Circuits Question: Thevenin Equivalent

In summary, Thevenin's equivalent circuit for this circuit with respect to a and b can be found by applying mesh currents and using Ohm's law to solve for the open circuit voltage (Vth). To find the short circuit current (Isc), the 10 amp source (Is) can be removed as it does not affect the circuit to the left of Vs2. Using a closed surface can also be a method to solve for Isc, but it is not necessary.
  • #1
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Homework Statement



Find thevenin equivelent of this circuit with respect to a and b

part0.jpg


Homework Equations


Mesh Currents

The Attempt at a Solution


Here I will try to find V(open circuit) = Vth = Vab ((I think this anyways))

I apply mesh currents (any faster methods are welcome if you know them)
part1.jpg


iA: -100/V/ + 10/ohm/(iA) + 40/ohm/(iA-iB)=0
iB: 40/ohm/(iB-iA) + 50/ohm/(iB-iC)=0
iC: 50/ohm/(iC-iB) + 20/ohm/(iC) + 200/ohm/(Ix) = 0
iD:-10/A/
iX: iX=iB-Ic

I get the folowing values using solve function on calculator:
iX=-40A/
iA=210/A/
iB=260/A/
iC=300/A/
iD=-10/A/

And using ohms law with parallel resistors having equal voltages..
Vab = 40/ohm/*(iA-iB) = -2000 = Voc = Vth

Now is where I begin to have trouble finding Isc...


I draw the short circuit between a and b (which I believe causes R3 and R4 to have no current through them thus making Ix = 0 and vs2 = 0 as well):

part2.jpg


Simplifying it you gain ((if my previous assumptions are correct reguarding the short circuit)):

part3.jpg



Now from here I am actually unsure what I should do. I do want to do mesh currents but I don't know if I will receive a correct answer. I also wondered if I could use a closed surface... something like this perhaps:

part4.jpg


If I used that I would get an equation like -100/V/ / 10/Ohms/ + Isc = 0

But then again it just doesn't seem quiet right to me. And i have no way to check my answer.
 
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  • #2
You could get rid of the 10 amp source (Is) right from the beginning; it does nothing to the circuit to the left of Vs2. It just adds 10 amps to the current in Vs2, which has no effect on Vs2's voltage.

So, Is contributes nothing to the current in the short you've placed across a-b because Vs2 absorbs the entire 10 amps from Is, and the current in the a-b short is just 10 amps.
 

1. What is the Thevenin equivalent circuit?

The Thevenin equivalent circuit is a simplified representation of a complex circuit that contains a voltage source and a single resistor. It is used to simplify circuit analysis and determine the behavior of a complex circuit at a particular point.

2. How is the Thevenin equivalent circuit calculated?

The Thevenin equivalent circuit is calculated by finding the open circuit voltage at the load terminals and the equivalent resistance of the circuit when all sources are replaced by their internal resistances. This can be done using various techniques such as Kirchhoff's laws and Norton's theorem.

3. What is the purpose of finding the Thevenin equivalent circuit?

The Thevenin equivalent circuit is used to simplify circuit analysis and determine the behavior of a complex circuit at a particular point. It is also useful for designing circuits and predicting their performance in different conditions.

4. How is the Thevenin equivalent circuit used in practical applications?

The Thevenin equivalent circuit is used in many practical applications, such as in electronic devices and power systems. It is used to determine the behavior of a circuit under different operating conditions and to design circuits that meet specific requirements.

5. What are the limitations of the Thevenin equivalent circuit?

The Thevenin equivalent circuit is only applicable to linear circuits, meaning that the behavior of the circuit must follow Ohm's law. It also assumes that the circuit is in a steady-state condition, and does not account for the effects of non-linear components or changing conditions. Additionally, it is only accurate for a single point in the circuit and may not accurately represent the behavior of the circuit as a whole.

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