Re: Internal resistance again.

[Bernard Cleyet <anngeorg@PACBELL.NET>]

I thought I'd sent this hours ago (no one had replied), but must have
been distracted by Blazing Saddles.   Did you know Brooks was married to
Ann Bancroft and Wilder to Gilda Radner?


It appears you are using a four terminal connection.  With such a method
the internal resistance of the P/S is irrelevant.  I presume, as you do,
that all metals have a + coefficient of resistance (w/ T).  So I wonder
if there is a problem with the EMF connection.  I recommend using a
fused method instead of mechanical.  It's difficult to weld * Al but
I've done it.  I haven't tried spot welding tho.  I don't know anything
about Ti.  What happens when you reverse the P/S connections?

or try amalgamating;  gallium or mercury might work, or low temp
soldering (tin alloys);  hard (silver) solder?

bc


p.s.    http://durafix.com/     better:
http://www.ccis.com/home/hn/page31.htm


Ludwik Kowalski wrote:

> 1) Sorry for spelling errors in my last message.
>
> 2) I am using a d.c. power supply to pass a
> current through a Ti foil. From the positive of the
> power supply the current goes through the
> ammeter and through the foil (firmly squeezed
> between two metallic plates). The voltage is
> measured between these plates. I expected
> voltage to be I*R but it is not.
>
> I=1 A --> 0.75 volts  (R=0.75 ohms)
> I=2 A --> 1.10 volts. (R=0.55 ohms)
>
> If anything I would expect R to go up with I, due
> to higher temperature. In reality the temperature
> does not change (to a touch) significantly.
>
> Two days ago I used an Al foil and observed
> similar results (same power supply).
>
> I=1.3 A  --> 0.55 volts
> I=2.0 A --> 1.06 volts
> I=2.9 A --> 1.30 volts
>
> How can this be explained? I am sure that R
> is essentially constant. The impedance of the
> voltmeter is certainly much larger than 1 ohm.
> Looks like the "internal r" of my power supply
> changes a lot with the current. Why is it so?
> Ludwik Kowalski