SYMMETRICAL THREE PHASE FAULT- NUMERICAL PROBLEM /KTU/ POWER SYSTEM ANALYSIS

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  • Опубликовано: 15 дек 2024

Комментарии • 9

  • @surendrakverma555
    @surendrakverma555 2 года назад

    Very good. Thanks 🙏🙏🙏🙏🙏

  • @yogeshgupta4577
    @yogeshgupta4577 3 года назад

    Hello Sir, please consider a 3 phase fault problem with fault impedance too. That will also help many students.

  • @powerful6815
    @powerful6815 3 года назад +3

    Sir, in your previous problem, 10% =.1pu only . But hear 10%=.1j pu .. Why here 'j' comes??

    • @praveenraj2370
      @praveenraj2370  3 года назад +4

      When ever you write reactance, mark it like j0.1 pu
      Because inductive reactance is jXL
      Even in the previous problem, you can use j0.1pu

  • @smithakv322
    @smithakv322 Год назад

    Sir IL also per unitil akkande

  • @spzajay2640
    @spzajay2640 Год назад +1

    20MVA=20000KVA right

  • @prashanths1488
    @prashanths1488 2 года назад +1

    If"=2.802+5.408 = 8.21 than varuthu but nenga 8.081 nu write panringa antha angle ah add panna -90 varaathu sir please replay me

  • @prashanths1488
    @prashanths1488 2 года назад

    Sir, Eg" value calculation wrong aha varuthu so calculation num sollikodunga