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Parallel Conductor Impedances

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kashew

Electrical
Jul 16, 2004
10
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 inductance value of 2 parallel cables is the inductance 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 inductance value being divided by 2 for a 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 inductance value for paralleled cables?
 
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Check your math, please.

For two parallel conductors of impedance (2+j2), I get a resultant of (1+j1), which matches the intuitive value of exactly half.

P.S. The last equation of Fig 21 is solved such that you don't have to manipulate any j^2 terms.
 
tinfoil is right

The correct impedance value is 1+j1

Parallel impedance is always calculated as the product over the sum of the two impedance values.

Z = (Z1*Z2)/(Z1+Z2)





 
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