Capacitor Voltage Equations: Explained

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SUMMARY

The discussion clarifies the two equations governing the voltage behavior in RC circuits: V(t) = Vo*e^(-t/RC) for discharging capacitors and Vc = Vs(1-e^-t/RC) for charging capacitors. The key distinction lies in their application; the first equation describes the voltage decay over time, while the second illustrates the voltage rise. Evaluating both equations at t = 0 and t → ∞ confirms their respective behaviors. The confusion regarding logarithmic functions is addressed, emphasizing that the equations involve exponential functions, not logarithmic ones.

PREREQUISITES
  • Understanding of RC circuit fundamentals
  • Familiarity with exponential functions and their properties
  • Basic knowledge of voltage and current relationships in circuits
  • Ability to evaluate limits in mathematical expressions
NEXT STEPS
  • Study the behavior of RC circuits under different configurations
  • Learn about the time constant (τ = RC) and its significance in circuit analysis
  • Explore the graphical representation of charging and discharging curves
  • Investigate the applications of RC circuits in filtering and timing circuits
USEFUL FOR

Electrical engineering students, hobbyists working with circuits, and anyone seeking to understand the dynamics of charging and discharging capacitors in RC circuits.

ace8888
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Homework Statement


Hi, I've run into two different equations for the voltage of a typical RC circuit, one resistor, and one capacitor.
Please explain the different between the two. One has a 1 - the natural log and the other one doesnt.

Homework Equations


1. V(t) = Vo*e^(-t/RC)

2. Vc = Vs(1-e^-t/RC)

The Attempt at a Solution

 
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One of the equations is for a charging capacitor and the other for a discharging capacitor. To figure out which is which, evaluate each expression at t = 0 and at t → ∞ .
 
kuruman said:
One of the equations is for a charging capacitor and the other for a discharging capacitor. To figure out which is which, evaluate each expression at t = 0 and at t → ∞ .
thanks!
 
ace8888 said:
One has a 1 - the natural log
It is not a log; it is a negative exponential.
Do you understand what these two curves look like?
 
haruspex said:
It is not a log; it is a negative exponential.
Do you understand what these two curves look like?

Typo, yes it is the natural exponent. I don't know why but i was thinking exponent and typed log. They are inverse of each other.
 

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