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Re: Association of coupled inductors (Dimensions of my flat spiral coil)



Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <twftesla-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>

Tesla list wrote:
> 
> Original poster: "Ben McMillen by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <spoonman534-at-yahoo-dot-com>
> 
> I thought inductance in parallel added like resistors in
> parallel? Or am I thinking of inductive reactance? Hrm..

Inductances add as resistors when they are not coupled. With coupling
what happens can be deduced from the expression that defines a
linear transformer (2 windings in this case):

v1=L1 di1/dt + M  di2/dt
v2=M  di1/dt + L2 di2/dt

With the inductors in series, di1/dt=di2/dt=di/dt, and the total
voltage is v1+v2=v:
v=(L1+L2+2M)di/dt
This corresponds to an inductor with inductance L=L1+L2+2M.

With the inductors in parallel, first invert the equations:

di1/dt=G11 v1 + G12 v2
di2/dt=G21 v1 + G22 v2

(The "G"s should be "gammas" in the usual notation.)
The voltage over the combination is v1=v2=v, and the derivative
of the current is di/dt=di1/dt+di2/dt:
di/dt=(G11+G22+G12+G22)v, and G12=G21.
This corresponds to an inductor with value L=1/(G11+G22+2 G12).

I leave as exercise the demonstration that these relations are
consistent
with the usual inductance formulas.

Antonio Carlos M. de Queiroz