How Do You Calculate Time in a Charging Capacitor Equation?

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Homework Help Overview

The discussion revolves around solving the equation for the voltage across a charging capacitor, specifically V = V0*[1 - e^(-t/RC)], with the goal of determining the time required to reach a specific voltage.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the algebraic manipulation of the equation to isolate time, with one participant questioning the validity of a logarithmic step taken by the original poster. Others request a step-by-step breakdown to identify potential errors.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's approach and suggesting that a more intuitive method may be acceptable for the problem at hand. There is no explicit consensus, but guidance has been offered regarding the algebraic steps involved.

Contextual Notes

One participant mentions that the lecturer may expect students to estimate values through inspection rather than solving the equation directly, indicating a possible instructional approach to the problem.

MidlandSoul
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Homework Statement


Solving the equation: V = V0*[1 - e^(-t/RC)]

Homework Equations


Need to solve this equation for time to calculate the time at which a charging capacitor achieves a certain voltage across it's plates.

The Attempt at a Solution


I solve it as: t = -RC*ln(V/V0) but this doesn't seem to work. Help!
 
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it looks like you might have done a step like this: ln(1+x) = ln(x) But this is not true.

Edit: Oh, welcome to physicsforums by the way!
 
Show your work step by step & we'll show you where you went wrong.
 
BruceW said:
it looks like you might have done a step like this: ln(1+x) = ln(x)
Or rather, ln(1-x) = ln(x)
 
Thanks guys. I think our lecturer intends for us to sort of 'have a guess' by inspection and using time constant ratios than actually solve this, so I can leave it for another day, phew!
Thank you for the feedback nonetheless.
 

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