# 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

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

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.

$er = 1
c = 3*10^8
Vp = \frac{c}{sqrt(er)} = 3*10^8$

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.