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Basic Series-Parallel circuit voltage problem

  1. Oct 31, 2012 #1
    1. The problem statement, all variables and given/known data
    Introductory Circuit Analysis 12th edt Boylestad, page 255, Example 7.8

    A3pTd.jpg

    This is not homework, I am just reading the text book. It says:

    "It is now obvious that:

    V2 = -E1 = -6V"

    Is it obvious? :S

    3. The attempt at a solution

    I don't really understand that, so I created the circuit in multisim and hooked up three voltmeters, and they all read -6. I understand XMM1 reading -6, and XMM2 is kinda measuring the same points, I can understand that.

    NvqJy.jpg

    But XMM3? -6? Seems.. strange

    If I visualize XMM3 measuring everything to the LEFT, the voltage is -6. Alright, but if I visualize it measuring everything from the right, surely it should be relevant that there is an 18V source there.

    Why is there no contribution in these three measurements from V2, the 18V source?
     
  2. jcsd
  3. Oct 31, 2012 #2

    lewando

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    Gold Member

    It does not help that there are two "V2" references. Also no reference to "E1". I think all they are saying is that since the V2 polarity indicators are opposite those of V1, the [edit-- measured voltage] is opposite.
     
    Last edited: Oct 31, 2012
  4. Oct 31, 2012 #3

    gneill

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    Staff: Mentor

    They all measure the same because they are all measuring the same thing; they are all connected to the same nodes in the circuit. The length of a wire, or its twists and turns, or branches or angles, mean nothing in a circuit diagram. The only thing that matters is what is connected to what. There is no (electrical) meaning to the "to the right" or "to the left" along a single wire: it is all the same node at the same potential.

    Meters connected across the 6V source can ONLY measure 6V; an ideal source will produce any amount of current required to maintain its specified potential. The 18V source has a resistor in series with it. The current through it will produce the required potential drop.
     
  5. Oct 31, 2012 #4
    Ah man, I'm sorry about that, the names in my drawing do not match the names in the book, my bad.

    Disregard the book's statement though, my concern is still wrapping my head around why it's -6V everywhere.
     
  6. Oct 31, 2012 #5

    lewando

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    Well it is not -6V everywhere. It is not -6V between R1 and R2 and neither between R4 and the 18V source. As gneill indicated it is only -6V at the "top" node with respect to the "bottom" node.
     
  7. Oct 31, 2012 #6
    I realize it is not -6V everywhere, I meant the three measurements I had taken.

    I also realize, and I try to remind myself, that what I am measuring with the two points of the voltmeter is the difference in voltage in those two points. And all three measurements I made are from the top to the bottom node, I understand that.

    But I still don't understand why V2 doesn't affect the voltages for XMM2. To me, that means all 18 volts from V2 must drop over R4. But I get -24 volts when I measure across R4.

    Of course, I could use mesh analysis, because I can use that method well, but I kinda need to understand this....
     
  8. Oct 31, 2012 #7
    There is an ideal source V1 that forces the potential measured by XMM2 to be -6. Ideal wires do not have resistance so all along the wire is the same voltage. All along the top of V1, into XMM1, XMM2, XMM3 are at the same voltage.

    If we did have a (real) wire with resistance, we would model it with a resistor.

    V2 is forcing one side of R4 to be +18 with respect to ground (the bottom of your circuit has a ground symbol on it -- this is purely convenience as it doesn't matter what the absolute potential there is. We choose 0 to make it simpler). V1 is forcing the other side of R4 to be -6 with respect to ground. The voltage across R4 is 18-(-6) = 24 volts. The sign you got depends on how you connect your voltmeter across R4.

    The voltage sources are shoving electrons into the circuit to make the voltages across them to be as advertised. This is changing the currents through the resistors.
     
    Last edited: Oct 31, 2012
  9. Oct 31, 2012 #8

    lewando

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    Right, so 18 + (-24) = -6. "all 18V from V2 must drop over R4" is not right.
     
  10. Oct 31, 2012 #9
    Just so you guys don't think I am just skimming over what you write, I've read all your posts several times and spent some time trying to digest it, still not quite there.

    It's weird to me that V2 is only relevant for the voltage over R4. If V1 is relevant for the voltage over R3, why isn't V2?
     
  11. Oct 31, 2012 #10

    gneill

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    Staff: Mentor

    V1 is an ideal source; It CANNOT change its potential, and any points its connected between MUST have that potential difference. V1 will absorb or produce ANY AMOUNT of current in order to maintain this fixed potential difference across its terminals. The same goes for V2, but since it has a resistance in series with it, any required potential drop to match V1's fixed potential will occur across it. The current WILL be such that the sum of V2 and the drop across R4 WILL be the potential presented by V1.
     
  12. Nov 1, 2012 #11
    Just to add:

    If you were to attach an ideal source V1 directly across an ideal source V2 of different voltage, there would be fire and explosions. This is because you have two unstoppable forces willing to supply any amount of current to get the potential between the same two points to be different. Placing a resistor between them makes the problem go away because the potential can be different across the resistor. Establishing that potential difference across the resistor requires current to flow and both ideal voltages must be able to sink or source that current (one side sources it and one side sinks it). In your circuit there is a resistor between V1 and V2. The result is current flows but the potentials at each end of the resistor are maintained at their places by the ideal sources.
     
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