Bifilar pancake coil overunity experiment

[ayeaye]

P = V * I

This is true about any part of the circuit, no matter whether there are coils or no coils. Both for input and output power of that part of the circuit.

Now it may look like that when the coil conducts more, the power consumption of the whole circuit may look like more. Thus one may think measuring power consumed by the coil by a dissipation of power on a resistor in series with the coil, etc. This does not measure the power consumed by the coil, not what concerns physics, different from some possible practical purposes.

Why? First because this is how power is always calculated, this is how electric power is measured, any electric power. Because this is the physical meaning of electric power. Several people who here doubted whether it's right, are therefore wrong. They are wrong in terms of physics, and this is what matters for research. They may not be wrong what concerns how some or other circuits are made, but this is not about the physical meaning of power.

So ok for their particular circuits, when more current goes through the coil, then their circuit consumes more power. But this is only about how their circuit is made, it is about their circuit and not about the physical power consumed by the coil. How so? Simply speaking, the current that goes through the coil, is not consumed by the coil. Like when the coil had zero resistance at any moment of time, the power it did consume at that moment of time were zero, zero and not a bit more.

Because again simply speaking, the current that goes through the coil, is most often dissipated in some resistance, and released as heat energy, thus none of that energy is lost in the coil. How the current that goes through the coil is used in the circuit, that depends on the circuit, whether it is used somehow or just wasted.

But the purpose of this experiment is to measure the input and output power of the coil, coil and coil only. Not the power in some circuit that can be incredible wasteful as well. But how much power some wasteful circuit consumes, gives no useful information about the coil, none and nothing, thus it makes no sense to measure, and has no physical significance whatsoever about the coil.

Please understand these things first. It makes no sense to do the experiments to measure the power efficiency of a coil otherwise.

[TinselKoala]

You may want to review this PDF slideshow on power measurement, from IEEE:

http://ieee.rackoneup.net/rrvs/13/fundpwrmea.pdf

[ayeaye]

You may want to review this PDF slideshow on power measurement, from IEEE:

http://ieee.rackoneup.net/rrvs/13/fundpwrmea.pdf

First of all, in this experiment the power is not measured by whole waveforms, so what at all has this to do with it?

This pdf is correct, but it is about something different. If you disagree and if you insist, i will talk directly to the IEEE guy, i'm sure he does understand.

But no matter what, you may notice even from that document, power is always  P = V * I  at any moment of time, this doesn't depend on what circuit or a part of it is measured.

If we measure power at an instant of time, then it's always  P = V * I . If we don't measure power at an instant of time, but measure like the amplitude of a sine waveform instead, then we have to multiply it by cosine fii, this comes from the waveform, not that power is any different. And even when we measure sine waves, no matter what circuit or part of circuit we measure, no matter whether it contains coils or contains no coils, power is always measured the same way, not different when it is a coil or the circuit or a part of it contains a coil, as you insist. It is never said in that pdf either, that a different equipment or a method of measurement should be used depending whether the part of circuit measured contains a coil or not, read by yourself if you don't believe.

[citfta]

P = I * V is not correct if the current and voltage are not in phase.

Ayeaye it is very clear you have a great ability to write programs and make graphs.  But unfortunately it is also very clear you lack some basic understandings about inductors.  Every first year electronics student learns in the Elementary AC and DC class about ELI the ICE man.  That is a little saying to help you remember the difference between an inductor and a capacitor.  The ELI part stands for an inductor. That is what the L means.  You can also see that the E (voltage) is before the I (current).  That is because the voltage always leads the current in an inductor.  In other words they are NEVER in phase in an active inductor.  Therefore the formula P = I * V is incorrect.  By active I mean while the current is changing.  Once the inductor current reaches peak there is no more change and the inductor is then just a resistor.


In a capacitor the current leads the voltage and therefore we get the ICE part.

Respectfully,
Carroll

[ayeaye]

P = I * V is not correct if the current and voltage are not in phase.

Holy cow, phase is only important when we measure a whole waveform, like when we measure the amplitude of a sine waveform and know that the form of the signal is a sine. But in this experiment we measure the power at every *instant* of time, during the waveform, not the whole waveform. And at every instant of time  P = V * I , at every instant of time, this is not true like for the amplitude of the sine waveform during all its duration.

I know what you are talking about, i learned about it in the university too. But i didn't learn electrical engineering, i learned automatic control, and there is often important to do measurement at every instant of time, and do calculations from it. Say you regulate a dc power voltage, it can change in every kind of ways, sure you cannot measure power by sine or any waveform, this and everything else has to be calculated at every instant of time.

To show you how the power calculated from the amplitudes of sine, comes from the instantaneous power, let's calculate a power of a sine signal, with amplitude 1, so that the voltage and power are shifted by 20 degrees. Using the following bc script, you can try it. We calculate it during a time 10000 seconds, and then calculate the same for the sine waveforms by  Vrms * Irms * cos(20 degrees) . The amplitude of both our voltage and current is 1, but by that equation both have to be root mean square, so it becomes  rms * rms * cos(20 degrees) .

pi = 3.1415926535897932384626433
pi2 = 2 * pi
#20 degrees to radians
r20 = 20 / 360 * pi2
rms = 1 / sqrt(2)
energy = 0
for (t = 0; t < 10000; t += 0.1) {
    current = s(t)
    voltage = s(t + r20)
    power = voltage * current
    energy += power
}
energy *= 0.1
power = energy / 10000
print "Power when measured at instants of time: ", power, "\n"
power = rms * rms * c(r20)
print "Power when calculated from the amplitudes of sine: ", power, "\n"
quit

The output of that script is the following. As you see, the results are quite close, and the results become closer the longer is the time over which we calculate it.

Power when measured at instants of time: .46983333927318779782
Power when calculated from the amplitudes of sine: .4698463103929541\
9202

So as you see, the instantaneous  P = V * I  is correct, different from that said.