I'm trying to solve and exercise (see picture) where I'm supposed to find the voltage drop between 2 points in a circuit, using Thevenin theorem. I removed the resistance between those two points.

Now, my professor told me that we may not apply KVL through a current source. How, then, can I build and equation to find VAB (or Vth)?

enter image description here

  • \$\begingroup\$ Notice that we know the current value that is flowing in this series circuit. So you can use the ohm's law and find all the voltage drops across the loop and apply KVL. Can you do it? \$\endgroup\$
    – G36
    Commented Mar 10, 2020 at 14:44
  • \$\begingroup\$ R3's value doesn't matter, since it's in series with a current source. Replace it with a short...now you have R2 in parallel with your current source. Find the Thevenin equivalent... \$\endgroup\$ Commented Mar 10, 2020 at 14:45
  • \$\begingroup\$ Are you allowed to apply KCL? \$\endgroup\$
    – jonk
    Commented Mar 11, 2020 at 1:06
  • 1
    \$\begingroup\$ Daniel, you shouldn't just leave an open question and not respond. I asked if you were allowed to apply KCL. Another question is what exactly you are allowed to apply. Another question is, why did you remove \$R_2\$ in the second circuit? Was that some approach you were trying to take? If so, please explain yourself better about that step. \$\endgroup\$
    – jonk
    Commented Mar 11, 2020 at 18:35
  • \$\begingroup\$ Only saw it now. Yes, my professor taught us that, in order to find Thevenin's equivelent, we remove whatever branch connects the two terminals we're trying to analyse. \$\endgroup\$ Commented Mar 11, 2020 at 19:05

2 Answers 2


One thing that may help you a lot in the future is to realize that you can label one (and only one) of your "nodes" (read as "wires") as being exactly zero volts (\$0\:\text{V}\$.) So your schematic can be re-drawn like this:


simulate this circuit – Schematic created using CircuitLab

On the left, I simply called the bottom wire "ground" and broke the wire connections, which aren't really needed. (They are all labeled as "ground" so they connect together, whether or not I draw a wire to show that fact.) But it is otherwise the same circuit.

On the right, I just slightly moved things around and then drew a box around an ideal voltage supply with a voltage divider. The only reason for moving things around was to make the voltage divider aspect just a little more obvious, in case it wasn't beforehand. But again, this is otherwise the exact same circuit.

Most textbooks and classwork quickly gets to the point of teaching about voltage dividers. For example, in "The Art of Electronics," 3rd edition, they discuss it on page 7. Which is right after they introduce the idea of micro- and milli- and such prefixes. In other words, very early on. So I'm pretty sure you know what one is. In the same book on page 12, the authors are already telling you about the Thevenin equivalent for it, as well. So I think you should be able to work out the following:


simulate this circuit

Here, I've replaced the voltage divider circuit on the left side with the new Thevenin equivalent on the right side. Please note that this does NOT look like what you drew as your second, lower drawing.

Now, you have a current sink and since this circuit has all its two-lead components in series with each other, you know what the circuit's current is. It's given to you on a silver platter.

You should be easily able to start at the Thevenin voltage on the left side of the right-side circuit shown above and compute the voltage drop across the Thevenin resistance to work out the voltage that must be at node A. Do you think you can do that?

  • \$\begingroup\$ Great answer and great advice! I realized I was missing a vital piece of information: that when a current source and voltage source are in series, the current is that of the current source. I'm actually a CS student and this class is sort of thrown in the curriculum. All the material I have are the exercises given by the professor. \$\endgroup\$ Commented Mar 12, 2020 at 0:45
  • \$\begingroup\$ @DanielOscar I used to teach CS 2nd and 3rd year students at the largest university in my state. (1990's.) (My classes were on computer architecture, concurrent programming, operating systems, and assembly language.) I had no idea you were CS, though. In a class of 75, I'd usually have 6 or 7 EE students. But at the time, our CS program didn't send CS degree students to take any electronics classes. It's interesting to hear your situation! (And I'm glad to hear what I wrote helped. Thanks for letting me know.) \$\endgroup\$
    – jonk
    Commented Mar 12, 2020 at 1:16

You can use KVL to solve for the voltage across the current source, but you can't assume that there is any particular voltage across the current source.

You still have other tools available. How about Ohm's Law and KCL? You can also apply your understanding of how series and parallel circuits behave.


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