Electrical Engineering/resistance


QUESTION: Measuring a resistor with a common multimeter (MM) if I heat only one of the terminals of the resistor, this happens:

-heating the terminal connected to the positive probe of the MM, the resistance increases
-heating the terminal connected to the negative probe of the MM, the resistance decreases

How to explain this situation?

ANSWER: Well, if the resistor is still connected to the circuit you will find the same effect even without heating the terminal.  This is because the parallel circuit components are causing loading of the resistor under test.  And if there is a diode or other non-linear element the battery from the MM is acting as a circuit operation.

To measure resistor's resistance it should be disconnected from the circuit.

Let me know if you need more.

---------- FOLLOW-UP ----------

QUESTION: Hi, thanks for prompt reply...
The situation I described is with the resistor disconnected from the circuit. It happens with carbon, metal film or wirewound resistors.

ANSWER: Then it must be in the instrumentation or placement of the probe.  If it is a very high resistance value it could be caused by repositioning the probes and leads or your hands being in a different position if you are touching anything during the test.

And, of course, it could relate to the connection of the probes, especially so if the resistor is a very low value where the contact resistance with the probe or lead enters into the picture.

Hope this helps.

---------- FOLLOW-UP ----------

QUESTION: I found an explanation: it's a Seebeck effect
Thanks a lot for your attention

The Seebeck effect could explain it if their were different materials used in composition of the resistor under test.  Normally, we think of the Seebeck effect exhibiting in thermocouples, thermopiles, special resistive bridges and such.  In measuring just a linear resistor of homogeneous materials it probably would produce only a negligible effect.  Or, if the resistor physical size and connections to the MM probe/leads represented enough of a dissimilar material effect so show up....  but why wouldn't it do the same on both ways of measuring?  It would require a greatly dissimilar connection material to the resistor leads, wouldn't it?


Electrical Engineering

All Answers

Answers by Expert:

Ask Experts




All technical areas of Electronics Engineering.


BSEE, MBA, Design, R&D, University Research.
Senior Life Member of IEEE. Life Fellow of AES.

IEEE, Consumer Electronics Society, Audio Engineering Society.
Broad teaching experience; work experience mostly in consumer electronics and conversion from analog to digital technologies. Pioneer in digital audio at all levels.

BSEE (Equiv) BYU BSEE University of North Dakota MSBA (MBA) Illinois State University Graduate Studies in Computer Science - Bradley University Graduate Studies - Ohio University Graduate Studies - University of Missouri Kansas City DeVry Tech - Electronics

©2017 About.com. All rights reserved.