Finding the open ckt voltage (Thevenin equivalent problem)

In summary, the conversation discusses finding Vth and Rth in a circuit and the disagreement between the results and the LTSpice simulation. One person suggests simplifying the circuit and properly treating the voltage source to obtain Rth. They also mention that finding the equivalent resistance of the entire circuit can also help in finding Rth. The other person mentions using different methods to solve the problem and suggests checking the calculations to find the mistake. It is then confirmed that there was a mistake in the calculations and the correct value for Vth is 11.32 V.
  • #1
rugerts
153
11
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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  • #2
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.
 
  • #3
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.
 
  • #4
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)
 
  • #5
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.
 
  • #6
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
 
  • #7
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?
 
  • #8
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.
 
  • #9
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.
 

FAQ: Finding the open ckt voltage (Thevenin equivalent problem)

1. What is the Thevenin equivalent problem?

The Thevenin equivalent problem is a method used in electrical circuit analysis to simplify complex circuits into a single voltage source and a single resistance. This makes it easier to analyze and understand the behavior of the circuit.

2. How do you find the open circuit voltage in Thevenin equivalent problem?

To find the open circuit voltage, the circuit is first simplified by removing all loads and shorting all voltage sources. Then, the voltage across the open terminals is measured using a voltmeter. This voltage is the open circuit voltage and is equivalent to the Thevenin voltage.

3. Why is the open circuit voltage important in the Thevenin equivalent problem?

The open circuit voltage is important because it represents the maximum voltage that can be obtained from the circuit when no load is connected. It is a key parameter in understanding the behavior of the circuit and can be used to calculate the maximum power that can be delivered to a load.

4. Can the open circuit voltage be calculated using Ohm's law?

No, the open circuit voltage cannot be calculated using Ohm's law because Ohm's law only applies to resistive circuits. The Thevenin equivalent problem involves both resistive and non-resistive elements, making Ohm's law insufficient for finding the open circuit voltage.

5. How does the value of the open circuit voltage affect the behavior of the circuit?

The value of the open circuit voltage does not affect the behavior of the circuit itself, but it is a crucial factor in determining the behavior of the circuit when a load is connected. It helps determine the current flow and power dissipation in the circuit, and can also be used to find the Thevenin resistance.

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