Initially charged capacitor disharged through a resistor

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Homework Help Overview

The discussion revolves around a capacitor discharging through a resistor, specifically focusing on the time required for the capacitor to lose certain fractions of its initial charge. The subject area includes concepts from circuits and exponential decay related to capacitors.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the mathematical formulation of the discharge process, specifically using the equation for current and charge over time. There is an exploration of the time required for the capacitor to lose one-third and two-thirds of its charge.

Discussion Status

Participants are actively engaging with the mathematical expressions and checking each other's reasoning. Some have confirmed the correctness of the derived expressions while others are questioning the signs in the equations. There appears to be a productive exchange of ideas, with guidance being offered on the mathematical details.

Contextual Notes

There is an emphasis on the correct interpretation of the equations and the importance of signs in the calculations. The original poster expresses uncertainty about the next steps in their reasoning.

Mike88
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Homework Statement


A capacitor with initial charge q0 is discharged through a resistor.
(a) In terms of the time constant τ, how long is required for the capacitor to lose the first one-third of its charge? answer x tau
(b) How long is required for the capacitor to lose the first two-thirds of its charge?
answer x tau



Homework Equations


i = -(q0/RC)e(-t/RC)


The Attempt at a Solution


I tried solving for time t and got -RC*ln(q/q0)= t
I'm lost i don't know what to do
 
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Mike88 said:
I tried solving for time t and got -RC*ln(q/q0)= t
Looks good to me (except for the minus sign). So what's the time when q = q0/3?
 
Doc Al said:
Looks good to me (except for the minus sign). So what's the time when q = q0/3?

i then get t = RC*ln(1/3) where RC = tau

is this correct?
 
Last edited:
Mike88 said:
i then get t = RC*ln(1/3) where RC = tau

is this correct?
Yes. But put that minus sign back in. (I didn't see that you had q0 on top! :wink:)
 
Doc Al said:
Yes. But put that minus sign back in. (I didn't see that you had q0 on top! :wink:)

thanks for the help i figured out everything.
 

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