**1072**] ??

**Go Down**Interested in more conceptual errors ... err ... theories?

**Here it comes:**

Let's say I have two capacitors each 1?F charged with 10 volts each, so the overall stored energy equals 100? Joules. Then I discharge each capacitor into a coil, so I have a magnetic field in two coils accordingly. Next I connect these two coils in series so that the back voltage adds up to 20 volts. Now I take these 20 volts and charge a 1?F capacitor with it. Then as a result I should have a capacitor of 1?F at 20 volts, so the stored energy should equal 200? Joules.

Therefore: 2?F/10V becomes 1?F/20V which equals an **energy gain of one hundred percent** (in an ideal circuit).

But let me guess: There has to be something wrong with this conception, hasn't it?

#### Free Energy | searching for free energy and discussing free energy

...I have a magnetic field in two coils accordingly. Next I connect these two coils in series so that the back voltage adds up to 20 volts.The coils do not store energy as voltage. Ideal inductors store energy as electric current/magnetic field ( W= ?LI

^{2}) and they don't lose that energy only while they are shorted.

What do you mean by "back voltage" ?

Two magnetic fields in two coils are generated by two charged capacitors simultaneously. As long as there are no changes made in the circuits then each magnetic field of the coils will create a back voltage (Back EMF) and will recharge its own capacitor in an oscillation (with slight losses). That would be the behavior of an ordinary LC circuit. So far, so good.

But as soon as the voltage of the capacitors is zero (zero crossing) the coils are disconnected from these capacitors and connected in series to a single capacitor of the same capacity as one of the initial two capacitors. Hence the series back voltage (Back EMF) of the coils should have enough power to charge this single capacitor to the combined voltage (with slight losses) previously received from two capacitors each of the same capacity. Since the energy (Joules) stored in a capacitor is a linear function of the capacity but a square function of the voltage, half the capacity but double the voltage means twice the stored energy (as shown by means of the Capacitor Charge Calculator).

In other words: Nothing is gained by connecting two parallel charged capacitors in series. The voltage doubles but the capacity is divide by four so the stored energy stays the same. But transferring the energy of two capacitors to the magnetic fields of two coils and then connecting these coils in series while receiving the Back EMF the outcome could be somewhat different.

The remaining question is: How to do this practically?

Has anyone ever gotten these kind of sparks before.? I took these off of the video from the homepage of our site here.? I don't know if its the metal he is using, but they are very much like the energy sparks that Tesla observed from his HF Pulsed DC experiments with Aether Energy in his magnifying transmitter.? It was Cold electricity that was totally different from regular hot currents.? I don't think this is, but they are nice sparks between his gap and grab pretty far.? Of course he is using like 6 amps..? He doesn't say why he made this device or what use it has...

Interested in more conceptual errors ... err ... theories?Here it comes:Let's say I have two capacitors each 1?F charged with 10 volts each, so the overall stored energy equals 100? Joules. Then I discharge each capacitor into a coil, so I have a magnetic field in two coils accordingly. Next I connect these two coils in series so that the back voltage adds up to 20 volts. Now I take these 20 volts and charge a 1?F capacitor with it. Then as a result I should have a capacitor of 1?F at 20 volts, so the stored energy should equal 200? Joules.

Therefore: 2?F/10V becomes 1?F/20V which equals an

energy gain of one hundred percent(in an ideal circuit).But let me guess: There has to be something wrong with this conception, hasn't it?

Capacitors are non linear devices ie. if we take a 1000uF capacitor and charge it to 100 volts we have 5 Joules of potential energy.

If we then charge the 1000uF capacitor to 200 volts it will have 20 Joules of potential energy, which is 15 joules more for the same voltage increase.

Then to charge the 1000uF to 300 volts it will have 45 Joules of potential energy which is 25 Joules for that 100 volt increase.

One 2uF capacitor at 10 volts is 100 microjoules so two capacitors equals 200 microjoules and a 1uF capacitor at at 20 volts is 200 microjoules so that's unity.

Cheers

I think this is the best representation of the capacitor question you posed.http://www.csun.edu/~gsl05670/labs/cap_plate_sep_volt.htm

It works with voltage and charge. Not energy persay.

Energy in my opinion has an active definition and not a standing definition..

?Active meaning to do something and standing meaning potential to do something or hasn't done anything yet.

?Anyways see how they talk about charge and voltage. Where they store charge in the electric field?

?My take on this is that the charge is based? in the structure of the object and represented on the surface of an object. The electric field allows those charges to be represented on the other surface. I don't think the field stores anything. The field organizes and allows charges to pose or influence the other side of the space via the electric field between the plates.?The easiest representation of a capacitor is a diaphragm. Nothing actually moves from plate to plate this is accomplished by the ability of the electric field to originate from one charge structure and motivate the other plate via the electric field. It is a static device. No current moves through the device and if it does odds on are if it does conduct it is a dielectric break down and ruins the capacitor. Some capacitors have a self healing ability, such as in the case of the electrolytic capacitor. These breakdowns can be seen if you take the cap apart. There will be little scars on the surface of the plates where the oxide was burnt away. The healing type of caps like the electrolytic caps use the liquid inside to repair the breach and re-oxidize the breach.

?As you can see there are a Great many ideas that are not agreeing with one another in our scientific theories. This waffling back and forth of accepted theories only goes to confuse us and most are designed to do just that.This is the crux to our problem. And hence is the problem we all face. No wonder many are confused about this stuff.

Commonly that is called

**Potential**energy for the non active, I think. No Biggie. Semantics. Displacement current is how capacitors work.

Cheers

#### Free Energy | searching for free energy and discussing free energy

In Tesla's Apparatus for producing currents of high frequency and potential, he charges the capacitors in parallel and then discharges them in series.

#### Free Energy | searching for free energy and discussing free energy

The coils do not store energy as voltage. Ideal inductors store energy as electric current/magnetic field ( W= ?LI^{2}) and they don't lose that energy only while they are shorted.

What do you mean by "back voltage" ?

I believe that Tesla's Bilfilar pancake coils used as transformers can store and/or magnify the voltage in the coils..

Capacitors are non linear devices ie. if we take a 1000uF capacitor and charge it to 100 volts we have 5 Joules of potential energy.Of course capacitors are non linear (voltage/energy). So this does not contradict anything regarding my calculations.

If we then charge the 1000uF capacitor to 200 volts it will have 20 Joules of potential energy, which is 15 joules more for the same voltage increase.

Then to charge the 1000uF to 300 volts it will have 45 Joules of potential energy which is 25 Joules for that 100 volt increase.One 2uF capacitor at 10 volts is 100 microjoules so two capacitors equals 200 microjoules and a 1uF capacitor at at 20 volts is 200 microjoules so that's unity.

Cheers

Regards

Yes, Tesla pancake coil when used in a transformer, can magnify voltage. It is very unlikely, though,? that it would produce additional energy. Anything that relates to the previous sentence, written by Tesla, is not made public.I believe that Tesla's Bilfilar pancake coils used as transformers can store and/or magnify the voltage in the coils..

#### Free Energy | searching for free energy and discussing free energy

Two magnetic fields in two coils are generated by two charged capacitors simultaneously. As long as there are no changes made in the circuits then each magnetic field of the coils will create a back voltage (Back EMF) and will recharge its own capacitor in an oscillation (with slight losses). That would be the behavior of an ordinary LC circuit. So far, so good.Yes.? Also, it should be mentioned, that after the voltage zero crossing, the capacitor will be recharged to an opposite voltage (-100%) at the next current zero-crossing.

But as soon as the voltage of the capacitors is zero (zero crossing) the coils are disconnected from these capacitors and connected in series to a single capacitor of the same capacity as one of the initial two capacitors.That would be difficult to engineer, because any disconnection and reconnection of the coils in different configuration means interruption of the current flowing in the coils.? The coils will fight this interruption by creating such high emf across the break that the current is maintained.? A lot of energy will be dissipated in the resistance of the break very quickly.? This will persist until all of the energy stored in the coil is exhausted.?

However, in theory it is possible to quickly switch the coils to a series connection without incurring losses.

Since the energy (Joules) stored in a capacitor is a linear function of the capacity but a square function of the voltage, half the capacity but double the voltage means twice the stored energy (as shown by means of the Capacitor Charge Calculator).True.

Hence the series back voltage (Back EMF) of the coils should have enough power to charge this single capacitor to the combined voltage (with slight losses) previously received from two capacitors each of the same capacity.The combined inductance of the two series coils (L

_{2}) is twice the inductance of a single coil (L

_{1}) in other words L

_{2}=2*L

_{1}.? The current (I) flowing through the coils does not change after connecting them in series . (if it did - we'd have a huge problem).

The energy stored in the two series-connected coils (W

_{2}) is also twice the energy stored in a single coil (W

_{1}) because the energy stored in an inductor is a linear function of its inductance but a square function of the current.? The relevant equations are:? W

_{1}= ?L

_{1}I

^{2},? W

_{2}= L

_{1}I

^{2}, W

_{2}=2*W

_{1}

When you go through the instantaneous differential equations, that energy stored in the two series-connected coils recharges the capacitor to -141.4% of the original voltage, not to -200% of the original voltage

The conceptual error here is in the phrase "combined voltage".? The combination is not a sum.?

This can be understood by considering that emf is proportional to current change rate (dI/dt) and two series-connected coils collectively possess twice the inductance (L_{2}=2*L_{1}). Inductance resists the change in current, thus the current in those series-connected coils changes slower than in the single coil (because L_{2} > L_{1}), even if the capacitor is the same (has the same capacitance).

Slower rate of change of current means smaller emf - that's why the emfs do not add up

Of course capacitors are non linear (voltage/energy). So this does not contradict anything regarding my calculations.Yes. Also, inductors are nonlinear (current/energy).? These nonlinear relationships with stored energy, do not contradict your calculations.

Unfortunately, the addition of series-connected inductances and emfs, does

The energy stored in the two series-connected coils (W2) is also twice the energy stored in a single coil (W1) because the energy stored in an inductor is a linear function of its inductance but a square function of the current.Nice. Then all the energy is stored in the coils. That's better than nothing so far.

Inductance resists the change in current, thus the current in those series-connected coils changes slower than in the single coil (because L2 > L1), even if the capacitor is the same (has the same capacitance).This sounds quite not complicated enough. Therefore I'm daring to ask what could happen (in terms of differential equations) if both coils would share the same core? What could happen if the coils are bifilar wound? What could happen if the core goes into saturation? What could happen if these are air coils without a ferromagnetic core?

But anyway I would not rely too much on differential equations because something must be wrong with these equations, otherwise there would be neither the presentation of the Kapanadze device nor of the Stepanov transformer nor of the Steven Marks TPU (not mentioned the story of Tesla's electric car).

So we should keep looking for the one differential equation that is wrong or missing.

Any hints are welcome.

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My take on this is that the charge is based? in the structure of the object and represented on the surface of an object. The electric field allows those charges to be represented on the other surface. I don't think the field stores anything. The field organizes and allows charges to pose or influence the other side of the space via the electric field between the plates.http://www.d3face.com/

Posted by: 璇 周 | 03/26/2013 at 02:19 AM