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Re: AC wire resistance with proximitry effects

Original poster: "Barton B. Anderson" <bartb@xxxxxxxxxxxxxxxx>

Hi Gerry,

The equations are identical, but there are considerations. When using Q=sqrt(L/C)/R, assuming R is the same for both forms, the final value will only be identical when Wo=w (1/sqrt(LC)=2pi*F). To ensure this is true, it's best to measure L and calc C or vice-versa. I simply find it easier to implement as Q=wL/ESR. This removes an incorrect input situation if L and C are actual inputs and if C is found from L, than it saves a step.

Take care,
Bart



Tesla list wrote:

Original poster: "Malcolm Watts" <m.j.watts@xxxxxxxxxxxx>

On 10 Oct 2005, at 11:47, Tesla list wrote:

> Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>
>
> Hi Bart,
>
> I understand now what you are saying.  One thing I might question is
> your equation for Q.  I dont think this is the right equation for a
> series RLC circuit:
>
> The RLC Laplace impedance is sL + R + 1/sC
>
> I(s) = V(s) / sL+R+1/sC
>
> to put in standard form:
>
> I(s)/V(s) =  (s/L) / (s^2 + sR/L + 1/LC) = (s/L) / (s^2 + s Wo/Q +
> Wo^2)
>
> Wo =  sqrt (1/LC)
>
> R/L = Wo/Q
>
> Q = WoL/R = sqrt (L/C) / R
>
> Gerry R
>
>
> >Original poster: "Barton B. Anderson" <bartb@xxxxxxxxxxxxxxxx>
>
> >First, regarding Ldc. Your right, it's not in the equation for
> >Fraga, but it is used when predicting Q. Here is what I'm doing, at
> >least at this time.
> >
> >ESR = wL/Q
> >Q = wL/ESR
> >I am replacing ESR with Fraga's resistance because it is a combined
> >R.

I'm struggling to see a difference apart from Bart defining "R" as
"ESR" which is quite correct.

??
Malcolm

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