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

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SUMMARY

The discussion focuses on simplifying the logarithmic expression in the charging equation of a simple RC circuit. The original equation presented is \( C = \frac{-t}{R \ln\left(\frac{v_s - v(t)}{v_s}\right)} \). Participants confirm that the logarithm in the denominator can be simplified using the property \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \), leading to a more manageable form of \( \ln\left(1 - \frac{v(t)}{v_s}\right) \).

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  • Understanding of RC circuits and their charging equations
  • Familiarity with logarithmic properties in algebra
  • Basic knowledge of calculus and differential equations
  • Ability to manipulate algebraic expressions
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  • Study the derivation of the charging equation for RC circuits
  • Learn about logarithmic identities and their applications in algebra
  • Explore advanced circuit analysis techniques
  • Investigate the impact of capacitance on circuit behavior
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Students studying electrical engineering, educators teaching circuit theory, and anyone interested in mastering algebraic manipulation in the context of electrical circuits.

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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


[/B]
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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