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

In summary, the problem involves finding the Thevenin equivalent circuit using the open/short circuit method. The goal is to verify the V(th)=V(oc) value, which was found to be 6.22V. However, when building the circuit in the software, a different voltage and current reading was obtained. It was determined that the Thevenin voltage is 4.22V and the circuit was set up incorrectly in the software. The correct method is to measure the potential at terminal b with respect to the reference node and subtract this value from the potential at terminal a, resulting in a Thevenin voltage of 4.22V.
  • #1
asdf12312
199
1
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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  • #2


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?
 
  • #3


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.
 
  • #4


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:
 
  • #5


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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  • #6


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.
 
  • #7


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?
 
  • #8


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:
 

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

1. What is a Thevenin equivalent circuit?

A Thevenin equivalent circuit is a simplified representation of a complex circuit that contains a voltage source and a series resistance. It is used to model the behavior of a circuit for analysis and design purposes.

2. How do I find the Thevenin equivalent circuit?

To find the Thevenin equivalent circuit, you first need to identify the load resistance and remove it from the original circuit. Then, you calculate the open-circuit voltage by measuring the voltage across the load terminals. Finally, you calculate the Thevenin resistance by short-circuiting the voltage source and measuring the resistance across the load terminals.

3. What are the benefits of using a Thevenin equivalent circuit?

Thevenin equivalent circuits simplify complex circuits, making them easier to analyze and design. They also help in determining the maximum power that can be delivered to a load and the effects of changing load resistance on the circuit.

4. Can I use a Thevenin equivalent circuit for any circuit?

Thevenin equivalent circuits are suitable for linear circuits, which means that the relationship between voltage and current is proportional. They are not applicable for non-linear circuits, such as those containing diodes or transistors.

5. How accurate is a Thevenin equivalent circuit?

The accuracy of a Thevenin equivalent circuit depends on the complexity of the original circuit. In general, it provides a good approximation of the behavior of the circuit for most practical applications. However, for highly complex circuits, the Thevenin equivalent circuit may not accurately represent the behavior of the original circuit.

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