# How can I experimentally check that the voltage in a thermocouple is generated be the gradient in the wires and not by the junction themselves?

I just found out on Wikipedia (also here and there) that the voltage in a thermocouple in not generated by the cold & hot junctions, but by the temperature gradient in the wires which generates a wire type + temperature dependent voltage gradient.

This quite surprised me because I completely ignored it, but it makes sense since in the Peltier effect, it is even reversed between P-type and N-type bars.

However, the question is how could this be experimentally demonstrated ? because I see no way to make voltage measurements without introducing other junctions and temperature gradients with the voltmeter wires and probes to the experiment…

Set up a circuit like the one in the Wikipedia article. Using a sensitive (high resistance ) voltmeter, measure (with brief contacts) the voltage between a point on one of the non-similar wires and a corresponding point on the other. You should find that the voltage difference decreases as you approach the junction.

• How can I deduce that this is not simply the consequence of changing the temperature of the cold junction ? And How can I know in what way my measure isn't impacted by the temperature difference between the wires and the voltmeter probes ? Aug 24, 2021 at 3:25
• Don't change the reference temperature, and keep the meter lead contacts brief so they do change the temperatures on the dissimilar wires. Aug 24, 2021 at 14:22
• I understand that, but on what basis should I know if there is one or no potential generated because of the temperature difference at the junction between wire and probe ? Aug 25, 2021 at 19:52

Instead of trying to show that the voltage is not generated in the junction, show that it can be generated without dissimilar metal junctions. Construct a flat parallel plate capacitor with two plates of the same material (e.g. aluminum foil) separated by an insulator (e.g. Saran wrap). Connect the plates with a piece of aluminum wire thin enough to provide reasonably high thermal resistance. Squeeze the plates and the dielectric material together between two nonconducting objects of different temperatures (preferably differing by a few hundred degrees Celsius). While the temperature gradient is being applied across the capacitor, use aluminum-bladed scissors to cut the thin aluminum wire. Allow the plates to reach the same temperature, then measure the voltage between them with a high-ohmic voltmeter.

The Seebeck effect should result in a net charge transfer between the plates of the capacitor, with electrons migrating from the hot plate to the cold plate to cancel out the Seebeck voltage. Then the wire connecting the two plates is cut. When the hot side of the capacitor cools down, the Seebeck potential disappears, but the transferred electrons remain trapped on the (formerly) cold plate. The formerly-hot plate, then, will have a slight positive voltage relative to the formerly-cold plate.

Disclaimer: I have not tried this, and therefore cannot say whether the results support, disprove, or are inconclusive with regard to the original claim.