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Re: Internal resistance



Now I calculated how you did it. You found the R(int) as a function of
the EMF out. I think I'll try it w/ my expensive (was) donated P/S. But
only at a few "points".

bc

p.s. Just as I feared (within 4 1/2 digits), the P/S has Zero R

Current (A.) EMF (V.)

zero 6.998 (couldn't decide 8 or 9)
.201 6.999
.700 6.998 as above

zero 19.991
0.595 19.991


zero 29.95
0.472 29.95
0.861 29.95 (beyond claimed range, and CC mode
not quite initiated)

not a suitable P/S for lab instruction!

p.p.s This reminds me of a magnet power supply (Alpha) that the E-Shop
thought was e-junk. I measured its internal R (perhaps w/ a 3 1/2) and
found it has ~ Zero R
then I realized the shop didn't understand it's purpose. It is constant
current, so w/ no load the EMF went skyward. (Lethally so!)



Ludwik Kowalski wrote:

On Thursday, April 1, 2004, at 04:56 PM, John Denker wrote:



Ludwik Kowalski wrote:


... r ... was found to decrease
monotonically from 1.4 ohms, at I=0.85 A,
to 1.0 ohms at 3.8A. .... (V1-V2)/I


The question is ill-posed.

In the key formula, V1 is not measured "at"
3.8A so the quotient (V1-V2)/I cannot be called
the resistance "at" (V1-V2)/I.



The value of r was expected to be the same for
all currents, in the range for which the power
supply was designed. But it turned out to be
different for different currents. Nowhere in our
textbooks was this possibility mentioned.
Many numerical problems would make no
sense if r were current-dependent. What is
wrong with testing this experimentally? What
is wrong with asking how r depends on I?
How to explain the observed dependance?
Ludwik Kowalski