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KCC's Quizzes AQQ254 about Thevenin & Norton Dipôles enigma

And here is a new quiz!

Many existing courses, articles, etc represent the Thevenin-Norton equivalent dipoles as shown above. 

However, there is something wrong somewhere. For example, the Thevenin dipole as represented does not consume any energy while the Norton equivalent does! Theory said they must be pretty equivalent. 

Question: What's wrong and how will you rectify the schematics?



Updated text.
[edited by: emassa at 2:08 PM (GMT -5) on 27 Feb 2024]
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  • What's missing is the load resistance!

  • 1) I believe it's only an equivalent transformation specifically with respect to a load or in a circuit that is intended to be loaded. Redefinition, not really a rebuttal. Currently at Node A-B, you can only see the same voltage.

    2) Adding a load to node A-B will make both circuits effectively equivalent assuming that the Thevenin/Norton were derived from the same circuit and choosing the same load to remove for analysis. In this case, both circuits will yield the same power/energy consumption/whatnots.

  • Is there something wrong? For example, the open circuit Thevenin dipole as represented does not consume any energy while the open circuit Norton equivalent does but the short circuit Thevenin dipole as represented does consume energy while the short circuit Norton equivalent does not. Theory says they are equivalent with respect to an external load.

  • The equivalents behave similarly for  a given resistive load. The equivalent itself is not "equivalent" but produces equivalent current and voltage for a resistive load.

    One limitation of these models is that they are limited to resistive loads. One way to improve the models is to change the source to a real frequency variable source and to change the resistors to an RLC network that produces equivalent impedance over the frequency range of the load.

  • I would add Rload between A and B.  

  • Nothings wrong with the schematics, both are equivalent when understood in context. When the same external load is applied to either of the circuits, the power consumed is equivalent. If you truly wanted to rectify the power consumed by the representation of this circuit, you could draw the thevenin equivalent as a closed circuit, but keep in mind that the internal resistance must be in series with the voltage source when applying a load.

  • The two are identical as black boxes, meaning we can't tell them apart externally. However, the Norton one will internally dissipate power if left open and the Thevenin one will internally dissipate power if shorted.

  • Many thanks Martin for your so prompt reply (once again)! Well the Thevenin Dipole is well an ideal voltage in series with a resistance. The 2 access points A and B are in fact connected to an open circuit. If one wants to connect a load, you simply replace the open circuit (infinite resistance) by the load. Or even better, your load has to be connected in parallel with the open circuit! The Norton dipole is in fact an ideal current source in parallel with an admittance. The missing item is the 2 access CANNOT left open: they must be terminated by a short! If one wants to connect a load, this latter has to be connected in series with the short. The symmetry of the 2 model representations is then perfect: Thevenin dipole terminated with an open circuit and Norton dipole terminated with a short circuit! The paradox of the original Norton consuming power disappeared!

  • Thanks  for your explanation! I think you are close with the perfect answer. The observation is the Norton dipole is incorrectly displayed. It has to be terminated by a short; exactly like the Thevenin dipole has to eb terminated with an open circuit! Connecting a load means to put it in parallel with the open-circuit for Thevenin and put it in series with the short-circuit for Norton!

  • Many thanks  ! I think you are are approaching the perfect answer! The observation is the Norton dipole is incorrectly displayed. It has to be terminated by a short; exactly like the Thevenin dipole has to be terminated with an open circuit! Connecting a load means to put it in parallel with the open-circuit for Thevenin and put it in series with the short-circuit for Norton! There is thus a common mistake made by many references to show a Norton dipole without having shorten the access A and B