Thevenin theorem, can't understand the circuit

Thread Starter


Joined Feb 25, 2017
I'm stuck on this Thevenin question and it's really frustrating as after lots of studying and practices I feel I am back to square one not knowing anything about volatge and current in a circuit. Would really appreciate some guidance along the way of me trying to solve this question so I can identify my many flaws in my understandings about Voltage, Current, references to the ground and everything els. So here is the question:

So I have found the Rth:

So the first thing I do see in this circuit is that Vab = Eth, Va can be any spot on the upper node and Vb= 0 because there is nothing between Vb and the ground, now I think that's wrong but can't convince myself at all why.


Joined Mar 31, 2012
It's strongly preferred to post your images on AAC so that people don't have to go out to third party sites and so that the images get archived along with the post.



You are doing just fine, so far. You need to get Vab and, yes, Vb is 0 V because that node is defined as your 0 V reference. So you just need to use any analysis technique you want to find Va.


Joined Feb 19, 2010
Usually Thevenin equivalent circuit is represented by Thevening voltage source and Thevenin resistance.

Your circuit has one current source and one voltage source. I think it would be convenient for you to convert the current source into a voltage source. You have 8 mA independent current source in parallel with 5.6 kOhm resistor, you can readily convert it into voltage source:



Joined Jun 17, 2014

Talking about convenience, it looks to me that this would be ultra simple if we convert the voltage source to a current source first. Then we have two current sources in parallel so the current adds (or subtracts), and two resistors in parallel which combine easily to make one resistor. Since then we would have one current source in parallel with one resistor, we could leave it that way or convert to a single voltage source in series with a single resistor. If the values were simpler we could do this in our heads very quickly. For example, try it with two 4k resistors instead of what is there now.

Thanks to WBahn for posting the actual pics. I hate going to a third website for the image as sometimes it doesnt show up properly or might even be missing.


Joined Mar 31, 2012
Any of the normal analysis methods will probably work about the same on this one. The values aren't "simple", so you're going to have to do a bit of calculation no matter what.

Moving the current source to the outside lets you do mesh analysis with two meshes, one of which is trivial as it is 8 mA.

Nodal analysis is likewise very simple as you get set up the single node equation, which also happens to be the voltage that is the Thevenin voltage, by inspection.

Superposition is quite viable, though my guess is that it probably involves the most work of the normal techniques, though not by much.

Converting either the voltage source and the 2.2 kΩ to a current source or the current source and the 5.6 kΩ to a voltage source are about equal effort when all is said and done, is my guess.

I'd probably do nodal analysis, as it's one equation, written by inspection, whose unknown is the value I am looking for.