How can the log in the denominator of the charging equation be simplified?

AI Thread Summary
The discussion focuses on simplifying the logarithm in the denominator of the capacitance formula derived from the charging equation of an RC circuit. The equation presented is C = -t / (R ln((vs - vt) / vs)). Participants confirm that the manipulation is correct and suggest using the property ln(a/b) = ln(a) - ln(b) to further simplify the logarithm. This approach could lead to a more straightforward expression for the capacitance. The conversation emphasizes basic algebraic manipulation in the context of electrical engineering equations.
Dr. Ficus
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Homework Statement



Basically, I'm just trying to verify that I can do basic algebraic manipulation correctly. The objective: solve the charging equation of a simple RC circuit for (C), capacitance.

Homework Equations


v\left(t\right)={v}_{s}\left(1-{e}^{{-t}/{RC}}\right)

The Attempt at a Solution


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I ended up with the following.
C=\frac{-t}{R\mathrm{ln}\left(\frac{{v}_{s}-{v}_{\left(t\right)}}{{v}_{s}}\right)}
Is this correct? Also, I have a suspicion that the log in the denominator can somehow be simplified more than just putting \mathrm{ln}\left(1-\frac{v\left(t\right)}{{v}_{s}}\right) but I can't think of how. Thanks in advance!
 
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That is correct. You can, if you so prefer, use the principle
\ln(\frac{a}{b})=\ln a - \ln b
for the log in the denominator.
 
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