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3 phase system

  1. Oct 10, 2014 #1
    vbn.png

    Why is Vab = Van -Vbn? I'm doing this after a while so maybe I'm getting something basic wrong but surely the potential difference between lines A and B is the voltage drop across phase a (Van) plus (rather than minus) the voltage drop across phase b (Vbn)?? So Vab = Van + Vbn?

    Thanks
     
  2. jcsd
  3. Oct 10, 2014 #2
    Notice orientations of '+' poles of source voltages Van and Vbn as you go through the loop a-n-b-a
     
  4. Oct 11, 2014 #3
    I thought it had something to do with the signs but I've kind of forgotten. Could you elaborate please?

    Cheers
     
  5. Oct 11, 2014 #4
    Ok let me illustrate this with DC circuits:
    Untitledcbe56.png
    Can you see the difference between case I and II ?
     
  6. Oct 11, 2014 #5

    NascentOxygen

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    Staff: Mentor

    In a simple DC circuit, if one point is at +6 volts, and another point is at +6 volts, what is the potential difference between the two? Is it 12 volts, or is it zero?

    If one point is at +6 volts and another is at -6 volts, what is the potential difference?

    Do you add the readings, or do you subtract them?
     
  7. Oct 13, 2014 #6
    Case I: Going from a to b is Van + Vbn ?
    Case II: Going from a to b is Van - Vbn?
     
  8. Oct 13, 2014 #7
    The first is 0V
    The second is 12V?

    How does that apply to this problem?
     
  9. Oct 13, 2014 #8
    :)
     
  10. Oct 13, 2014 #9

    NascentOxygen

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    Staff: Mentor

    There is no difference. In either case the P.D. is given by the same equation.

    Van is voltage at a referenced to n. Vbn is voltage at b referenced to n, that convention is clear.
    So the P.D. is always Vab = Van - Vbn, and consistent with your output arrow direction.

    Did you intend something different?
     
  11. Oct 13, 2014 #10
    Arrow denotes voltage Vab (missing in the drawings). Vbn changes polarity so the outputs Vab aren't same in both circuits
     
    Last edited: Oct 13, 2014
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