .
Part A
What is the capacitance?

Part B
If the peak voltage is held constant, what is the peak current at 500 kHz

Homework Equations


Xc=1/wC
Vc=IcXc

The Attempt at a Solution


I rearranged equation 2 to solve for Xc = 2.7V/0.000240A = 11250
I then rearranged equation to to solve for C =1/wXc = 3.56x10-4

This is not correct. Why is this not correct and what is the proper way to go about solving it?
Thank you.

Correction: I re-did the second calculation and got 3.56x10-10F by properly converting kHz to Hz but it's still wrong.
 
Last edited:
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Hi ReidMerrill.
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Did you convert kHz to radians/sec?
 
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NascentOxygen said:
Hi ReidMerrill.
aerobanner.gif


Did you convert kHz to radians/sec?
Oh you know what I did not. I wasn't aware I needed to. My professor did not explain this well at all.
I would just divide the frequency by 2pi, correct?EDIT: That worked. Thank you!
 
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