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Re: Internal resistance



We did allow currents to last for couple of seconds
each time (just enough to read constant I and V).
That is how the effect of temperature was made
negligible. I will do other suggested things when
I have time. And I hope that others will do these
things, before me, as Bernard did.
Ludwik Kowalski


On Thursday, April 1, 2004, at 07:29 PM, Brian Whatcott wrote:

At 04:28 PM 4/1/2004, you 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


An interesting aspect of these observations,
is the aspect that you certainly mention to students, but
perhaps didn't recognize here: the implied zero crossing
of the voltage drop versus load current graph.
It is not a given that the supply has zero [internal] current
with no output load connected.
To account for this possibility, you could work with two
load resistors. To establish non-linearity, you would use
three load resistors. Cooling the supply between each
application, to account for temperature effects with current...



Brian Whatcott Altus OK Eureka!