In article , TokaMundo wrote:
On Tue, 2 Aug 2005 22:58:33 +0000 (UTC), (Don
Klipstein) Gave us:
Going by "High Frequency Resistance", pages 3323-3325 of the 43rd
edition of the "CRC Handbook":
They give a formula X=pi*d*SQR((2*u*f)/rho)*SQR(1000)
u is magnetic permeability, unity for copper.
rho is resistivity in microohm-cm.
They simplify this for copper, to x=10*d*.01071SQR(f)
d is diameter in centimeters, and f is frequency in Hz.
So, for 17 mm diameter copper wire at 60 Hz this "x" is 1.41.
Next is a table that gives ratio of AC resistance to DC resistance as a
function of this "x".
This table has an entry for 1.4, giving AC resistance 1.020 times DC
resistance.
The table I saw shows the AC and DC resistance as being exactly the
same for both.
Maybe your table rounds? Please cite source as well as I did.
Your flaw is where you failed to note the topic given in the CRC
handbook.
I did note this, as you quoted above.
60 Hz is NOT high frequency... at all.
Try some calculations at 100 kHz and you'll see that those
frequencies down near zero (ie 60Hz) yield very nearly nil difference.
The formulas are functions of wire diameter, wire resistivity and
frequency, and do not lose validity merely because a thick wire has AC
resistance greater than DC resistance at a frequency that is easy to label
"NOT high".
And as you asked... Ratio of AC resistance to DC resistance of 17 mm
diameter copper wire at 100 KHz is about 21.5. This does not invalidate
the calculation for 60 Hz.
You would have been better off claiming that resistance 2% higher at 60
Hz than at DC is a negligible increase.
- Don Klipstein )