Find the thevenin equivalent circuit (just need my values checked )

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Discussion Overview

The discussion focuses on finding the Thevenin equivalent circuit using the open/short circuit method. Participants are verifying calculations related to the Thevenin voltage and addressing discrepancies between theoretical calculations and software simulation results.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for the Thevenin voltage, V(th), as 6.22V based on a node voltage equation but questions the validity of this value against software readings.
  • Another participant suggests measuring the open circuit voltage directly between terminals a and b to determine the correct value.
  • There is a discussion about the reference node used in node voltage calculations, with one participant asserting that the reference node is not node b.
  • Clarification is provided that the voltage of interest is the potential difference between terminals a and b, which involves considering the 2V source in the circuit.
  • A participant confirms the calculation of the Thevenin voltage as 4.22V after accounting for the 2V increase due to the voltage source.

Areas of Agreement / Disagreement

Participants express differing views on the correct Thevenin voltage, with one participant asserting 6.22V and another confirming 4.22V after adjustments. The discussion remains unresolved regarding which value is correct.

Contextual Notes

There are indications of confusion regarding the reference node and the impact of the voltage source on the calculations. The discussion highlights the importance of correctly identifying the terminals and the reference point in circuit analysis.

Who May Find This Useful

Students and practitioners involved in circuit analysis, particularly those learning about Thevenin equivalents and node voltage methods.

asdf12312
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find the thevenin equivalent circuit (just need my values checked!)

Homework Statement


find the thevenin equivalent circuit (using open/short circuit method is how i did it):
5lq58j.png

Homework Equations


n/a

The Attempt at a Solution


I just need to verify that my V(th)=V(oc) value is right:

Vc/4 + Vc/6 + (Vc-2)/3 - 4 = 0
Vc = Voc=Vth=6.22Vi am pretty sure i did it right..however, when i build my circuit in the software my class uses (multisim) i get a different voltage and current reading when i correct it like the picture above (with point B above the 2V source). i have to connect to point B is BELOW the 2V source, like this, to get the rite reading:
2gt28eg.png


i know why this is. when i connect it to point B above the 2V source, i get a reading of 4.22V with the software...but i built the circuit like it looks in the diagram. and when i did it below the voltage source, i just add 2V to that value. so my actual question is, is the thevenin voltage 4.22V or 6.22V? and did i set up my circuit in the software wrong somehow?
 
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If everything behind the a and b terminals was hidden inside a black box, what would you determine the open circuit voltage to be with your meter?
 


huh?? so you're saying i should only measure open circuit voltage where a and b are in the diagram? i got 4.22V when i did that, but it disagrees with my node voltage calculation.
 


asdf12312 said:
huh?? so you're saying i should only measure open circuit voltage where a and b are in the diagram? i got 4.22V when i did that, but it disagrees with my node voltage calculation.

Your node potential was calculated with respect to the reference node, which is NOT not node b :wink:
 


sorry but i don't understand. the only node i labeled was Vc. are you saying i did my node voltage thing wrong? that i should only look at what is between terminals a and b and ignore the 2V source entirely?
 
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asdf12312 said:
huh?? I am sorry i don't understand. are you saying i did my node voltage analysis wrong? that i should only look at what is between terminals a and b and ignore the 2V source entirely?

No, you did your node potential calculations fine. You chose a reference node and found the potential of the node Vc with respect to that reference node. But the voltage you're interested in is the potential difference between terminals a and b.

For that you should find the potential at terminal b with respect to the reference node and take the difference: Vc - Vb, as you found out by subtracting the 2V potential at node b from Vc.
 


ah i see..so working my way from the ground node to the right path, I find there's a 2V increase when it reaches terminal b, because of the voltage source. so I take the voltage at a and subtract the voltage at b:

6.22V-2V = 4.22V.

is that right?
 


asdf12312 said:
ah i see..so working my way from the ground node to the right path, I find there's a 2V increase when it reaches terminal b, because of the voltage source. so I take the voltage at a and subtract the voltage at b:

6.22V-2V = 4.22V.

is that right?

Yes, that's right :smile:
 

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