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3 phase circuit

  1. Mar 19, 2009 #1
    https://www.physicsforums.com/attachment [Broken]....

    how do i measure the current flow threw the neutral wire if phase R is open circuited? the impedance per load is 3+7j
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 19, 2009 #2
    Your link gives a 404 error...
     
  4. Mar 19, 2009 #3
  5. Mar 19, 2009 #4
    Do you need to physically measure it by using an ammeter or calculate it? And you'll have to show your attempt before we can give you a push in the right direction.
     
    Last edited: Mar 19, 2009
  6. Mar 19, 2009 #5
    both. i attached an ammeter to the neutral wire and i put an access wire from the input of the ac power suplly to the out R and i have got a reading, but i need to hand calculate it.
     
    Last edited: Mar 19, 2009
  7. Mar 19, 2009 #6
    Okay, I've never heard of access wire (I'm non-english) but I assume it open circuits the R phase, as you described in your first post.

    But this calculation should be fairly simple to do with only the basic knowledge of AC circuits and KCL. Just think what happens to the current if a circuit is opened and what happens when currents meet in a node. I assume you have had theory on 3-phase and you know the phases of the power supplies in such a system.
     
  8. Mar 19, 2009 #7
    as far as i am aware it is now an unbalanced load meaning the current of R, Y and B is Ir+Ib+Ic???and im sure the impedance 3+7j is needed to work it out to.
     
  9. Mar 19, 2009 #8
    You're right, but if the R-phase is opened, is there a current trough that wire? And you can substitute the one voltage supply with three different ones in each loop with the same voltage, but different phase since they are in star formation. Can you tell me what the phases are?

    If you are having trouble figuring the phases out, here's a good reading to find them out.
     
    Last edited: Mar 19, 2009
  10. Mar 19, 2009 #9
    do i need to know the phases to work out the current in the nuetral wire?
     
  11. Mar 19, 2009 #10
    Yes you do. The phases of the voltage supplies affect the phases of the currents. The phases of the voltage are standardized, though.
     
  12. Mar 19, 2009 #11
    sorry jus realised the impedance is 7+3j. i have worked out the phases to be Z1 is -23.2, Z2 is -143.2 and Z3 is 263.2
     
  13. Mar 19, 2009 #12
    Do you mean the phases of the currents? And is the impedance 3+j7 or 7+3j, you have given two different values?
     
    Last edited: Mar 19, 2009
  14. Mar 19, 2009 #13
    yes and it is 7+j3
     
  15. Mar 19, 2009 #14
    Right. The phase of I2 is correct, but I guess there's a typo in I3. Shouldn't it be -263.2? And the current for I1 would be right if the wire wasn't open circuited. I have given you hints earlier, of what the value of I1 should be.

    Once you find out the I1, you are all set to calculate as you have said, I1+I2+I3. You probably have the magnitude as well, if you have determined the phases.
     
  16. Mar 19, 2009 #15
    yes it is a typo it is -263.2. I only need the magnitude do i not? just to find the amps in the neutral wire though and the magnitude is the same for each phase i get it to be 7.62A?
     
  17. Mar 19, 2009 #16
    That could be, but usually when they ask for current they want the phase as well. You are right, though, the magnitude is the same for each phase and its the same for the current in the neutral wire. But I get about double the magnitude you get. How did you calculate it.
     
  18. Mar 19, 2009 #17
    squareroot(14^2 + 6^2) = 15.23
     
  19. Mar 20, 2009 #18
    Um, now you have calculated the magnitude of the impedance and doubled it. To find out the current, you have to use Ohm's law.
     
  20. Mar 20, 2009 #19
    so its 115/squareroot(7^2+3^2) = 15.1? but that is the same as the line current :S:S
     
  21. Mar 20, 2009 #20
    In this case their magnitude are the same, since only two wires are carrying a current.
     
    Last edited: Mar 20, 2009
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