Can we use superposition for Thevenin equivalent with multiple sources?

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Dethrocutionx
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Hi guys,

I have a question that is nagging me quite a lot. Let us say that we have a circuit who has two independent sources (any combination of voltage and current sources). In the determination of Vth, can we choose whether or not to use superposition to find this Vth? Because my understanding was that you could use superposition to simplify the analysis or choose to leave both the sources in and find the equivalent voltage. Am I correct?

Thanks guys :).
 
Yes you can use superposition to find this Vth voltage.
 
Thanks! Appreciate it. I have another question related to the determination of Vth. I have an example I am doing where the A and B terminals are in the middle. What does one do to determine this Vth then? Because when the terminals are on the left or the right, it makes sense how we are taking a look at this circuit's inner workings. But the confusion is there when it is in between.

Here's a picture of the problem:
https://www.dropbox.com/s/e3tfnt05unozupw/20121207_164124(0).jpg
 
Well In your diagram A,B terminals are not in the middle.
Because we always can redraw it to this form

Hand-drawn circuit schematic: a 1 kΩ resistor in series with a 1 mA current source labelled V, with terminals A and B at the right, a 1 kΩ resistor on the right branch, and VAB marked across the terminals.


And Vth = - 1mA*1KΩ = -1V and Rth = 1KΩ
 
Attachments
  • Handwritten circuit schematic: a 1 kΩ resistor in series with a 1 mA current source, labelled V on the left, with a 1 kΩ resistor on the right between terminals A and B, and VAB marked.
    bez tytułu.JPG
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Dethrocutionx said:
https://www.dropbox.com/s/ctix99089gdmzqm/20121207_171919.jpg?m

I somewhat understand how we can get Vth to be 6V using the left side of the circuit. Why can't we use the 2mA and the 2kΩ resistor to determine the same Vth?
The answer is that the second Kirchhoff's law doesn't allow us to do so.
Why? I hope that this diagram explains everything

Two circuit schematics with a down arrow between them: the top shows a 4K resistor, 3K resistor, 2mA current source and 2K resistor with voltages V1, V2, V3 and Vth marked; the bottom shows the redrawn circuit with the 3K, 2K and 4K resistors and the 2mA source.


As you can see Vth voltage is equal to

Vht = V2 + V3 and also Vth = V1 (In a parallel circuit, the voltage across each of the components is the same)

therefore

Vth = 2mA * 3KΩ = 6V
 
Attachments
  • Two circuit schematics with a down arrow between them: top shows a 4K resistor, 3K resistor, 2mA current source and 2K resistor with Vth, V1, V2 and V3 marked; bottom redraws it with the 2mA source in series with 2K and 4K resistors and 3K resistor.
    1.PNG
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That makes perfect sense! Thank you so much :). I truly appreciate it