Practical Guige to Free Energy Devices

capacitor voltage will not be 100 volts DC but instead it will be 141.4 volts DC and you need to remember that
when choosing the voltage rating of the capacitor. In that instance I would suggest a capacitor which is made to
operate with voltages up to 200 volts.
You probably already knew all of that, but it may not have occurred to you that if you use a standard AC voltmeter
on a waveform which is not a sine wave, that the reading on the meter is most unlikely to be correct or anywhere
near correct. So, please don’t merrily connect an AC voltmeter across a circuit which is producing sharp voltage
spikes like, for instance, one of John Bedini’s battery pulsing circuits, and think that the meter reading means
anything (other than meaning that you don’t understand what you are doing).
You will, hopefully, have learned that power in watts is determined by multiplying the current in amps by the
voltage in volts. For example, 10 amps of current flowing out of a 12 volt power supply, represents 120 watts of
power. Unfortunately, that only holds true for circuits which are operating on DC, or AC circuits which have only
resistors in them. The situation changes for AC circuits which have non-resistive components in them.
The circuits of this type which you are likely to come across are circuits which have coils in them, and you need to
think about what you are doing when you deal with these types of circuit. For example, consider this circuit:
This is the output section of a prototype which you have just built. The input to the prototype is DC and measures
at 12 volts, 2 amps (which is 24 watts). Your AC voltmeter on the output reads 15 volts and your AC ammeter
reads 2.5 amps and you are delighted because 15 x 2.5 = 37.5 which looks much bigger than the 24 watts of
input power. But, just before you go rushing off to announce on YouTube that you have made a prototype with
COP = 1.56 or 156% efficient, you need to consider the real facts.
This is an AC circuit and unless your prototype is producing a perfect sine wave, then the AC voltmeter reading
will be meaningless. It is just possible that your AC ammeter is one of the few types that can accurately measure
the current no matter what sort of waveform is fed to it, but it is distinctly possible that it will be a digital meter
which assesses current by measuring the AC voltage across a resistor in series with the output, and if that is the
case, it will probably be assuming a sine wave. The odds are that both readings are wrong, but let’s take the case
where we have great meters which are reading the values perfectly correctly. Then the output will be 37.5 watts,
won’t it? Well, actually, no it won’t. The reason for this is that the circuit is feeding the transformer winding which
is a coil and coils don’t work like that.
The problem is that, unlike a resistor, when you apply a voltage across a coil the coil starts absorbing energy and
feeding it into the magnetic field around the coil, so there is a delay before the current reaches it’s maximum
value. With DC, this generally doesn’t matter very much, but with AC where the voltage is changing continuously,
it matters a great deal. The situation can be as shown in this graph of both voltage and current:
At first, this does not look like any great problem, but it has a very significant effect on the actual power in watts.
To get the 37.5 watts output which we were talking about earlier, we multiplied the average voltage level by the
average current level. But these two values do not occur at the same time and that has a major effect.
As this can be a little difficult to see, let’s take the peak values rather than the averages as they are easier to see.
Let’s say that in our example graph that the voltage peak is 10 volts and the current peak is 3 amps. If this were
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