Finding the open ckt voltage (Thevenin equivalent problem)

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The discussion centers on finding the open circuit voltage (Vth) and Thevenin equivalent resistance (Rth) in a circuit. Participants suggest simplifying the circuit first to streamline calculations, emphasizing that this approach often saves time and effort. One user mentions using a delta-wye transform to find the equivalent resistance, while others advocate for simpler series and parallel combinations. A calculation error is identified, with a corrected Vth value of 11.32 V aligning with simulation results. The conversation highlights the importance of different methods in circuit analysis while encouraging users to verify their calculations.
rugerts
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Homework Statement
Find v_oc.
Relevant Equations
Mesh or Node analysis. => KVL and KCL.
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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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