Parallel cable impedance

[kashew]

In the IEEE Buff Book (ANSI/IEEE std 242-1986) more specifically page 94, figure 21 shows the basic formula for calculating parallel impedance.

Z= (R1+jX1)*(R2+jX2)/(R1+jX1)+(R2+jX2)

If you substitute some simple whole numbers for the values such as R=2 and X=2 (for ease of calculation) the resulting impedance is

Z= 1/2 +j1

This illustrates that the resistance value is divided by the number of conductors (in this case 2) and the reactance value of 2 parallel cables is the reactance value of the one cable.

In the Buff Book on the next page 96, however, there appears a contradiction in the example where it shows the reactance value being divided by 2 for paralleled cable.

I must be missing some basic understanding here. Can anyone explain the difference?

The basic question is: What do you do with the reactance component for paralleled cables?

The calculation is being used for fault current and voltage drop.

 

[jwerthman]

Do your math again. I used R=2 and X=2 and came up with the parallel combination of Z=1 +j1 which is correct. Be careful of all the j^2 terms in the equation - don't forget that j^2 is -1.