Calculating Peak Current in a Microfarad Capacitor Circuit

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SUMMARY

The discussion centers on calculating the peak current in a capacitor circuit, specifically a 0.30 microfarad capacitor connected to an AC generator with a peak voltage of 10.0 V and a frequency of 100 Hz. The correct formula for peak current (Ic) is Ic = Vc / Xc, where Xc is the capacitive reactance calculated as Xc = 1 / (2 * π * f * C). The user initially miscalculated the peak current as 53,051 A, while the correct answer is 1.8 * 10^-3 A, highlighting the importance of correctly applying the formula for capacitive reactance.

PREREQUISITES
  • Understanding of AC circuits
  • Familiarity with capacitive reactance (Xc)
  • Knowledge of the relationship between voltage, current, and reactance
  • Basic proficiency in using mathematical equations involving π
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  • Study the concept of capacitive reactance in detail
  • Learn how to calculate peak current in various AC circuits
  • Explore the implications of frequency on capacitive circuits
  • Investigate the differences between AC and DC circuit calculations
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Electrical engineering students, hobbyists working on AC circuits, and anyone looking to deepen their understanding of capacitor behavior in alternating current applications.

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Homework Statement


A .30 microfarad capacitor is connected across an AC generator that produces a peak voltage of 10.0 V. What is the peak current through the capacitor if the emf frequency is 100 Hz?


Homework Equations


Ic (peak current)= Vc/Xc, Xc= 1/(2*pi*f*C)


The Attempt at a Solution



I solved the equation as: 10V/(2*pi*100 Hz*.30*10^-6 F)= 53051 A, however the book answer is completely different (1.8*10^-3 A), so any hints toward a better equation to use (or better method) would be great. Thank you!
 
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Look at XC again!

VC/XC = VC/(1/(2·π·f·C))

What you computed was VC·XC
 
Ohh, that makes a lot more sense! Thank you! :)
 

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