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I am thinking of the thermocouple as just two wires twisted together and the twisted section being heated (hot junction). If one of the wires has a larger amount of the wire heated, but the same temperature difference, the voltage reading is the same for a smaller area.

From my understanding of a thermocouple, the voltage is generated from the Seebeck effect where the charge carries with more energy from the temperature increase will migrate to the cooler end. Surely then, if more of the wire is heated, the electrons will have a higher kinetic energy and therefore there would be an increased voltage? I can't see why this isn't the case however.

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  • $\begingroup$ For a similar reason that changing the area of the plates in a battery does not change the potential difference between the battery terminals. You are observing the effect of a property of two substance "in contact" being different. That difference in property is not a function of the size of the contact area. $\endgroup$ – Farcher Jan 28 '18 at 10:36
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Seebeck Effect (1821): two dissimilar metals, with junctions at different temperatures, produce a voltage. It was empirically discovered that "the voltage produced is proportional to the temperature difference between the two junctions."

Voltage is proportional to temperature difference; the voltage does not depend how much of the wire is hot, but the current does.

For example, for practical heating and cooling Peltier devices based off of the Seebeck effect and the related Peltier effect, adding these junctions in series increases voltage; adding them in parallel increases current, maintaining the same voltage. Heating up more of the wire is equivalent to adding junctions in parallel. The voltage generated by the Seebeck effect depends only on the difference in temperature. Does not matter how much wire is heated. It's the current that would increase if more of the wire is heated.

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  • $\begingroup$ I understand now why the voltage would remain the same, but I am not too certain why the current would increase. As in, if we consider the equation I = nAQv , surely none of the variables there are really changing so the current would not change. ( I am thinking not in terms of the Peltier device but the thermocouple I described in my question) $\endgroup$ – Terminus Est Feb 24 '18 at 17:28
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    $\begingroup$ Works same way a battery does. Batteries produce a voltage, and when you put them in parallel the current increases and voltage stays the same $\endgroup$ – pentane Feb 25 '18 at 4:28
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The electromotive force created by the Seebeck Effect ($E_{emf}$) is expressed as:

$$E_{emf}=-S\nabla T$$

where: "where $S$ is the Seebeck coefficient (also known as thermopower), a property of the local material, and $\nabla T$ is the temperature gradient" (Wikipedia). This would be the difference in emf between points 2 and 4 in the figure below, caused by a difference in temperature $T_1-T_2$.

Now, let's consider that making the heated region longer (region 1-2 in figure below) does not increase the temperature gradient component nor the Seebeck coefficient, in the equation above, and therefore does not increase the $E_{emf}$, the emf due to the Seebeck effect. A longer heated region is merely more 'deadweight': the electrons in this region have nothing causing them to move in a particular direction. The heated region could stretch a kilometre, but would add nothing to the voltage.

The Seebeck Effect

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