# ballast Inductor Choke coil (fwd)

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From:  Alfred A. Skrocki [SMTP:alfred.skrocki-at-cybernetworking-dot-com]
Sent:  Friday, April 10, 1998 3:50 AM
To:  Tesla List
Subject:  Re: ballast Inductor Choke coil (fwd)

On Friday, April 03, 1998 5:39 PM Erik
[SMTP:ESchulz531-at-aol-dot-com] wrote;

> WHAT,
>
>     N = SQRT (38200 * ((9 * 3) + (10 * 12))/3 * 3)
>
> Then:
>
>     N = 1475 turns would be required
>
> Sorry but you are wrong it is 789.895

Referring back to the original post on Tue, 31 Mar 1998 by Erik Schulz
<ESchulz531-at-aol-dot-com>

> Matt,
>       When an inductor is placed in series with an AC source it
> causes resistance governed by this equation.  R = 2 * Pi * f * L
>
> So for a transformer that uses about 1kw you would need 14.4
> ohms with a 120 volt wall socket, so the inductance should be

So we have 1 kilowatt at 120 volt and since P = V * I we can divide P by V
to get I, thus 1000/120 = 8.33... amps. The equation for reactance in terms
of applied AC RMS voltage and a given current is;

XL = E/I = 120/8.33 = 14.4 ohms

The equation for determining inductive reactance XL = 2 * pi * F * L
solving for L gives;

L = 1 / ( 2 * pi * F / XL)

Then plugging in our values, assuming F = 60 Hz.;

1 / ( 2 * 3.1415926 * 60 / 14.4) = 0.0382 Henrys

> so on a 12 inch cylinder with a 6 inch diameter you would need
> about 780 turns with air as the core.

Using the standard equation for a single layer solenoid;

N = SQRT ( L * (( 9 * R ) + ( 10 * H )) / ( R * R ))

N = Number of turns needed.
L = inductance in microhenrys.
H = height (inches).

Notice that this equation requires that H be in microhenrys, so to convert
0.0382 Henrys into microhenrys we multiply by 1,000,000 which give us 38200
microhenrys. Plugging this value and the given dimensions for the cylinder
into the above equation gives;

N = SQRT ( 38200 * (( 9 * 3 ) + ( 10 * 12 )) / ( 3 * 3 ))

Then:

N = 789.8945077 then rounding off = 789.895 turns would be required.

I stand corrected! Apparently I mis-entered the values in the last equation
on the original post. I apologize to all who were inconvienced. I do
however still stand firm on my conclusion that an inductance of this size
is impractical as a single layer aircore inductor, since it's D.C
resistance is still about 5 times greater than it's reactance and since;

Z = SQRT (( R * R ) + ( XL * XL ))

The inductor would never allow more than about 200 watts to pass at 120
volts! When designing a current limiting inductor (or the primary of a
transformer) it is important that the inductors DC resistance be
significantly smaller than the inductors inductive reactance for it to pass
the required current.

Sincerely

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