Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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 $d\tau \over \tau$?

share|cite|improve this question

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


share|cite|improve this answer
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 – Jon Nov 25 '11 at 20:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.