# RE: Current Limiting and Impedence

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• Subject: RE: Current Limiting and Impedence
• From: "Tesla list" <tesla@xxxxxxxxxx>
• Date: Tue, 26 Apr 2005 14:35:11 -0600
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• Resent-date: Tue, 26 Apr 2005 14:35:51 -0600 (MDT)
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`Original poster: "Mark Dunn" <mdunn@xxxxxxxxxxxx>`

`Gerry:`

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

```Z   <>  Z1   + Z2   + ...
Z^2 <>  Z1^2 + Z2^2 + ...
(Expressed as magnitude values, not vectors)```

`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`

```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```

`Original poster: "Gerald  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>`

`Hi Mark,`

`This can be a confusing area.`

```Series resistances add:
Rtotal = R1 + R2```

```For parallel resistances, the conductances (G) add:
Gtotal = G1 + G2```

```or expressed as resistance:
1/Rtotal = 1/R1 + 1/R2```

`Now the tough part:`

```Impedance, in the general case, has resistive (R) and reactive (X)
components (sometimes refered to as the real and imaginary parts).  For
series impedances, the resistive components add up and the reactive
components add up (keeping in mind that capacitive reactance is negative```

`and inductive reactance is positive) so you get the following:`

`Ztotal =  Rtotal + jXtotal  = Rtotal + j(XLtotal -XCtotal)`

```You can't linearly add the resistive and reactive components together.
The
impedance is a complex number denoted by the j prescript on the reactive```

```part.  What you can do instead is determine the magnitude of the
impedance
using:```

`Z = sqrt(R^2 + X^2)`

```You can think of the R and X terms being two sides of a right angle
triangle and the Z being the hypotenus (sp?).  Series impedances add
similar to resistances in that:```

```Ztotal = Z1 + Z2
but one needs to keep to the rules of complex math.```

`Parallel impedances also behave similar to parallel resistances in that:`

```1/Ztotal = 1/Z1 + 1/Z2
and again one needs to keep to the rules of complex math.```

`Hope this helps more than being confusing.`

`Gerry R`

```>Original poster: "Mark Dunn" <mdunn@xxxxxxxxxxxx>
>
>I still don't quite get it.
>
>1.  Not that it matters for this question, but I thought I could sum
>series impedence.  Could you re-confirm? Are you sure you haven't
>confused with  Z^2 = R^2 + X^2?  I don't mean to question and I am not
>an expert so I am just making sure.
>
>If it is Z^2 then my formula for parallel Z must be wrong.  I'm using
>1/Z = 1/Z1 + 1/Z2 + ... .  Just like resistance.```