Re: Calculating resistance

[Bernard Cleyet <anngeorg@PACBELL.NET>]

I've been discussing w/ LK for almost a week about the theoretically and xptly
dtmnd resistance between conductors imbedded in a poorly conducting plane.  I
still have a problem with correctness of the assumption that a conductive slice
is the same as either a guarded bi (di) semi infinite cylinder capacitor or
co-planar disks.  However, evidently, the two methods agree, vide below.

Harnwell (Principles of E & EM  1949) argues [as does Eisberg & Lerner PHYSICS
(1981) problem 22-33] that the relationship between the capacitance of a pair of
conductors and the resistance between them when imbedded in an infinite less
conductive medium. is:  R = kappa sub zero / (conductivity * the capacitance).
[eq. one]

L.  measured the thickness of an aquadag plane and it's conductivity (or claimed
by the mfgr.?) and found the resistance of silver-dag dots.  The formula given by
Harnwell (and DB, by another method) for the capacitance of a pair of parallel
infinite cylinders separation c and radii a is:

C / l = Pi * kappa sub zero * cosh (c / 2a)   (cosh reduces to ln (c / a) when a
<< c.  (n.b. this is the capacitance per unit ength)

substituting above in  eq. one.  R = kappa * ln (c / 2a) / [sigma * l * Pi *
kappa]

L's values:  l = 0.013 mm;  rho (= 1/ sigma) = 0.32 ohm m   (or 3.125 mho / m); a
= 0.5 cm; c = 20 cm

R = ln (20 / 0.5) / [ 3.125 * 0.013E-3 * Pi ] = ~ 29 k ohm, which, I assume,
agrees with L's measurement with the Pasco apparatus.

bc

P.s.  The argument:  the current that flows from an embedded electrode is =
surface integral of the current density "dot" area ds = surface enclosing the
electrode (except the ignorable current supply wire) integral E dot ds.  which
(the integral) is, recognizably, the charge enclosed (Gauss' law).  Considering
the two integrals as identical, then i  = sigma * q / kappa,

Taking the electrodes as a capacitor with capacitance C = q / V ,    then  i =
[sigma / kappa] * C * V ,

as R = V = i  ...  "QED"

P.p.s As DB insisted, Harnwell does indeed use conformal mapping to solve
Laplace's eg.  Rereading Churchill's Intro. to Complex Variables, 1948 (first
time > 40 years ago!) convinced me.


Ludwik Kowalski wrote:

> I think this is worth sharing with the list. Give the reference
> but do not assume people have access to this book. I suggest
> you write the formula and describe how you used it to get
> the R of the infinite size sheet of a conductive film of known
> thickness. Show numerical calculation for Pasco sheets.
> Outlining the derivation is likely to be appreciated by many,
> provided it is explained in simple terms.
> Ludwik.
>
> Bernard Cleyet wrote:
>
> > Damn! I threw out that scrap of paper and must begin again.
> >
> > I claimed ~ 0.1 ohm  you got 0.06 .  My calc. remembers entries, so I pushed
> > the key a jillion times and found more exactly your number 0.0587.....
> >
> > with conductivity 1/ 0.32 instead of 1 E+5 mho/m   and thickness 0.013
> > instead of 0.2 mm    (your revised thickness is because you scraped off the
> > aquadag and found the diff. between both and the paper only?)
> >
> > 28.9 k ohm
> >
> > bc
> >
> > P.s. It's Ln (separation / radius)
> >
> > Ludwik Kowalski wrote:
> >
> > > You are probably referring to what is on page 42. Note that
> > > in my case a=b (figure 1.29). If I use your formula then
> > > R=0 because ln(b/a)=ln(1)=0. You probably made a typing
> > > error, or something of that kind. My sigma was 1/0.32
> > > and the sheet thickness was 0.013 mm. (What R do you
> > > get after correcting the mistake?)
> > > Ludwik
> > >
> > > Bernard Cleyet wrote:
> > >
> > > > Correct.  However, the method is applicable to 2-D.
> > > >
> > > > Find the capacitance per unit length of two cylinders (parallel of
> > > > separation l)  replace kappa sub zero with the conductivity / unit area
> > > > and invert.
> > > >
> > > > e.g.  resistance between two coaxial cylinders:   C = (2 Pi kappa * l) /
> > > > [ ln (b/a)];   R = ln (b/a) / (2 Pi * sigma * l)
> > > > where sigma is the volume conductivity
> > > >
> > > > I have all ready to mail.  I think you'll appreciate it any way.
> > > >
> > > > bc
> > > >
> > > > P.s. there's another method:   a square mesh of resistors with the
> > > > number of resistors increased => in the limit to area conductivity.
> > > >
> > > > Ludwik Kowalski wrote:
> > > >
> > > > > Hi again, Bernard:
> > > > > It occurred to me that perhaps the R formula on page 102 refers
> > > > > to a 3-dimentional system (current flowing between two spheres
> > > > > and not between two flat electrodes in a conducting sheet). This
> > > > > would explain why R is nearly 200 smaller than what was
> > > > > calculated with Bowman's formula. Verify this before sending
> > > > > me anything. I am interested in 2-dim only, as you know.
> > > > > Ludwik