## Transcending Reactive vs Real Power Concepts.

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• As most of us know the REASON that we cannot simply multiply the voltage times the amperage, (VI); to obtain the power expressed in an inductor placed across
Message 1 of 1 , May 29, 2005
As most of us know the REASON that we cannot simply
multiply the voltage times the amperage, (VI); to
obtain the power expressed in an inductor placed
across AC, is that there is a time lag between the
cause and effect; where voltage being the cause, and
its amperage being the effect. For the AC inductor,
because of this time lag, that for the external AC
source we can speculate that if the current were truly
90 degrees out of phase with the voltage, (this often
stated premise is not completely true, it depends
whether the inductor has a large enough inductive
reactance X(L)= 2pi*F*L to form a phase angle near 90
degrees); but if the amperage were near 90 degrees out
of phase with the voltage, by the time the AC voltage
input has made a complete pulse in one polarity, and
thus gone back to its zero crossing point; at that
moment in time the amperage in the coil will have
reached its maximum, again using the 90 degree phase
angle example. Up until this starting point 180
degrees into the AC cycle, a magnetic field from the
inductor is established that agrees with its polarity
input. But in the third quadrant of the AC cycle, the
voltage input reverses polarity, but the current has
not yet followed its example, because of the time lag
the current is going in the opposite direction then
what its voltage source would cause it to act. In the
third quadrant then the magnetic field of that
inductor is then causing a self induction by virtue of
its magnetic field collapse that establishes a
polarity OPPOSITE to the polarity of the source. This
is why back emf, or a backwards working voltage source
established by mistimed magnetic fields exists. The
net result of this backward working voltage during
portions of the AC cycle is that the forward working
voltage does not conduct it's OHM's law amount of
amperage, because of this extra effect of AC
resistance. We call this AC resistance inductive
reactance. The effects of both the actual resistance
and its inductive reactance is reffered to as
impedance, noted as Z where the effects of both
principles are obtained by summing the squares of both
quantities, and taking the square root of this result
shown as the impedance equation Z= sq rt[R^2+
X(L)^2].

Now some commentators note that because the coils
magnetic field establishes its own self induction, we
are actually dealing with the summation of two power
sources. They explain that because of the storage of
energy manifested as a magnetic field, in certain
portions of the AC cycle the coils collapsing magnetic
field is establishing a polarity opposite to its
source, and in terms of the AC source being a
generator, that generator as its VI apparent power
input also experiences borrowed and returned energy.
When the coils magnetic field collapses a portion of
the power input is returned to the generator. Now the
real amount of energy expeneded on the coil is just
its conversion to heat by I^2R heat losses, and the
rest is just borrowed and returned energy. The
essential question then becomes does this source also
act like a banker and charge interest on its borrowed
and returned assets? This would seem the sensible
thing to do, but perhaps nature doesnt always model
mans actions.

So we have two aspects, and this is where the third
aspect comes into being, where the concept of phase
angles comes into play. We have the apparent power
input which is just the voltage times the amperage
across the load,or (VI) of the generator as the
apparent power input. Then we have the real power
input which is undeniable because we can measure the
heat release as the I^2R quantity. Because we have a
difference between the real and apparent power inputs
is merely the % of borrowed and returned energies vs
real energy output. But we also have a method to
compute what this heat release should be by methods of
trigonometry. We can draw out the AC voltage and
amperage AC cycles, and find out the % portions where
they are in agreement, where this portion will be
expressed as a cosine angle. The real power input
therefore becomes VI* cos( acting phase angle of the
inductor), and we conclude that this also equals I^2R
as the heat loss. Recall that in trigonometry we have
a y coordinant, [expressed as X(L)in phase angles]:
where y/radius is the sine quantity of the angle, and
conversely x/radius is the cosine. (In trigonmetry we
allow the radius of the circle to be the unit 1 to
simplify these calculations.) To find the phase angle
of an inductor we let the inductive reactance to be
the positive y quantity, and the actual resistance R
to be the x quantity. But a little more trig is need
to find the actual phase angle which can be calculated
with any calculator having trig functions. The tangent
of the angle is expressed as y/x. So if we have a coil
where X(L) and R is both known, the tangent is easily
known; it is just X(L)/R, which is also the
theoretical Q factor of the coil. But what we NEED to
find out is the inverse of this, which will tell us
the radians, which can further be reduced to an angle
by multiplying by 360 degrees/ 2pi radians, which is
the conversion factor to degrees. Suppose I have an
inductive reactance 5 times the resistance; what is
the phase angle? We know the tangent is merely 5, but
this tells us nothing. Instead what we NEED to find
out is the answer to the question; the tangent of what
use the inverse function, called the co-tangent, or
tan^-1 to determine the answer. And tan^-1 (5) = an
and this is the method for finding the phase angle.
Now if X(L)is very large vs the R quantity of
resistance, X((L) forms the y axis, R forms the x
axis, and the hypotenus, (ordinarily the radius
quantity in trig functions) becomes the impedance Z.
We already know by the Pythagorean theorem that the
sum of the squares of the right angle triangle sides
is equal to the square of the hypotenus, and by then
taking the square root of both sides of the equation
we reach the same equation for the impedance Z. And if
X(L)>> R then X(L)~= Z. The length of the radius and
its y reflection are almost equal, but the x
reflection as the additive vector is very small
compared to the other two quatites by definition since
we have initially made the condition X(L) is far
greater then R. The x reflection is also the cosine of
the phase angle, and it is this no. as a percentage
that is used for the real power calculations. Thus
with this condition the apparent power will be far
greater then the real power. These things are all
rudimentary for the electrical engineer.

Now we come to electrical resonance, where X(C) forms
the negative Y axis, the opposite reactances in series
can cancel, leaving R as the predominant factor for
conduction, and in this condition then we suspect that
now the apparent power will be the same thing as the
real power, and the amperage has now come closely in
phase with the impressed voltage. However because of a
factor known as "internal capacity" we may not always
arrive at a conduction that equals ohms law
requirements. The depends on the construction of the
inductor we are resonating. However a VERY
misunderstood thing can happen here. We might assume
that since now the apparent power equals the real
power, which also occurs in DC circuits after the
magnetic field is established, that % wise the
generator is no longer having a significant portion of
"borrowed and returned" energy allocations. We have
cancelled the reactive state of the circuit, so have
we not also cancelled the "borrowed and returned"
field energies? The generator would seem to indicate
so, because the apparent power equals the real power,
and we are not seeing differences between those
definitions. However in actuality the opposite thing
has occured, the borrowed and returned field energies
will have gone up q times, even though the generator
does not seem to indicate this % wise in its
allocations. A simple proof of this is the definition
of stored energy for the L and C components. The
energy stored in C is defined as [CV^2]/2, and for L
it is [LI^2]/2. In series resonance each L and C
component developes an internal voltage rise against
each other, in opposite directions with respect to
each other, so that on the outside of the circuit, the
generator only sees this net cancellation. For the C
case, since the V term is squared, and we are
obtaining series resonant internal voltage rise, the
stored energy has gone up Q times, where the acting Q
factor is the ratio of the inside voltage rise to the
outside voltage source. We then have increased the
borrowed and returned energies, but now a different
relationship exists. Formerly the energy was borrowed
and returned to the generator, but now the borrowed
and returned energies only exist between the L and C
values... It is literally as if the expanded electric
and magnetic field energies were being obtained in
power another load? That isnt so easy a proposition.
If we divert the expanded voltages to another circuit,
this in turn destroys the properties of the resonance
itself, so that the actual resonant rise of voltage is
wiped out.

A little over a year ago I set up some circuits where
the resonances were obtained from a 3 phase alternator
functioning @ 480 hz. These resonances were special in
that they were designed to be maximum energy transfer
circuits, generally defined as a circuit that drops
the open circuit voltage of the source in half, but on
the only side of the coin, the maximum amount of power
to load is obtained from that generator. Between these
3 AC phases, set up as Delta Series Resonances,
(DSR's), the voltage rise q factor of 5 was noted.
What this means is that 5 times more amperage is
procurred then what exists in the reactive state, and
also the internal rise of voltage was 5 times that of
the source. Now between these three 120 degrees of
voltage rises, the AC was turned into DC by means of
full wave rectifications, which because of the poly
phased inputs, thus fills up the DC ripple without use
of a filter capacity. The load for these DC currents
was a 3/8ths inch width of a ferrite block, normally
considered an insulator. Now ferrite is actually
considered a class of semiconductor, in that it looses
resistance with heat. The ferrite in this case
gradually grew hot, its resistance decreased
phenomenally going from 30,000 ohms to around 7 ohms,
until it began to glow on a corner, the temp around
900 degree fahrenheit. Meanwhile the outside coils,
where this internal current was obtained from on the
rectifications also experienced a drop of current from
what existed in its resonant state. The addition of
its amperage input to drop, in fact it drops low
enough so that only half of the current that the coils
would consume in its reactive state was noted. We
might say that X(L) of the coils appeared to double.
What this also means is that once again we should be
dealing with apparent vs real power input ratios. Now
the inside load, being DC has no power factor
arguments associated with it, but the outside coils
do, because they are AC reactances, placed into
resonance, and then reduced from resonance by addition
generation maximum power transfer principle, where the
internal impedance or reactance of the outside coil
systems have been matched by an equal resistive load.
In terms of this Z(int) of the source has been given
an equal R(load) in ohmic values, the requirement for
maximum energy transfer. The outside spiral coils
themselves, constructed as maxium energy tranfer
resonances, (METR components), use the same principle
in the R(int) of the alternator stator coils equals
R(load) of the coils resistance R value.

Now in the cited circumstance, the stator lines each
serving two phases contains ~ 2.0 -2.4 Amps, and the
division into phase currents should reduce this by
about 1.7 times the stator line currents. The outer
phase currents derived from these stator line currents
are somewhat inbalanced, and the pic I took of the
power input shows 1.4, .85 and 1 amp currents on this
outer triangle, a sum of 3.25 Amps. On the DC current
procurred from the outside misphased AC currents I
arrive at a sum of 3 DC amps, obtained from this
outside AC current oscillation. Both the alternator
input voltage and the ferrite DC voltage are about
equalized at 17 volts. As such no resonant rise of
voltage is taking place, the voltage input is about
the same as the voltage output to the load. Now if we
average the phases currents to ~ 1.1 amps, with 17
volts across them, the apparent power input is 17* 1.1
= 18.7 watts per phase, or the generator is inputing
an apparent power of 56.1 watts. But the REAL power of
the inside triangle load being shown by the 900 degree
heat of the ferrite is also 17VDC*3A = 51 watts. The
coils themselves that lay on the delivery lines to the
ferrite have a combined 3 watts heat loss, close to
what the apparent power shows itself as, where the
combined loads of 51 + 3 watts almost equals the
apparent power input of 56 watts.

Here is where the kicker comes in; the apparent power
in is yeilding almost the same amount as real power
out!. We surmise that since no resonant rise of
voltage is occuring on the outer triangle coils, that
we should treat this as a apparent power input, and
not a real one. Now we have about a 84 degree phase
angle, made as tan^-!(10), from a situation that
started out with a phase angle dictated by the
tan^-1(5), from the coil systems having a q of 5.
about double the impedance on their conductions, which
is where the 10 figure in tan^-1 estimations comes in.
If this were true, the mechanical energy required to
turn the alternator should only reflect a real power
load of only 10 watts, by the conventional phase angle
analysis, but yet we have a bonafide real power load
of 5 times this amount!

Have we indeed turned an apparent power input into a
real one? Perhaps the apparent vs real power arguments
do not apply for this special case. What I would like
to be able to do is power the alternator with a geared
up bicycle gearing rig, and drive the alternator by
bicycled leg power. For only a 10 watt power
requirement to make 900 degrees, this seems possible
with a human energy input. Then I could stick in real
resistive power loads, and compare the respective
drags for both situations. In any case the above
arguments either show that the current electrical
phase angle theory is inadequate to explain the above
situation, or conversely we have used the "free" field
oscillations inherent in resonance to create a
situation of overunity. The calculations also show
that ~ 6 times the energy transfer occurs in the LC
field energy transfers then do the actual real power
being inputed as I^2R heat losses on the DSR's
themselves, which are just under 1 ohms resistance.

Sincerely
Harvey D Norris

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