I have seen that if we have a thermocouple made up from some metals AB, and we are also able to construct two more thermocouples made up from metals AC and BC, then the emf in thermocouple AB equals \$emf_{AC}-emf_{BC}\$. Why is this? Also, what would be the resistance in such circuits (like thermocouples with a voltmeter)? I'm confused since the conductors are made up from different materials.

  • \$\begingroup\$ Please clarify this question, what do you mean by emf in a thermocouple? Are you trying to minimize noise? \$\endgroup\$
    – Voltage Spike
    Aug 29, 2016 at 20:06
  • \$\begingroup\$ @laptop2d: Thermocouples are a voltage caused by a joint between two dissimilar metals (the Seebeck effect). The emf that he is referring to is due to that effect. He isn't talking about noise. \$\endgroup\$ Aug 29, 2016 at 20:55
  • \$\begingroup\$ Wait--Voltage, Ok got it. Just took me a while. I haven't heard someone refer to voltage as EMF for a while \$\endgroup\$
    – Voltage Spike
    Aug 29, 2016 at 21:02
  • \$\begingroup\$ Great reference from Omega at omega.com/temperature/Z/pdf/z021-032.pdf --- you are dealing with the Law of Intermediate Metals. \$\endgroup\$ Aug 29, 2016 at 21:09

2 Answers 2


The transitive property of the Seebeck effect can be understood by thinking about a ring of metals at constant temperature:


Where the A on the right and the A on the left are connected. Since in a loop we must have zero voltage drop absent external forces:

\$ V_{AB} + V_{BC} + V_{CA} = 0 \$

Which some rearranging will give you the transitive property.

The 'resistance' of a thermocouple is slightly more difficult to talk about. When you force a current through a junction of two dissimilar metals (by hooking up a voltage source), you heat up the junction, or remove heat from the junction. This is the Peltier effect.

Because the junction creates a voltage of its own, the voltage and the resistance don't have a linear relationship: if you reverse-bias the junction, the junction can have 0 voltage across it, while the current might be non-negative, giving rise to a resistance of 0, if we use Ohms law.


Different metals A and B (at the same temperature) have a different density of electrons. When they are brought into contact electrons will flow from the metal with the higher density to the metal with the lower density. The metal that loses electrons will become more positive the other more negative. The resulting field will move electrons in the opposite direction and a thermodynamic equilibrium is possible, which results in a certain voltage between the two metals.

In electronic devices a lot of metal connections are present. Pins, solder, copper, connectors, etc. Therefore a lot of voltages exist, but luckily this is not a problem.

When there is a connection like ABA, the resulting potential between A and B appears twice from A to B and from B to A but with the opposite sign, because the direction is different! Therefore the voltage cancels out and the metal B drops out of the equation. It is possible to go on and put as many metals in between, like ABA, ABCA, ABCDA, ... and so on. As long as the first and the last metal is the same the net voltage is zero.

For this reason the voltage between AB can't be measured, because we would need to make contact again with probes of, say, metal M and we would have MABM.

For your case with contacts AC and BC (brought into contact again) the same reasoning leads to the result that only the outermost metals are relevant and ACBC becomes AC.

The voltage between two metals depends on the temperature so if we have ABA and one junction is at a different temperature, the net voltage is no longer zero and we can measure this difference.

So, a thermocouple can't measure the temperature, but only the difference of the temperature between two junctions. A thermometer based on thermocouples therefore has to have a temperature sensor built in to determine the absolute temperature of the thermocouple, because the voltage difference only represents the temperature difference between the thermocouple and the meter.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.