Re: double wound secondary (fwd)

["Tesla list" <tesla@xxxxxxxxxx>]

---------- Forwarded message ----------
Date: Wed, 03 Oct 2007 16:28:52 -0700
From: Jim Lux <jimlux@xxxxxxxxxxxxx>
To: Tesla list <tesla@xxxxxxxxxx>
Subject: Re: double wound secondary (fwd)

Tesla list wrote:
> ---------- Forwarded message ----------
> Date: Wed, 3 Oct 2007 12:42:50 -0500
> From: Terry Oxandale <Toxandale@xxxxxxx>
> To: Tesla list <tesla@xxxxxxxxxx>
> Subject: RE: double wound secondary (fwd)
> 
> I believe the rule is that two tightly coupled mutually inductive coils
> of same polarity (which this appears to be the case) would have a sum
> inductance equal to about the inductive value of one of the coils. So...
> 
> Terry
> 

Nope.. 4 times...

Two separate inductors, uncoupled, would have double the inductance of 
one inductor.   Consider that each inductor is subject only to the field 
of itself.

If you have them tightly coupled, each inductor is subject to the field 
from itself, plus the field from the other inductor.  You've got 2, so 
it's 4 times total.

For three it would be 9, etc. N^2



>