Re: Re Secondary Resonance LC and Harmonics

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: FIFTYGUY@xxxxxxx 

In a message dated 6/29/05 5:49:15 PM Eastern Daylight Time, 
tesla@xxxxxxxxxx writes:

We can examine the classic equation for an air cored inductor: L = u
Nsqrd A / l
Multiply the numerator and denominator by 4pi  we get:

L = u Nsqrd 4pi sqrd  r sqrd / 4 pi l  =  u (2pi r N)sqrd / 4 pi l

Since 2pi rN equals wire length we can write:

L = u (wire length)sqrd / 4 pi l

Letting the solenoid height ( l ) equal the wire length,  we find
that the classic equation predicts that the inductance of a straight
wire is simply:

Lstwire = u wire length / 4 pi

    But using the above "classic equation" also collapses the inductance 
to zero as the radius "r" and the turns "N" go to zero.

-Phil LaBudde

>True if R and N are both zero then we have no wire and hence no
>inductance.

    My intent was to show quite the opposite: If R and N are both zero 
then we have a STRAIGHT wire with minimum inductance. Actually, minimum 
inductance is obtained by doubling the wire back on itself so one half 
cancels the other's inductance.

> I think what you meant to ask is what happens as R
> approaches zero for a constant length of wire.

    Sure...

>As R approaches zero the solenoid length (l) approaches the wire
>length. The substitution we used where (l)= wirelength does not lead
>to a contradiction, it should be a legitimate step.
>To help wrap your mind around this geometry consider what happens when
>you take wire from a fat inductor and wind it around progessivly
>skinnier forms. Sure enough as the radius gets smaller the length (l)
>aproaches that of the straight wire.

    With you so far.


>Of course the really interesting implication is that at R=0, N is
>infinite.

   I disagree. Because we have not changed the thickness of the wire, as 
R=0, at some point we must GIVE UP turns to lay the wire on top of itself. 
At some very small R, the wire must physically "corkscrew" and have spaced 
turns in order to complete the turns. In this way, we have used the same 
wire to form less turns.
   Which brings me back to my original point: as both R and N go to zero, 
so does the inductance. The only way you get a straight wire (R=0) to have 
any turns ("N") is via eddy currents.


>Jared and Larry

    Also:

> Furthermore classic description also predicts that exactly half of the
> inductance of a straight wire resides within the wire itself.

    I don't understand this statement at all. How do we observe this 
hidden inductance? How does inductance "reside" somewhere? Where does the 
other half of the inductance "reside"?

-Phil LaBudde