Should Vout be negative or positive?

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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 between 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.