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I need to calculate systematic error for $\tau$ in capacitor's charging formula( $V_c(t)=V_s(1-e^{-t\over\tau})$ )

I converted it to : $\tau=-{t \over ln(1-{V_c \over V_s})}$
and continued by doing: $ln(\tau)=ln(-t)-ln(ln(1-{V_c \over V_s}))$
then tried to derivative: ${d\tau \over \tau}={dt \over t}- ...$
I can't go ahead any more!

What about discharge equaltion ( $V_c(t)=V_0 e^{-t\over\tau}$ )?
How should i continue and get result for $d\tau \over \tau$?

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1 Answer 1

up vote 2 down vote accepted

You should work out this

$$d\tau=-\frac{dt}{\ln\left(1-\frac{V_c}{V_s}\right)}$$ $$+\frac{t}{\ln^2\left(1-\frac{V_c}{V_s}\right)}\left(-\frac{dV_c}{V_s}+V_c\frac{dV_s}{V_s^2}\right)$$

The next step is to divide by $\tau$ and you will get


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Hi, Thanks , But how you got $+\frac{t}{\ln^2\left(1-\frac{V_c}{V_s}\right)}\left(-\frac{dV_c}{V_s}+V_c\frac{‌​dV_s}{V_s^2}\right)$ from $d\tau=-\frac{dt}{\ln\left(1-\frac{V_c}{V_s}\right)}$ ? and can you give a help about discharge equation that I added? –  Snigger Nov 25 '11 at 14:01
This is quite simple. I have derived $1/\ln$ and then I have exploited the differential of its argument through partial derivatives of $V_s$ and $V_c$. –  Jon Nov 25 '11 at 14:07
Thanks can you help about discharge equation too? –  Snigger Nov 25 '11 at 15:10
This is fairly well explained at en.wikipedia.org/wiki/Capacitor. –  Jon Nov 25 '11 at 20:40

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