In the Ti case the resistance values show a deviation of about +/- 15 % from
the average value. For Al the deviation is about 13%.
A 7.5% deviation in each (V and I) meter reading would account for this.
Take many V vs I readings and see if a straight line fit is at all
reasonable with sizable, but random, deviations.
Was the pressure squeezing the foils always the same?
Bob Sciamanda
Physics, Edinboro Univ of PA (Em) http://www.velocity.net/~trebor/
trebor@velocity.net
----- Original Message -----
From: "Ludwik Kowalski" <kowalskil@MAIL.MONTCLAIR.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Friday, May 07, 2004 1:50 PM
Subject: Internal resistance again.
| 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