Practical Guige to Free Energy Devices

necessary to get it charged up to 1200 because the average voltage during discharge is V / 2.
If the energy produced by the transformer secondary is assumed equal to 321 watts and the voltage required for
the charge of the capacitor is 1200 volts, then the current delivered by the secondary will be 321 watts / 1200
volts which is 0.267 amps.
On the other hand, as the frequency of the alternator is 1000 peaks per second, then in one thousandth of a
second the secondary has to deliver to the capacitor 0.000267 coulombs which is 267 microcoulombs.
Hence, the capacity of capacitor 48 must be adjusted to store that amount of electricity, and its value can be
determined by Q = K x V where K is the capacity of the capacitor in microfarads when Q is given in
microcoulombs and V is the maximum voltage, and so, K = 267 / 1200 which is 0.222 microfarads.
It is also well known that if you want to produce an oscillatory discharge through a circuit, the capacitance,
resistance and inductance of the circuit must be arranged so that the square root of 1000 x L millihenrys / K
microfarads is greater than the resistance of the circuit in ohms (K being the capacitance of the capacitor).
It has been supposed that the resistance of the oscillatory circuit is 15 ohms. Using 20 ohms in the above
equation will allow a suitable value of inductance to be calculated, one which will satisfy the required conditions
for the production of the oscillatory discharge in the circuit, and so the inductance in millihenrys is 400 x K / 4000
or 400 x 0.222 / 4000 which is 0.0222 millihenries, or equal to 22,200 centimetres of wire.
It is possible to determine the number of oscillations per second which can be achieved in any such circuit, and
that is given by Hz = 5033000 / the square root of L x K where L is in centimetres and K is in microfarads. And so,
knowing the inductance and the capacitance we get Hz = 5033000 / sqrt(22200 x 0.222) which is 71900 Hz. That
represents 72 oscillations in each of the 1000 sparks per second.
The current strength developed by these oscillations can be determined from the formula below which uses the
Voltage V in volts, the resistance of the circuit R in ohms, the inductance L in henrys, and the capacitance K in
farads:
Current = V / sqrt(R x R + (6.28 x Hz x L – 1/(6.28 x Hz x K) 2 ) or in our case:
Current = 600 / √ 15 x 15 + (6.28 x 71900 x 0.0000222 – (1/(6.28 x 71900 x 0.00000222))) 2
Which works out as 40 amps, meaning that the impedance of the oscillating circuit is equal to the resistance in
ohms of that circuit, since the inductive reactance and the capacitative reactance are so combined that the
resultant value of the total reactance is equal to zero, and the only voltage required to produce the current of 40
amps is that needed to overcome the ohmic resistance of the circuit which is 15 ohms. This also means that the
E.M.F. is in phase with the current, and therefore, the watts are a maximum.
Hence, there can be no doubt that battery 3-4 will be fully charged during the period when battery 1-2 is
discharging, especially since the charging current can be further increased at will, even without taking more power
from the discharging battery. In fact, it is easy to increase the number of turns in the transformer secondary 46-47
in order to increase the voltage. Obviously, as the amount of power delivered to the primary 44-45 of this
apparatus is always 321 watts, if the voltage is increased, the quantity of electricity which the capacitor 49
receives will be reduced accordingly. Consequently, the value of that capacitor must also be reduced and so the
number of oscillations per second will also be increased. Finally, by increasing the voltage, the current intensity is
increased proportionately.
Therefore, it is always possible to combine, in the manner described, the values of resistance, inductance,
capacitance and voltage in the oscillating circuit, so as to obtain the required current strength to fully charge one
of the batteries during the period when the other battery is discharging.
Once this battery charging has been attained, if the plant is to continue in operation, then the batteries need to be
swapped over by altering their connections to the circuit. To accomplish this, cylinder 5 is rotated until the
conductive paths 11 and 12 come into contact with the brushes which are connected to the negative poles of the
batteries, and then, battery 3-4 which is fully charged will be connected with the motor 27-28, and its discharge
will now be produced by this path: battery terminal 25, ammeter 24, commutator poles 20 and 18 (now connected
together due to the rotation of the cylinder 5 through 90 degrees), wire 26, motor 27-28, variable resistor 29, wire
30, apparatus 40 and 41, and commutator path 11 which closes the circuit to the negative pole of battery 3-4.
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