ac 3 phase

Discussion in 'Homework Help' started by yoamocuy, Oct 27, 2009.

  1. yoamocuy

    Thread Starter Active Member

    Oct 7, 2009
    80
    0
    L=0.1 H
    R=10Ω
    C=150uF

    calculate the rms value of the line current.

    Here's my attempt at a solution.

    L=XL/(2*pi*f)
    0.1=XL/(2*pi*50)
    Xl=31.4

    Xc=1/(2*pi*f*C)
    Xc=1/(2*pi*50*150*10^-6)
    Xc=21.2

    VL=sqrt(3)*VP
    Vp=415/sqrt(3)
    Vp=239.6 V

    total impedance of one line=14.28 at an angle of 45.5°

    IP=239.6/14.28
    IP=16.78A at an angle of -45.5°
    IP=IL

    at ths point I'm not sure if IL is the same as the rms value of the current or if its different.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Your answer looks fine.
     
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