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Should Vout be negative or positive?

  1. Jan 26, 2013 #1
    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?

    Attached Files:

  2. jcsd
  3. Jan 26, 2013 #2


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  4. Jan 26, 2013 #3

    Simon Bridge

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    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.
  5. Jan 26, 2013 #4
    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?
  6. Jan 26, 2013 #5


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    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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