Is My Solution Correct for Finding v0 Using Superposition Method?

AI Thread Summary
The discussion revolves around verifying the correctness of a solution for finding v0 using the superposition method. The initial approach involved applying Kirchhoff's Current Law (KCL) at a node with three branches, but the user received a significantly different result when compared to a circuit-solving program. Key mistakes identified include incorrect sign usage for the voltage source and an error in calculating the total resistance. Additionally, it was clarified that the current source should be replaced with an open circuit when analyzing the voltage source. The conversation emphasizes the importance of accurate circuit analysis techniques in solving for v0.
noppawit
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I'm solving by using superposition, I would like to ask that my solution is correct or not.

By the way, the question is "Find v0"

http://www.wisheyebio.com/uploads/Picture1.png
 
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I don't know about superposition, but if you use KCL at the top node, you have three branches:

The left branch has the current source and two resistors.

The right branch has the voltage source and two resistors.

The middle branch has the remaining resistor.

KCL says that:

current going into node = current coming out of node

which, in this case means:

current in left branch + current in right branch = current in middle branch.

I called the current in the right branch i0, which makes v0 = i0R. I get a significantly different answer (-1.00 V), which I confirmed using a circuit-solving program.
 
noppawit said:
I'm solving by using superposition, I would like to ask that my solution is correct or not.

By the way, the question is "Find v0"

http://www.wisheyebio.com/uploads/Picture1.png
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You made two mistakes in the second equation. Since i_x is entering the plus sign of the voltage source, it should enter the equation as +9V, instead of -9V.
Also, R + R + \frac{2}{3}R = \frac{8}{3}R, not \frac{32}{3}R
 
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IIRC, when treating the 9V source you would replace the current source with an open circuit, not a short. So the 2/3 R was incorrect to begin with.
 
Redbelly98 said:
IIRC, when treating the 9V source you would replace the current source with an open circuit, not a short. So the 2/3 R was incorrect to begin with.
You are right.
 

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