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# Re: Current Limiting and Impedence

• To: tesla@xxxxxxxxxx
• Subject: Re: Current Limiting and Impedence
• From: "Tesla list" <tesla@xxxxxxxxxx>
• Date: Tue, 26 Apr 2005 17:10:09 -0600
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• Delivered-to: tesla@pupman.com
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• Resent-date: Tue, 26 Apr 2005 17:10:19 -0600 (MDT)
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`Original poster: "Gerald  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>`

`Hi Mark,`

`Original poster: "Mark Dunn" <mdunn@xxxxxxxxxxxx>`

`So to conclude, the way I understand it and re-stated to be sure:`

```Z   <>  Z1   + Z2   + ...
Z^2 <>  Z1^2 + Z2^2 + ...
```

Not sure what you are meaning by <> but doing the Z^2 = Z1^2 + Z2^2 would not be the correct thing to do. You can do the Z^2 = R^2 + X^2 for any one vector if you like to make the magnitude visible, but there is also a phase angle that goes with this magnitude that sometimes you dont care about. The pythagorean theorem only works if the two components are at right angles to each other (like resistive and reactive components), so you cant use it to combine arbitrary impedances together to find the resulting magnitude. When you are adding complex numbers, its best to keep the numbers in vector form to combine (adding up all the real components and then adding up all the imaginary components) and then determine the magnitude as the last step. If you are multiplying (or dividing) complex numbers, conversion to magnitude/phase representation can be easier as the magnitudes mulitply (or divide) and the angles add (or subtract).

`BTW, when you have a complex number of this form:`

`1/(a+jb)`

You can multiply both the numerator and denominator by is its complex conjugate (a-jb) to put it into the form of c+jd. Example:

1/(a+jb) * (a-jb)/(a-jb) = a/(aa + bb) - jb/(aa + bb) Note that j=sqrt(-1) and jj = -1

`Instead we must break down the Z's into R, and X`

```Z1^2 = R1^2 + X1^2
Z2^2 = R2^2 + X2^2```

```And must assign the correct sign for X in case both inductive and
capacitive reactance is present.```

`Then`

Z^2 = (R1 + R2 + ...)^2 + (X1 + X2 + ...)^2
`YES, this is correct`

```I'm not sure this solves my problem, because in my case the R's were
near zero and the X's were inductive.  I will go back and recalculate.```

```Makes sense because all examples I find in books compute impedence for
the entire circuit and never for an individual components and this is
likely why.```

```Finally, the only time Z = Z1 + Z2 +... is when the Z's are vectors and
then this expression represents vector addition.```

```Thanks.
Mark
```

If you have a reference explaining the complex number system, you might find this helpful. It has many applications (working with complex impedances is just one application).

`Gerry R`