--- In email@example.com
, "cavetronics" <cavetronics@...> wrote:
> Below is a post you have made and i happin to have a system running i
> have built and it works as your post has stated. i run a load of
> about 58 watts it cost me about 5.5-5.8 watts to run .
This of course would be remarkable, however one must be more specific
as to what the input consists of, and what the output consists of,
where here I will try supplying the records for my cited case, and the
associated wonderings about how this resonant circuit functions, as
this was the example cited in my past post you have cited. I notice
that yahoo has added letters to the prefix URL, so I dont know if the
URLs will go through.
Here is the Output;
SrFe 800-900 degree F heat output@ 17 DCV yeilding 3 Amps/ ~equal
stator voltage input @ ~1.1 A phase delivery.
(A dc volt meter in fluxuation between 17.1 and 17.2 volts shows the
voltage across the ferrite, and a DC amp reading shows 3.08 A. This DC
current is obtained by two full wave rectifications across three AC
resonant voltage rises. The ferrite load producing heat is acting with
an apparent resistance of 17 volts/ 3 A = 5.6 ohms
Now between this load and its diodes acccomplishing the dual phased DC
rectification are equal and opposite reactances of 7 ohms each. The
stator line output wires from the alternator are "ballasted" by these
7 ohm reactances. What this means is that the solitary reactances
alone when placed across the three phases of the alternator with the
voltage input set at 7 volts, each phase will conduct near one amp.
If the reactances of L and C are placed in series, five amps will
result with a 5 fold internal increase of voltage to accomplish the
higher amperage conduction, thus the resonant circuits show a q of 5.
However here the circuit is "loaded" so that no resonant voltage rise
is supplied to the load as the input jpeg here shows.
Input to SrFe heating; ~17.7 stator volts yeilding ~1.1 A on phases
Stator line 1 supplying 2.2 amps splits into phase 1 bearing 1.44 A and
phase 3 bearing 1 A
Stator line 2 supplying 2.07 A splits into phase 1 @1.44 A and phase 2
@ .85 A
Stator line 3 supplying 2.24 A splits into phase 3 @ 1A and phase 2
@ .85 A
The circuit is acting in the direction of a parallel resonant circuit
by virtue of the fact that the impedance of the load(s) is
approximately double what the solitary reactances of 7 ohms provide. In
this particular instance if the load was replaced by a short across all
inner voltage rises, the circuit having a q of 5 would appear as 5
times the ordinary 7 ohm reactance value or 35 ohms, so that given the
17.7 volt stator, only .5 A would be admitted into the circuit, where
we would assume that the stator delivery lines would contain 1.7 times
this amount or ~ .86 A. Inside the circiuts themselves we should find
fives times the .5 A input or 2.5 A, due to the tank q of 5.
Postnote, computer arrived at new house today, will be in better