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PSC, current and Node Voltage Analysis

  1. Oct 19, 2009 #1
    First off, I apologize if this belongs under homework help. It’s kind of a grey area I suppose...

    Anyway, I just finished my linear circuits midterm. One of the questions on the midterm had a circuit we needed to solve using node analysis. That’s not the question, though. I’ve crudely redrawn the circuit:
    Circuit.png
    I forgot to draw on the common node, but it’s the bottom one.

    Now I thought, due to the passive sign convention, that ib would be equal to -v2, since ib is in the direction of a voltage rise (v=-bR == v2=-ib). The thing is though, ib=-v2 gives really ugly fractional results and ib=v2 gives really nice integer results. Also, the one person I’ve talked to so far said that ib=v2.

    I may have gotten one or two of the resistors wrong redrawing it, but I know for a fact that the v2 to ground resistor was 1Ω.

    So anyway, what is correct? (and if you could point out where I went wrong, that’d be great too =P)Usually the only things I get wrong are mathematical errors, but when it’s something fundamental like this I like to take a step back and make sure I have my concepts right. Thanks in advance.
     
  2. jcsd
  3. Oct 19, 2009 #2

    berkeman

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    Staff: Mentor

    As drawn, -iB * 1 Ohm = V2.
     
  4. Oct 19, 2009 #3
    Yeah, that’s what I thought. (obviously)

    I guess we’ll see what happens when I get my midterm back...
     
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