Simplifying circuit for Thevenin theorem

Thread Starter


Joined May 18, 2020
Hello! I've got this circuit:

I need to apply Thevenin's theorem on A B nodes, so far I am having hard time simplifying it to calculate Rth.
Here is photo with what I did. 8 is R6+R7. Can someone give me a clue on how to proceed?

Mod: Posted photo.
AAA 338 15.02.gif
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Joined Jul 24, 2018
First thing you can do is recognize R3 and (what you call) R8 are in parallel, so you can replace them with an equivalent R3 || R8 resistor. Then, once you do that, if should be easier to solve, especially how you have it drawn. As a hint for after that, see that the node you connect the left side of R4 to is the same node as you connect R1 to on the left. Try redrawing it after that (what I would do is connect the left side of R4 to the left side of R1 instead of directly to R8 and R3, as a suggestion - it'll make it a little clearer) and see if you can figure out the next step.

Hope that helps!


Joined Mar 31, 2012
Try to redraw the circuit with Node A at the top and Node B at the bottom. Then add the resistors that are directly connected to Node A at the top and the resistors that are directly connected to Node B at the bottom. Now connect in the resistors that are on the other ends of the resistors you've already drawn. Eventually the top and bottom will come together. Then go through and make sure that every node connects to the correct resistors -- color coding the nodes in the original diagram and the redrawn diagram helps enormously for this. You might need to tweak your diagram a few times to get it drawn in a way in which it is clear how to combine things.