CKT impedance

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arnettda

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I want to calculate the total impedance for a wire run. Is there a simple way to do this? If I use ohms per ft x 2 would that be close?
 
that's how I'd do it. check the tables for conductor properties, or go online. haven't done this in a while. Table 8 gives reisistance for DC and I'm not sure how AC impedance would factor into this, and Table 9 gives AC effective Z for three phase. Neither of these gives the actual impedance value per 1000 feet that you need.
 
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take a look at this paper:

Ametani, A., “A General Formulation of Impedance and Admittance of Cables,” IEEE
Transactions on Power Apparatus and Systems, vol. PAS-99, no. 3, pp. 902–10,
May/June 1980.
 
CKT impedance

The formula you can use to find the length of a circuit is as follows:
D = Cmil X VD divided by 2 X K X I
Then:
D = Is the distance from the load to the power supply or panel.
Cmil = The circular mil area of the circuit conductor as listed in the NEC Chapter 9, Table 8.
VD = The voltage drop of the circuit expressed in volts.
K = The constant value, which is 12.9 homs for copper and 21.2 ohms for aluminum.
I = The load in amperes at 100%.
 
I want to calculate the total impedance for a wire run. Is there a simple way to do this? If I use ohms per ft x 2 would that be close?
Probably close for most small conductors.
In UK the standard is BS7671.
It gives voltage drop as mV/A/m
m being metres not miles...;)
For larger conductors it gives r, x, and z so you can calculate voltage drop for either ac or dc.
 
I want to calculate the total impedance for a wire run. Is there a simple way to do this? If I use ohms per ft x 2 would that be close?
How accurate an impedance value do you need?

Other parameters influence any ohms/ft value. Do a forum search on the term "voltage drop formula". All the formulas that you see posted are based on ohm's law: E=IR (or E=IZ). Some you have to separate the variables and constants for the R (or Z) value. For example, in the formula cripple posted...
E=VD
I=I​
So rearramge the equation to...
E=2?K?I?D/cmil​
...thus...
R=2?K?D/cmil​
...for the total length of wire in the circuit (i.e. both out and back).
 
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The impedance of the circuit includes everything that limits current flow through that circuit. This includes the connected load, and the connected load should dominate.

It is often really useful to figure out the impedance of everything _except_ the load, because this helps determine things like voltage drop at the load. This includes the impedance of the source (transformer, battery, or generator), the impedance of any distribution equipment, and the impedance of the circuit conductors.

Most times you can ignore the impedance of the source and the inductance of the wires, and just look at resistance. Under what circumstances are you trying to figure out circuit impedance?

Resistance can be calculated many ways, including ohms per foot times total circuit length (distance out plus distance back), or resistivity * {length} / {conductor cross sectional area}. cripple's formula comes from the latter; the constant 12.9 is the resistance of a wire 1 circular mil in cross section and 1 foot in length. Note that this 'constant' actually changes with temperature, quite significantly. If you are doing voltage drop calculations for long outside circuits, the change of resistance versus temperature can make a difference of a couple of wire sizes!

-Jon
 
I'm not sure why you want the impedence, but one suggestion would be to use one of the many voltage drop calculators to find the voltage drop.

Then divide the voltage drop by the current to find the impedence. That way, you have a more accurate AC impedence. Some of the calcualtors even let you included the effects of the conduit type, and circuit power factor.

Don't know what current to put into the voltage drop calculator? Just assume something reasonable, like 10 amps for #12 wire. You are going to divide your vdrop by that current, so it will cancel out.
 
I want to calculate the total impedance for a wire run. Is there a simple way to do this? If I use ohms per ft x 2 would that be close?

Arne,

I calculated the differences D.C. and 60 cycle AC "impedance" once,
resulting from a 'discussion' with a master electrician.

The difference between D.C. and 60 cycle AC "impedance"
was in the fractions of a percent.

If you are measuring the length of the cable in milli-meters,
then you might need to calculate the "impedance"!

:)
 
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