I have this circuit:

Initially it was all on blue. Then I opened the circuit in A and B (on red). The exercise asks the value of resistor R knowing that there passes a current of 1 A, using the Thevenin's Theorem. My question is, when I open the circuit on A and B, I remove the resistor R to the Thevenin's equivalent circuit. So, does the current I2 becomes zero, and I becomes equal to I1 ? It would be, since I'm opening the circuit there. But when ill find Thevenin's equivalent resistor, I can't say the resistor of 2Ohm is in series with the resistor of 6Ohm, so, the node between them keeps dividing the current to terminal A. How come?

If current I2 becomes zero when I open the circuit, that node wouldnt divide current anymore, so the current that passes on the resistor of 2Ohm would be equal to the current that passes on resistor of 6Ohm, so both resistors would be in series.

  • \$\begingroup\$ $$I=\frac{V}{R}$$ \$\endgroup\$
    – MrPhooky
    Apr 11, 2017 at 12:51
  • \$\begingroup\$ @MrPhooky sorry, I didn't understand. Doesn't I2 becomes zero? \$\endgroup\$ Apr 11, 2017 at 12:53
  • \$\begingroup\$ R = infinty when R is removed \$\endgroup\$
    – MrPhooky
    Apr 11, 2017 at 12:55
  • \$\begingroup\$ @MrPhooky so why aren't in series resistor of 2Ohm and 6Ohm? \$\endgroup\$ Apr 11, 2017 at 13:03
  • 1
    \$\begingroup\$ This is your third question about Thevenin's theorem in less than 24 hours. You might want to improve one or more of your existing questions before churning out more. \$\endgroup\$
    – Dampmaskin
    Apr 11, 2017 at 13:20

2 Answers 2


The goal of what you're trying to do is find the equivalent resistance in respect to the points where you removed the resistor. You are correct in that those 2 are in series, however by doing that conversion and turning it into a 8Ohm you lose one of the nodes connected to where the resistor used to be, and so it defeats the purpose of the whole calculation in the first place.

I would recommend re-drawing the circuit such that A and B are not in the middle of the diagram. That should make things easier to comprehend.


The exercise asks the value of resistor R knowing that there passes a current of 1 A, using the Thevenin's Theorem.

You may want to consider why the authors of this problem is giving you this information. Perhaps they are trying to reduce the amount of work you need to do?

  • \$\begingroup\$ ah! I understood your first paragraph! yes, besides they are in serious, we have to preserve the fact that the terminals A and B keep in that place, and doing the 8ohm's, they would disappear. That's easier to understand. The possible redraw that I can make is putting A and B between the 2 resistors, it is easier to see that we can't sum each pair of resistors in series. The bonus tip is irrevelant, I know how to solve these exercises, I just want to get an answer to this specific doubt that I have. \$\endgroup\$ Apr 11, 2017 at 14:11

As J.Jhon said adding the 2 resistors together is valid as they are in series but it does defeat the purpose of the question, so don't do that.

From my understanding of your question I think the problem wants you to use Thevenin's Theorem to simplify the circuit so you can eventually find the voltage drop across R then use Ohm's law to find the resistance value.

These pointers should get you started

  1. Where is ground, what are all your voltages relative to? I am assuming the bottom node but I am not sure.

  2. When solving Thev/Nor problems it is aways a good idea to either only have voltage or current sources, never both. In this problem I would suggest transforming the current source on the left to a voltage source, this will help with simplification.

  3. This problem is best solved (I think) with a nodal analysis, so identify each node and figure out the voltage, then move to the next node until you reach the nodes with the A and B terminals.

  4. If you find the voltage on the left side (HINT: 18V), you will notice some branches branches in parallel, ie all those branches will have a drop of 18V. Use that fact as a starting point.


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