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Finding voltages/currents in a 3-phase circuit

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data
    http://img268.imageshack.us/img268/8981/71713784.png [Broken]
    http://img268.imageshack.us/img268/8981/71713784.png [Broken]

    In a 3-phase circuit, the switch SA is opened, disconnecting the grounding wire. Due to the powerful generator, the voltage between phases remains the same - 400 V. The impedances are given. Find all the missing voltages/currents.


    3. The attempt at a solution

    First of all, as the voltage between the phases remains the same, that means that when the grounding wire was connected, the voltages in the phases were 400/sqrt3=231 V

    The problem Im having here is finding the voltages. If those were given, it would be rather easy to find the rest. I usually tend to solve these by using the vector diagram and solving the given triangles (I dont know is this is the proper way to do it, but so far it has given me right answers).
    Now that we are only given the impedances and the voltage between phases, it leads me to believe that we are supposed to find the currents first, and then find the voltages by multiplying the found currents and impedances. But Im not really sure on how to find them.
    I tried adding the according impedances together: Za+Zb and then Za+Zc. Then find the currents which which go from A>B and A>C (as the voltages between both of them is 400 V.) Then add the A>B and A>C vectors together to find Ia.
    This seems to be incorrect though, giving me answers off by about 0.3-0.5 A.
    So why didnt this way work? How to approach the problem?
    And what exactly causes the voltages to change when the grounding wire is disconnected?

    Thanks in advance
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 3, 2012 #2

    gneill

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    Staff: Mentor

    Redraw the circuit with the phase voltage supplies included. You should know the phase angles and potentials for these sources. The rest should be straightforward circuit analysis.

    attachment.php?attachmentid=52585&stc=1&d=1351948993.gif
     

    Attached Files:

  4. Nov 3, 2012 #3
    But thats what I posted to be the problem. I dont know how to find the voltages. They are certainly not 231 V as they would be if the grounding wire was connected. And the phase angles between the voltages are not 120 degrees either.
     
  5. Nov 3, 2012 #4

    gneill

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    You should assume that the generators supplying the load are ideal so that their voltages and relative phases remain fixed. What will change are the currents (magnitude and phase).
     
  6. Nov 3, 2012 #5
    I dont think I follow what you're trying to say. Are you suggesting the voltmeters Ua, Ub and Uc will all show 231 volts? Or am I understanding you wrong?

    Because they dont. Here's a random example with the results:
    http://img706.imageshack.us/img706/4700/57510807.png [Broken]
    http://img706.imageshack.us/img706/4700/57510807.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  7. Nov 3, 2012 #6

    gneill

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    No, that's not what I'm saying. Note that the voltmeters are NOT measuring the inter-phase voltages. They're measuring the potential between the phase and the 'floating ground' where the branches interconnect on the right hand side.

    I'm saying that the power supplies providing voltage to the branches remain ideal in terms of potential and phase relations.
     
  8. Nov 3, 2012 #7
    I guess you are not saying it without a reason, as it seems like a hint, but I still cant see a way to find these voltages. In fact, I still dont understand why my initial try which I described in the original post wouldnt work. Where's the logic flaw in there?
     
  9. Nov 3, 2012 #8

    gneill

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    When the neutral line from the power supply common ground is severed by opening the switch, the common ground for the loads becomes a "floating" ground; it's no longer pinned to the reference point of the power supply's common ground. That allows the voltage measurements between the phase buses and this floating ground to vary depending upon the given loads.

    You end up with a circuit that consists of two loops with three independent power supplies as I've drawn. These power supplies have specific voltages and phases between them WHICH DO NOT CHANGE NO MATTER WHAT, and you should be able to state what they are. Solve the given circuit for voltages and currents --- you have the voltage source values and the load impedances so all voltages and currents can be calculated.

    A convenient approach might be to take the floating ground as a reference point and use nodal analysis to find the relative potential of the power supply neutral node. Then the individual voltages and currents for the branches can be found.
     
    Last edited: Nov 3, 2012
  10. Nov 3, 2012 #9
    The phase angle between the sources is 120 degrees. The voltage between phases should be 400 V as well, right?
     
  11. Nov 3, 2012 #10

    gneill

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    Yes, the phase angle is 120° between sources. The standard practice is to assign the first phase an angle of 0°, the second an angle of -120°, and the third an angle of -240°.

    I think you worked out the individual source voltage previously, yielding 231V for each source (in particular, the value is ##400V/\sqrt{3}##).

    If you take two of the sources using this voltage and their phase angles and take the difference, you'll find that the magnitude is 400V.
     
  12. Nov 3, 2012 #11
    But I still dont understand where do I derail...
    I first tried to find the current in wire A. That means, I looked at the two loops (A->B and A->C) separatedly. The impedance in AB ought to be ZA+ZB=140-20j, and in AC, ZA+ZB=140-70j.
    As the voltage between AB and AC is 400 V, we could calculate the currents from A to B and A to C separatedly. So for example,
    IAB=400/ZA+ZB
    Then add the current vectors together to get the real current in A. This proved to be wrong, though.
     
  13. Nov 3, 2012 #12

    gneill

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    Sorry, I'm not following your description. You need to write KVL loop equations or KCL node equations to solve for the currents. Because the branch ground is floating there will be interactions between the voltages and currents in the branches, so you must write the circuit equations to sort them out.

    Note also that the 400V between buses will also have associated phase angles! So you can't just assume 400V with 0 phase angle!

    My suggestion is to assume a reference point at the branch grounds (the right-hand side where all the branches come together) and use nodal analysis to solve for VN, the source's neutral voltage with respect to the branch grounds. Then you'll easily be able to write expressions for the voltages and individual currents.
     
  14. Nov 4, 2012 #13
    An gneill says, you cannot ignore the phase relationships especially when you add the currents together. Maybe if you redrew the circuit, sliding the sources Va and Vc into their arms with Za and Zc, and getting rid of the N wire, it will be easier to see.

    In a 3-phase system, if all three loads are exactly balanced (ie all equal) then no current flows through the N wire. This means you can disconnect it without any changes in voltages or currents occurring.

    This problem is showing that if the loads are not balanced, current does flow in the N wire and disconnecting it will cause changes to the voltages and currents in the circuit.
     
  15. Nov 5, 2012 #14
    Now that I dont follow. Maybe Im slow, I dont know, but seem to miss how to find VN that way.
    As I've already said, I usually solve these with triangles, which in their own way, follow Kirchhoff's laws. Havent really solved much of anything which involves phase angles with Kirchhoff's laws directly. So my initial plan was to find VN after everything else had already been found.

    And I didnt actually assume the phase angle between the 400 V's to be 0. I took it as 60 degrees (solving with the triangles), but it gave me a worng outcome.
    Im pretty sure Im missing some important phase angle here. It would just be easier if one of you could point it out, or name the actual phase angles between those voltages so I could compare them with my results.
     
  16. Nov 5, 2012 #15

    gneill

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    The only fixed phases angles will be those of the individual power supplies feeding the branches. Because the neutral line has been severed, the phases and voltages between the branches are going to sort themselves out by mutual interactions of the currents and potentials. That's why you need to apply Kirchhoff's laws and analyze the circuit.
    As stated above, removing the neutral tie point from the common junction of the voltage generators is going to change the current paths and phases angles in the branches.

    Write the circuit equations and solve for the currents and potentials.

    I mentioned one approach (solving for VN using nodal analysis) which I think is practical, but you could also write and solve loop equations.
     
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