enter image description here

I was taught to do it with a longer method which I understand, by moving the Va and Vb terminals up to the source and gradually moving further back while finding equivalent circuits. I want to do it the way I was taught in my notes though.

What am I doing wrong when I try to calculate Voc and Isc here?

for Voc I just used the voltage divider formula 12*r4/(r1+r3+r4) This is incorrect.. but why?

For Isc when we invalidate the path for current to flow through r4 and this resistor is effectively gone. I use the method I was taught to calculate the circuit resistance by also shorting the voltage source.After r1 the current splits up between R2 and R3 which are in parallel due to the short at Va and Vb,the current splits in two evenly here, so that gives me the (wrong) Isc.

I'm extremely confused. what am I doing wrong? Thanks.


1 Answer 1


Final edit note: In the following discussion, I'd assumed you understood more about Thevenin than appears true. My apologies for that. The following still applies, but only after you understand what I wrote. I won't change what follows, but see the last section for my final edit additions. Hopefully, they finally help enough. And thanks for the discussion. At times, it's a good way to get the help you need.

\$V\$ forms a Thevenin equivalent with \$R_1\$ and \$R_2\$ (which make up a voltage divider.) I'll assume for a moment that you can see that much, already. This means you can replace all three components with just two components, \$V_{_\text{TH}}\$ and \$R_{_\text{TH}}\$.

Now, \$R_{_\text{TH}}\$ is in series with \$R_3\$, so you can just add them together. That's simpler, still, now.

Now, you have, once again, a voltage divider made up of \$R_{_\text{TH}}+R_3\$ and \$R_4\$, with a source voltage of \$V_{_\text{TH}}\$. And now again, you should be able to work out the new equivalent Thevenin voltage and Thevenin resistance that results.

Do you follow this, okay?

Edited, given conversation below this question

Okay. Now that you've provided yet another concept,

enter image description here

involving an unexplained method applied to the above circuit, I think you need a conversation. I can't do much of that here. But I can walk through some basic ideas.


There's a lot of theory behind it, but I'm going to ignore all that and just focus on the idea of some "black box" that holds some unknown number of power sources and resistors wired up in unknown ways but exposing just two contact points.

You can take an idealized voltmeter (one that takes exact and accurate measurements without drawing any current from the circuit) and put its two leads onto the two contact points and measure a voltage. You will get some reading. Call it \$V_{_\text{OC}}\$, for "open circuit voltage."

You can take an idealized ammeter (one that takes exact and accurate measurements without removing any voltage from the circuit) and put its two leads onto the two contact points and measure a current. You will get some reading. Call it \$I_{_\text{SC}}\$, for "short circuit current."

You can now compute \$R_{_\text{TH}}=\frac{V_{_\text{OC}}}{I_{_\text{SC}}}\$, which must be the effective internal resistance of this box. It has to be there inside the box because, when you shorted its output pair in order to measure the current it was that resistance that limited the current provided by \$V_{_\text{TH}} = V_{_\text{OC}}\$ (I just snuck in the fact that the open circuit voltage is the same as the Thevenin voltage.) If \$R_{_\text{TH}}\$ inside the box was anything else, then you would have read a different current than you did.

There are lots of pages on the web about this. Here's one of them that I found more quickly than others. Skim over it until you feel somewhat better comfort with what's there.

In the meantime, here's a slide from that presentation to consider in thinking about what I just wrote above:

enter image description here


Let's take your circuit and chop it up six ways to Sunday:


simulate this circuit – Schematic created using CircuitLab

So, your question might be about the last case, Case 5, where the load is unknown at the time of the question and you are wanting to find the Thevenin equivalent of everything that's inside the black box, there.

That's what I thought you were talking about.

But perhaps you were actually asking about Case 4, which the diagram you linked (and I've included above) might instead be about. Here, the question is about the Thevenin equivalent of everything except for \$R_4\$, instead pushing \$R_4\$ into the load. In this case, you actually would ignore \$R_4\$ (because it's the load) and focus on the rest.

That might be what you were trying to understand.

If so, we are talking about different things, entirely. And details matter.

I've also added many other possible cases that could be considered. All of these make no changes whatsoever to the schematic. All they do is put the dividing line in different places.

So the details matter. I can't emphasize this enough. And, if these all feel like slippery ideas that you can't fully put your finger on, then you need to keep staring at them until something clicks for you. They aren't slippery. They are important.

As it turns out, you can start at Case 1 and come up with one answer for the Thevenin equivalent of the black box, then add \$R_1\$ to it and come up with a different answer for the black box, which is Case 2. And so on. Each time you change the boundary of the question, you will get a new answer for the Thevenin equivalent of the black box. Obviously, because what's inside the black box is changing. So the answer to the question changes.

(Also note that the stuff that gets ignored is also different as you change the boundaries.)

Does this help?

Norton and Thevenin

Taking Case 5, you may perform the following steps:


simulate this circuit

In short:

Step A: Perform Thevenin to Norton conversion: \$R_{N_1}=R_1\$ and \$I_{N_1}=\frac{V_1}{R_{N_1}}\$.

Step B: Perform Norton to Thevenin conversion: \$R_{{TH}_1}=R_{N_1}\mid\mid R_2\$ and \$V_{{TH}_1}=I_{N_1}\cdot R_{{TH}_1}\$.

Step C: Perform Thevenin to Norton conversion: \$R_{N_2}=R_3+R_{{TH}_1}\$ and \$I_{N_2}=\frac{V_{{TH}_1}}{R_{N_2}}\$.

Step D: Perform Norton to Thevenin conversion: \$R_{{TH}_2}=R_{N_2}\mid\mid R_4\$ and \$V_{{TH}_2}=I_{N_2}\cdot R_{{TH}_2}\$.

A recommendation note: It would be worthwhile, instead of working forward on a problem like this all the time, to try and work backwards from the above steps. This means starting at the end and substituting in prior results, simplifying as you go. See what you end up with, that way. There are a few insights to be had, if you do that. If all you ever do is follow recipes in the forward direction, it may be a lot longer before you uncover some interesting thoughts about the processes you are applying.

  • \$\begingroup\$ I solved it by going from thevenin to norton to thevenin to norton. But my book showed an example where the short circuit current and open circuit voltage are solved without any resistor simplification. Then the total resistance is just calculated from those and all three are used to create the equivalent circuits AFTER. I wanted to know how to do that with this circuit, I tried and failed badly, if we can't do it with this circuit then why not? \$\endgroup\$
    – ramose
    Aug 6, 2021 at 21:17
  • 1
    \$\begingroup\$ @ramose Well, tell me... how do you work out the open-circuit voltage without any resistor simplification, resistor divider analysis, or other means? You will need the open circuit voltage. Then, after you short the two nodes, you will need to compute the current. How would you do that? But, if you used what I describe above, then you would have the open circuit voltage and also the series impedance, as well. All at once. You can also solve all this without the above if you use either nodal or mesh analysis matrix methods. Is that what you are supposed to do here? \$\endgroup\$
    – jonk
    Aug 6, 2021 at 21:59
  • \$\begingroup\$ @ jonk I'm missing some basic knowledge I think. I attempted to get the Voc and Isc by the method I mentioned in the main body of my text which failed and I don't know why. Was just following a similar unexplained method in this circuit where they just ignore the last resistor and easily find the Voc. : postimg.cc/jwz9HNxd. \$\endgroup\$
    – ramose
    Aug 8, 2021 at 19:39
  • \$\begingroup\$ @ramose Okay. Thanks. That picture you referenced is not the same problem that you presented here (or, at least, as I understood your writing -- which may be as much my ability to interpret and/or stoip myself from projecting my views onto what you wrote.) Often, examples will have what is called a load and you are supposed to work out the circuit up to but not including the load, which is what that diagram appears to be showing -- looking at yours and theirs now. \$\endgroup\$
    – jonk
    Aug 8, 2021 at 20:56
  • 1
    \$\begingroup\$ @ramose I've added more to my answer. Hopefully, it helps. \$\endgroup\$
    – jonk
    Aug 9, 2021 at 0:02

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