Nominal II Equivlent Circuit

Discussion in 'Homework Help' started by Kayne, Apr 8, 2011.

  1. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    Hi All,

    The variables for the question are as followed.

    Length of power transmission line = 1000m
    Frequency = 50 Hz
    To be represented as symmetrical 'nominal II' equivlent circuit
    Series impedance Z = resistance of 12 ohms in series with inducatance = 1.2mH, and the parallel branches at each end consists of 3.3nF.

    Z = 12+j1.2*10^-3

    A) Calculate the four distributed parameters R,L,G,C

     R = \frac{12}{1000} = 0.012 ohm m^-1; <br />
L = \frac{1.2*10^-3}{1000}= 1.2 uHm^-1 <br />
C = \frac{2*3.3*10^-9}{1000}= 6.6nFm^-1

    G = I have assumed that is 0 becuase of the line distance and that it is in the Air... - Am I correct in assuming this or is there a way to calculate this?

    B) The approximate frequency beyond which this representation of the line would not be valid.

     er = 1 <br />
c = 3*10^8<br />
Vp = \frac{c}{sqrt(er)} = 3*10^8
    So the frequencys whos wavelength matches the lenght of cable is

     fo= \frac{Vp}{\lambda} = \frac{3*10^8}{100} = 300kHz

    Cable length must be less then 1/20th of wavelength therefore
     f1=\frac{300*10^3}{20}= 300Hz

    There is more to the question which I have done but wanted to check to see if I was on the right track before adding.

    Thanks for your help
     
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