# Re: Rewrite of Mutual Inductance Laws for Tesla List

```Original poster: "harvey norris by way of Terry Fritz <twftesla-at-qwest-dot-net>" <harvich-at-yahoo-dot-com>

--- Tesla list <tesla-at-pupman-dot-com> wrote:
> Original poster: "Paul Nicholson by way of Terry
> Fritz <twftesla-at-qwest-dot-net>"
> <paul-at-abelian.demon.co.uk>
>
> Harvey,
>
> >
>
http://groups.yahoo-dot-com/group/teslafy/files/RI/Dsc00184.jpg
>
> > L1 =10.8 mh, C1 =14 uf
> > L2= 60 henry, C2= 1 nf
>
> > Input = 1.67 volts *.35 A = .58 VAR
> > Output=(4/3.16*1000)*.004A = 5.06 VAR
>
> Unfortunately, the jpg doesn't make clear the
> circuit arrangement, so
> we're not sure just where the V and I measurements
> were picked up.
Only one element is engaged for the primary, an
amperage meter in line measures that, and also a
voltage meter across that element (primary LC input)
> But it's clear you've got a pair of coupled LC's
> both resonant near
>
> between the alternator
> and L1C1, in which case if L1C1 is tuned to the
> alternator frequency
> these values will represent *real* power, not VAR.
Exactly as intended, this only shows what is
inputed,but because of resonance that input is
equivalent to the real power input. This is only made
to counter the first argument that would be made that
reactive power input does not equal the real one, but
for resonance they are equal. So far so good...
>
> Whereas with the L2C2 measurements, the current can
> only be the reactive
> circulating current (since you've not mentioned a
> the 'output' reading is really the VAR of the stored
> energy in L2C2
> resonator.
Unfortunately there is no load, only the L2C2 quantity
as that load, where L2 contains 1000 ohms resistance
to be factored in as a load. If C2 has an arc gap
across it, then we can deal with that load of arcing
resistance itself circumventing the resonance, in
which case the delivery is reduced, and everyone
becomes happy with the reduced delivery.
>
> On this basis, we can estimate the Q factor of your
> combined L1C1/L2C2
> system.  Your 5.06 VAR represents a stored energy of
> 1680uJ (since this
> energy circulates back and forth between L2 and C2
> some 2 * pi * 480
> times per second).
>
> If 0.58 Watts of real power are necessary to
> replenish the leakage from
> your store of 1680uJ, then the Q is given by
>
>   Q = 2 * pi * frequency * stored energy / input
> power
>     = 2 * 3.141 * 480 * 1680e-6/0.58
>     = 8.7
Exactly as noted formerly, on the internal voltage
side of L2C2, that voltage will rise ~ 8.7 times, (or
lower)  then that inputed: but the point being made
here.(with 3 phase inputs) is that the q can go higher
than that circumstance. Let us again look at the
actual operation at 15 volts  stator input;
http://groups.yahoo-dot-com/group/teslafy/files/RI/Dsc00178.jpg
Here 15.03 volts stator enables a 364.8=~365 volts
across the interphasing. That q is 365/15= 24.3 which
sounds good, but is not that available if the
connections were measured with no load at all. In that
case if the voltage were measured without the
interphasing across it, the voltage reading would be
higher. THE CONTENTION THAT NO LOAD IS ACROSS THE
MIDPOINTS DOES NOT CONSIDER THE FACT THAT THE OHMIC
RESISTANCE OF THE ELEMENT BEING RESONATED ITSELF IS
THE ACTUAL LOAD. Again if we simply placed the ohmic
resistance, (impedance) of the voltage meter itself as
the load, a higher voltage would be delivered. We can
determine this quickly for the case of book values
approximated as a near "no load" situation. For X(L)~=
Z, .15 henry with 12.5 ohms resistance will be by
X(L)= 2 pi F L= 6.28* 480* .15 = 452 ohms. The q
factor determining internal voltage rise for each DSR2
side to accomplish conduction then equals 452/12.5 =
36.17.
The voltage across the interphasing should be 1.7
times this value at 61.5 times the input voltage. The
reason that only 24.3 times the input voltage occurs
is that the R(int) value of the load to be resonated
IS ALREADY TAKEN INTO CONSIDERATION. Therefore to say
that no load exists is quite redundant, any component
to be resonated has its internal resistance, which as
you have formerly aptly pointed out, that R (int)
value must also be taken into consideration in the
final resonance, here shown to be reduced at ~ 1/3 to
the  expected  voltage value if that component itself
had no resistance. Now let us look at the final q
value in real operation at 15 volts. The first stage
of resonance has increased the voltage input from 15
volts to 365. The second voltage rise increases this
to that made with 20.3 ma across the 1 nf capacity,
equivalent to 6400 volts. The inner Q normally only
exibits a 8 to 9 q factor. 9 * 365 = 3285 volts. but
6400 volts is appearing. This means a q factor 1.9
times that, almost double to your figurings... So then
can you revise your estimations...? As I have said the
method increases the available q factor that would
normally be available, and it is most logical that
when we make a magnetic field in opposition to
resonance, this method countermands the devilish
limitations of resonance inherent with internal
capacity. If a better method is available of
cancelling the internal capacity of a coil delimiting
a q factor, to make a better Q factor overall, I also
would like to know about that one. HDN, AKA Dr
Wrongway.
> So I'm afraid I don't see anything in your post
> which indicates a
> problem with mutual inductance.  At times like this
> you always have to
> ask yourself which is the most likely of two
> possibilities:  Either
> you've misunderstood the operation of the circuit,
> or some very well
> tried and tested laws of physics are wrong.
> --
> Paul Nicholson
> --

=====
Tesla Research Group; Pioneering the Applications of Interphasal Resonances
http://groups.yahoo-dot-com/group/teslafy/

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