Re: Elementary Inductance.

• To: tesla-at-pupman-dot-com
• Subject: Re: Elementary Inductance.
• From: Bob Misiura <misiura-at-nccoast-dot-net> (by way of Terry Fritz <twf-at-verinet-dot-com>)
• Date: Mon, 22 Mar 1999 18:50:52 -0700
• Approved: twf-at-verinet-dot-com

Jimmy,

All the things you mentioned, except perhaps the gauge of the wire, do
affect inductance.  

If you change only one variable, this is what happens:

Increase the length of the wire, you increase the inductance (and
decrease the frequency)
Increase the number of turns of wire and you increase the inductance.
Increase the diameter and you increase inductance.
Increase the spacing between turns, and you decrease the inductance
(change the height of a coil, without increasing the number of turns)

A simple formula is:

      d(sq)n(sq)
L=  ____________
      18d + 40l

Inductance equals . . .Diameter squared, times number of turns squared,
over eighteen times the diameter, plus forty times the length.  

This is a simplified formula.  There are other formulas that will be
more accurate that take into consideration the length to diameter ratio
and turns per inch.  The above formula is "for coils having a length
equal or greator than 0.4d."

take care
bob

Tesla List wrote:
> 
> Original Poster: "JimmyD" <jim_del-at-email.msn-dot-com>
> 
> I know this is an elementary question, and that it may have been answered
> before on the list but,
> 
> In _lay terms_ , what determines the inductance in a coil?
> 
> The diameter, the height, the guage of the wire, the number of turns or the
> length of the wire?
> 
> In tesla operation what is the relationship of the primary to the secodary?
> The guage? The number of turns? The diameters? The heights?

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