Power Factor with capacitor

Discussion in 'Homework Help' started by zorro_phu, Jul 11, 2018.

  1. zorro_phu

    Thread Starter New Member

    Jun 14, 2018
    17
    0
    In the question below part b, i'm not sure how am i suppose to sovle this. Is there formula or sth i can use to find the value for capacitor.
    Screenshot_20180711-154321_1.jpg
    Below is all i could do, i'm stuck. Any help is appreciated.
    Capture.PNG
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    What if you were given an impedance for the capacitor, say -j5 Ω, and asked to calculated the power factor for the total load. What would it be? From there, can you see how to turn it around and find the impedance needed to result in unity power factor?
     
  3. drc_567

    Senior Member

    Dec 29, 2008
    339
    38
    Just curious ... looking at the circuit as 230v being placed across the total load of 20+j4, excluding the capacitor ... is that interpretation correct?
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Yes, but only coincidentally.

    Write the two voltages, as phasors, explicitly. So write Van, Vbn.

    Now use the fact that Vab = Van - Vnb = Van + Vbn

    But even then you need to recognize that while Vab is the voltage across the total load, you can't just throw that voltage and that load at Ohm's Law because the two individual loads are NOT in series -- the current that flows in one does not have to be the same as the current that flows in the other because of the neutral connection.

    However, in this case, symmetry argues that the current in the neutral is zero. But it is critically important that you understand that if the top load had been different from the bottom load, that you could not just add them together. It's also important that you understand that you are only able to take twice the individual voltage because the two line voltages are 180° apart in phase.
     
  5. drc_567

    Senior Member

    Dec 29, 2008
    339
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    ... no neutral current if the phase impedances are equal and well balanced.

    One remaining question is about the compensation necessary to improve the power factor. Does it make more sense to put a compensating reactance in series with each phase? The problem definition shows that a single capacitive reactance is placed in parallel with the two series inductive impedances.
     
  6. WBahn

    Moderator

    Mar 31, 2012
    22,644
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    The problem specifies that the source is balanced. It also specifies where the capacitor goes for this problem. The TS has to work the problem they are given.
     
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