Finding the open ckt voltage (Thevenin equivalent problem)

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Homework Help Overview

The discussion revolves around finding the Thevenin equivalent voltage (Vth) and resistance (Rth) for a given circuit. Participants are comparing their analytical results with circuit simulation outcomes from LTSpice.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss different methods for simplifying the circuit to find Rth, including series and parallel combinations as well as delta-wye transformations. Questions arise about the effectiveness of these methods and their implications for finding Vth.

Discussion Status

There is an ongoing exploration of various simplification techniques, with some participants suggesting that simplifying the circuit first may lead to quicker solutions. Others express uncertainty about the calculations and seek clarification on specific steps and results.

Contextual Notes

Some participants mention discrepancies between their calculations and simulation results, indicating potential errors in analysis or assumptions made during the problem-solving process.

rugerts
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Homework Statement
Find v_oc.
Relevant Equations
Mesh or Node analysis. => KVL and KCL.
1.PNG
2.PNG
1572120055079.png


I'm trying to find Vth (and then Rth). My work and results don't seem to agree with what my circuit simulation (done on LTSpice) shows. Can anyone point out what may be wrong here?
 
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I didn't even try to follow what you did, since you did it the hard way. I suggest that you simplify the circuit (with the voltage source treated properly for getting Rth) and THEN do calculations.
 
phinds said:
I didn't even try to follow what you did, since you did it the hard way. I suggest that you simplify the circuit (with the voltage source treated properly for getting Rth) and THEN do calculations.
1572120924055.png
1572120941197.png

This is what I'm going based off of.
What exactly are you talking about?

Also, is it really that hard? Granted, what you propose might be much easier, but what I'm doing isn't very involved.
Thanks for the reply.
 
rugerts said:
What exactly are you talking about?
You've got a circuit that easily resolves down to one resistor if you just simply properly. Don't even need to DO any loop or node stuff.

If you're new to all this, you'll soon realize that simplifying the circuit FIRST is almost always going to save time. On this problem, it totally SOLVES the problem (of getting Rth)
 
phinds said:
You've got a circuit that easily resolves down to one resistor if you just simply properly. Don't even need to DO any loop or node stuff.

If you're new to all this, you'll soon realize that simplifying the circuit FIRST is almost always going to save time. On this problem, it totally SOLVES the problem (of getting Rth)
Could you be a bit more specific? You're saying it solves the problem, but how exactly? I've found Req to be 31.8473 k. It took a delta-wye transform.
 
rugerts said:
Could you be a bit more specific? You're saying it solves the problem, but how exactly? I've found Req to be 31.8473 k. It took a delta-wye transform.
No, just simple series & parallel 'til you get to a single R
1572124254353.png
 
phinds said:
No, just simple series & parallel 'til you get to a single R
View attachment 251885
Okay. Well that's one way of doing it. The delta wye works as well.

But, again, could you please follow up with what this would help with? If I'm interpreting you correctly, you're saying to find the single equivalent resistance of the entire ckt? Is this then equal to Rth? What about the voltage across the open ckt, Vth?
 
rugerts said:
Okay. Well that's one way of doing it. The delta wye works as well
But messier by far
you're saying to find the single equivalent resistance of the entire ckt?
yes
Is this then equal to Rth?
What do you think?
What about the voltage across the open ckt, Vth?
Separate issue in either case.
 
In any case, whatever method you personally find easiest and fastest, that's what you should use. I think my way is that for Rth but maybe that's just me.

Also, you can use the same type of simplification technique to get Vth
 
  • #10
phinds said:
In any case, whatever method you personally find easiest and fastest, that's what you should use. I think my way is that for Rth but maybe that's just me.

Also, you can use the same type of simplification technique to get Vth
Appreciate the feedback. I definitely think it's still important to solve the problem in different ways.

In the meantime, however, do you mind checking my work to see what may have gone wrong? It's bothering me quite a bit and I keep going over it, and can't seem to find what my mistake is.
 
  • #11
Sorry but that kind of analysis makes my head hurt when I know there's an easier way.
 
  • #12
I think you may have slipped up in your calculations here (finger error on your calculator?):

1572148289839.png


When I use your current values and resistance values in that calculation I get a ##V_{th}## of 11.32 V, agreeing with your spice simulation.
 
  • #13
gneill said:
I think you may have slipped up in your calculations here (finger error on your calculator?):

View attachment 251902

When I use your current values and resistance values in that calculation I get a ##V_{th}## of 11.32 V, agreeing with your spice simulation.
Yep. I get the same. Thanks for double checking.
 

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