Should Vout be negative or positive?

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The discussion centers on the determination of V_out in a circuit using Kirchhoff's voltage law. The calculations show that the output voltage V_out is -8V, indicating a negative value. Participants agree that the potential difference between the voltage sources leads to this negative output. Observations about the polarity of the voltage sources and the direction of current flow confirm that the bottom has a higher potential than the top. Overall, the conclusion is that V_out should indeed be negative based on the circuit analysis.
Clandry
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I attached the circuit.

I did kirkoff's voltage law (assuming current goes clockwise):
-15+5-10,000I-40000I=0 where I=current
I=-2*10^-4 A

V_out=40,000*-2*10^-4A = -8V
The way V_out is shown, it should solve to be negative, correct?
 

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Clandry said:
The way V_out is shown, it should solve to be negative, correct?

Correct.
 
V is the difference betwen potentials at two places in the circuit.
If you look at the voltage sources, the top is less positive than the bottom - so the bottom must have a higher potential than the top. Replace the voltage sources with an equivalent single voltage source and the answer to your question will become clear.

Note: when you do kirkoffs laws to analyze a circuit - draw voltage arrows on the components as well as current arrows at the nodes.
 
gneill said:
Correct.
Did you solve the problem?
I'm asking because if you didn't, can you tell just by looking it that it will be negative?
 
Clandry said:
Did you solve the problem?
I'm asking because if you didn't, can you tell just by looking it that it will be negative?

Yes, I can tell just by looking; But I've looked at a LOT of circuit problems over the years!

I took note of the two voltage supplies and determined that the top of them will be negative w.r.t. the bottom. I then looked at the polarity specified for "measuring" the output and drew the conclusion that Vout must be negative.
 

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