I have the following circuit and I'm asked to find the Thevenin resistance between points A and B.

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I tried to solve it by creating an open circuit between the points, as usual, thus removing the 10ohm, and 20ohm resistors but the solution I found was wrong. The circuit looked like that:

enter image description here

Is there any general advice as to how one should proceed when the points you have to find the Thevenin voltage and resistance are connected with a lot of other components?

  • 1
    \$\begingroup\$ You don't remove components to find the Thevenin/Norton equivalent circuits. Perhaps you're confusing that with setting the current and voltage sources to zero? \$\endgroup\$
    – vir
    Commented Jun 21, 2022 at 20:37
  • \$\begingroup\$ Replace the current sources with open circuits. Replace the voltage sources with short circuits. Then solve for your resistance between A and B looking back into the network. A lot of stuff will get shorted out and make the problem easier. \$\endgroup\$ Commented Jun 21, 2022 at 20:41
  • \$\begingroup\$ In the upper right hand corner you have 10 ohms in series with 20 ohms. That makes 30 ohms. That 30 ohms is in parallel with 20 ohms. However in your modified circuit, you have 20 ohms where the replacement for 20 || 30 should be. I'm pretty sure that 20 ohms in parallel with 30 ohms is not 20 ohms. \$\endgroup\$ Commented Jun 21, 2022 at 21:00

2 Answers 2


Determining the Thévenin resistance means looking for the small-signal resistance offered by the connecting terminals A and B when all sources are turned off: open-circuit the current source and short-circuit the voltage sources. Then, redraw the circuit to simplify it: resistors in series can be replaced by one element, paralleled resistors can be combined and so on. Then, inspect the circuit, meaning you can read the circuit in your head by looking at it and "see" the equivalent resistance offered in terminals A and B. The below sketch shows how to get there:

enter image description here

SPICE is a valuable assistant to verify your steps. The primitive .TF computes the small-signal resistance offered between terminals A and B and confirms the simulated value in the right side.

For the Thévenin voltage, as suggested by Mister Mystère - I like the accent, like in Thévenin :-), you can apply the superposition theorem which works considering the linear circuit. Again, SPICE can assist you in checking all the intermediate steps as shown below:

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I leave the symbolic determination of these values for the student.


Shut down all sources, i.e. replace current sources by open circuits, and voltage sources by shorts. Resistances which connect two identical nets can be replaced by a wire (you'll see many of them here), resistances with a floating end do not carry current at all and can be removed (the bottom right ones will).

Simplify the circuit by series/parallel association and you should end up with 9 Ohm.

Thevenin has nothing to do with this method, though you will end up with the resistance that happens to be in the Thevenin equivalent of this circuit. To get the Thevenin equivalent voltage, turn on each source one at a time solving for the open circuit voltage, then sum all results: just like in linear algebra, linear circuits follow circuit(Va+Vb)=circuit(Va)+circuit(Vb).


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