Nominal II Equivlent Circuit

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

  1. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    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