Peak Voltage across a capacitor.

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SUMMARY

The discussion focuses on calculating the peak voltage across a 4.0 μF capacitor in an AC circuit. The user successfully determined the peak current supplied by the emf to be 25 mA. To find the peak voltage across the capacitor, the relevant equation is V = I * XC, where XC is the capacitive reactance calculated using the formula XC = 1/(2πfC). The user is encouraged to explore the concept of a capacitive voltage divider for further understanding.

PREREQUISITES
  • Understanding of AC circuit analysis
  • Familiarity with capacitive reactance (XC)
  • Knowledge of the relationship between current (I), voltage (V), and capacitance (C)
  • Basic grasp of frequency (f) in electrical circuits
NEXT STEPS
  • Learn how to calculate capacitive reactance (XC) for different frequencies
  • Study the principles of capacitive voltage dividers in AC circuits
  • Explore the implications of peak current and voltage in capacitor circuits
  • Review the derivation and application of the formula I = V/XC
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone looking to deepen their understanding of AC circuits and capacitor behavior.

ReidMerrill
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Homework Statement


https://session.masteringphysics.com/problemAsset/1385140/3/36.P42.jpg
36.P42.jpg

Part A
What is the peak current supplied by the emf in the figure?
Part B
What is the peak voltage across the 4.0 μF capacitor?

Homework Equations


I = V/XC = E0/1/2πfC = 2πfCE0

The Attempt at a Solution


I got 25mA for part A, which is correct. I don't know how to do part B and my textbook is less than useful.

Any help would be appreciated. Thanks!
 
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