When I tried to use Source Transformation method on the 12 V voltage source the problem gave me a wrong answer. If I transformed the 2 A current source it works just fine. My question here, Is there any Cases in Deducing Norton and Thevenin equivalents where I cannot use source transformation method?

The right answer for I Norton is -0.4 A

enter image description here

  • 4
    \$\begingroup\$ Show your work. It should work just fine, my guess is you made a simple mistake somewhere. \$\endgroup\$ – alex.forencich Jan 1 '18 at 22:29
  • \$\begingroup\$ Not sure if this is what you got wrong, but notice that i in the diagram is not the same as the Norton source current. \$\endgroup\$ – The Photon Jan 1 '18 at 23:18
  • \$\begingroup\$ The transformations always work. They are not always helpful, without first transforming back, again. But they always work. There is pretty solid theory behind them. How exactly did you work things out doing the transformation on the 12 V source and its series resistance? Show what you did there. I'm curious how you applied it to get a result. \$\endgroup\$ – jonk Jan 2 '18 at 4:25

My question here, Is there any Cases in Deducing Norton and Thevenin equivalents where I cannot use source transformation method?

  • There is no Thevenin equivalent for an ideal current source.

  • There is no Norton equivalent for an ideal voltage source.

Any other linear one-port network (meaning a network of linear resistors, ideal voltage and current sources, and linear dependent sources) will have an I-V curve from which either a Thevenin or Norton equivalent circuit can be found.

  • \$\begingroup\$ Even in those cases it is possible to use such transformations, as long as you are comfortable dealing with limits to infinity as part of your circuit analysis Arsenal. \$\endgroup\$ – Edgar Brown Dec 28 '18 at 5:29
  • \$\begingroup\$ @EdgarBrown, sorry, who are you replying to? \$\endgroup\$ – The Photon Dec 28 '18 at 6:24
  • \$\begingroup\$ If this is a comment on my answer, then think about the Thevenin equivalent of an ideal current source. It would have to have both the voltage source and the series resistance going to infinity. While you could mathematically define such a thing, you'd basically have to use the Norton circuit to do so (taking V and R to infinity while maintaining V/R = I). And it wouldn't be a useful way to simplify a circuit analysis. \$\endgroup\$ – The Photon Dec 28 '18 at 6:26
  • \$\begingroup\$ Having several elements on a circuit going to infinity is not a problem, as long as you know what you are doing. I once had a student that had all the theoretical aspects down solid, but was too lazy to remember the more common rules of thumb for op-amp feedback. He had no problem manipulating all those infinite elements within the circuit and got a full grade in the exam. A schematic is just a shorthand form for a set of equations that makes it easier to manipulate them. \$\endgroup\$ – Edgar Brown Dec 28 '18 at 16:28
  • \$\begingroup\$ @EdgarBrown, That's true (and it's why I said "you could mathematically define such a thing" in my comment), but it only makes sense to make a Thevenin or Norton equivalent of your circuit, if it actually makes analyzing the circuit simpler. If it makes the analysis more complicated, why bother to do it? \$\endgroup\$ – The Photon Dec 28 '18 at 16:35

It's not possible to source transform a lone ideal source. Other than that, you can use it for pretty much anything. However, I will also note that the layout of the circuit will definitely determine what makes the most sense.

What you have to look at is what you get after performing the transformation, and whether or not that will help simplify the circuit. Thevenin equivalents are series while Norton equivalents are parallel, so you have to look at where changing series elements into parallel or vise-versa will help. Generally, if you have two blocks in series, try to transform them to thevenin equivalents so you can combine the sources and resistances. If you have two blocks in parallel, use the Norton equivalent so the sources and resistances can be combined in parallel.

In this case, transforming the 2A source and 4 ohm resistor, then combining with the 12V source and 6 ohm resistor, then transforming one more time to get a norton equivalent is probably the most straightforward sequence. You can source transform the 6 ohm and 12 volt sources first, but that doesn't really make combining elements much easier.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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