Capacitor Voltage Equations: Explained

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Homework Help Overview

The discussion revolves around two equations related to the voltage in an RC circuit, specifically focusing on the differences between the equations for charging and discharging capacitors.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the distinction between the two equations by evaluating them at specific time points, questioning the nature of the mathematical expressions involved.

Discussion Status

Some participants have provided insights into the nature of the equations, suggesting that one represents charging and the other discharging behavior of the capacitor. There is an ongoing clarification regarding the mathematical terms used in the equations.

Contextual Notes

There is a mention of evaluating the equations at t = 0 and t → ∞ to understand their behavior, indicating a potential gap in the original poster's understanding of the concepts.

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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