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Re: Internal resistance again.



The current may not be flowing uniformly at the voltage contacts. This
would be the case if the foil is not long, thin, and narrow or if the
voltage contacts are too close to the current contacts.

If the foil is rectangular and thin, you may need to use the van der Pauw
method.

Larry Woolf
General Atomics
3550 General Atomics Court
Mail Stop 78-110
San Diego CA 92121
Ph:858-526-8575
FAX:858-526-8568
http://www.ga.com
http://www.sci-ed-ga.org


-----Original Message-----
From: Ludwik Kowalski
Sent: Friday, May 07, 2004 11:49 AM
Subject: Re: Internal resistance again.

I am using a d.c. power supply to pass a current
through an Al 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=1.3 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.