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I conducted an experiment, where a capacitor is discharged over a (big) resistor for $t$ seconds and then the remaining charge is measured with a ballistic galvanometer. The data I have is:

t       phi
1.00    65
1.75    60
1.75    61
2.28    56
3.13    52
3.94    48
5.16    44
5.85    40
13.35   20

When I plot this data (gnuplot here) in a semilog plot, I get this:

As far as I can tell, those points lie on a straight line, which would then be an exponential function. And that is what I would expect as well, since this should be

$$Q(t) = Q_0 \exp\left(\frac{-t}{RC}\right)$$

However, when I fit it using this gnuplot snippet, I get fit parameters that make no sense to me.

f1(x) = a * exp(-x/g);
fit f1(x) "test.dat" using 1:2:3 via a,g


Final set of parameters            Asymptotic Standard Error
=======================            ==========================

a               = 49.5555          +/- 7.647        (15.43%)
g               = -3.43204e+07     +/- 3.281e+13    (9.561e+07%)

I tried this same thing in Octave and Grace as well, they all give me this constant function.

How can I get a meaningful fit for this, other than trying to fit it by hand?

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closed as off topic by Qmechanic, Sklivvz, Manishearth Dec 29 '12 at 15:58

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You could try a linear regression of ln(Q) versus t. – John Rennie Oct 18 '12 at 15:17
That worked. I am not sure why it works now … – Martin Ueding Oct 18 '12 at 15:24
I've never used gnuplot, but nonlinear functions are generally harder to fit than linear functions and unless you choose the starting point carefully the fit may not converge. – John Rennie Oct 18 '12 at 15:32
I'm not sure this is really on topic for us, but perhaps it could be migrated to Cross Validated or Computational Science. Thoughts? – David Z Oct 18 '12 at 17:19
up vote 0 down vote accepted

This isn't really an answer, but I can't put images in a comment.

I tried fitting your expression using the Excel Solver and got $Q_0$ = 71.1 and $1/RC$ = 0.0966. The resulting fit looked like:

Solver fit

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I got $RC = 10$ with the linear fit of the logarithmic data, so that looks pretty good. I wonder how Excel can fit it, and the three tool I used failed … – Martin Ueding Oct 18 '12 at 18:58

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