Initially charged capacitor disharged through a resistor

AI Thread Summary
The discussion focuses on the discharge of a capacitor through a resistor, specifically calculating the time required for the capacitor to lose one-third and two-thirds of its initial charge. The time constant τ is used in the calculations, with the equation i = -(q0/RC)e^(-t/RC) being referenced. A user initially struggles with the correct formulation but ultimately derives the time for losing one-third of the charge as t = RC*ln(1/3), confirming the need to include a negative sign in the equation. The conversation concludes with the user expressing gratitude after resolving their confusion. The thread effectively clarifies the mathematical approach to capacitor discharge in relation to time constants.
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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