Yes Wattsup.
This is the question that MH asked you and EMJ to answer.
MH answered the question(after a modification to the question),and got it wrong.

I really did not notice that question because at certain times in this thread things got out of hand and I just jump pages to not have an upset stomach. So sorry for missing it.

So feel easy Wattsup,as it was a question MH him self could not answer correctly,and all the bad karma he threw your way,and the gloating he made throughout many threads regarding this question,simply can now be dismissed as garbage.

Anyone can ask a wrong question. So let's try to fix the question. You cannot have an ideal coil if it is to be 5 Henrys, since at such a rather high inductance there has to be "some" resistance. Then you cannot use an ideal voltage because the voltage would always remain the same regardless. So both the voltage and coil cannot be ideal.

So let's  just say the voltage source is non-ideal at over the highest voltage reading in the question which is 4 volts at 3 seconds. So let's give a non-ideal voltage of 12 volts to start and we know this is DC.

So........ I can only answer this question under Spin Conveyance and not under EE Electron Flow. In SC, when you have a DC source, both polarities enter the coil as I had clearly shown in my HCS videos. So if the negative polarity was connected to the coil first, that negative polarity is already at the other end of the coil. When the 12 volts positive is then connected you should get a quasi voltage spike (semi-short or damped short) first then it could indeed drop to 4 volts for about 3 seconds (depends on the available amps) as the positive pushes the negative potential back up to the center of that fat coil. But then the voltage would never drop to zero during the next 2 seconds and show instead a gradual rise to 12 volts again (or almost 12 volts) as both polarities fight it out in the center of that coil. Eventually the thin wires at the center of this high inductance coil having no thermal release will just pop and you will then see 12 volts forever. If you dissect the coil it should show a break near the center.

So............ If anyone has a 5 henry coil and a 12 volt dc source, let's place our bets. hehehe

So........... this just shows only one great formula.......... "ideal" = "unreal". You can use these ideal "mentalations" for years and wind up in the looney bin. For me they have always been useless methods of thinking that you just learned something when in actuality they are designed to reinforce a construct by distracting your real force of logic. Saying "let's pretend" only produces fairy tales. I prefer reality.

But regardless, I think it is better to not make a mountain out of a mole hill and just leave it at that. You know, I still have full respect for @MH just for sticking it out with us OU crazies as well as you and just consider this whole quagmire a better proof of the effects of chemtrails. Maybe some chelation therapy would be in order. You guys both need to step out of the boxers ring and let those ring girls walk around a bit. Some quality time off would be great for all. Ding ding game over. hahahahaha

wattsup

Yes the game is really over and for me there are just two outstanding issues:




MileHigh

1.  Brad admits that he is wrong about my response to the harder question.

No,i will do no such thing,as your answers are wrong when an ideal coil/inductor,and an ideal voltage are used as per your description in your question.

2.  Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands what he is doing.

Perhaps some learning of your own MH.
1-Please describe what an ideal inductor is.
2-Please describe what an ideal voltage is.

What's been going on for the past six years?  Even The Shadow can't answer that one.  I intentionally answered a more difficult question in order to leave the original question up for discussion among the regulars.  Again, we are talking about a very simple circuit here consisting of a power source and one single solitary component.

First of all,you are talking about a fictional power source,and a fictional component.
But that aside,you question clearly states-an !ideal! voltage across an !ideal! coil/inductor.
This is the question you asked,and that is what my answers refer to.

My answer to your original question stands-you cannot place an ideal voltage across an ideal inductor--or coil,as you have chosen to use.

As the inductor is ideal,there is no resistance,and so T is instant.
An ideal voltage of 3 volts placed across a resistance of 0 ohms results in an instant current with an infinite value,as there is no resistance to the flow of that current,nor is there a drop in value of you ideal voltage.

For both Brad and Wattsup, I am not holding out much hope for either of you but there are always the high-odd payouts for the long shot bets.  You guys would be the winners, but nothing ventured nothing gained and no pain no gain.

The more rubbish we get rid of-the likes you have tried to peddle in this post,the further man will go when dealing with truth's.

Shall i post some links to ideal inductors and ideal voltages MH,or are you able to do some research of your own.?.
Will you post the resistance value for an ideal inductor?.


Brad

@Mag 
lol definitely not books =). I'm just not understanding the electron differential you speak of not out of disagreement, but sheer ignorance. I don't see it as counting electrons so to speak, but measuring imbalance in charge. but maybe this is what you are saying by differential. the electron ratio to positive charge which produces a voltage. Well I'm saying in reality, we shouldn't lose any electrons like we shouldn't lose any air or water. and we apparently don't. we do lose energy though, rather have to expend some more in the process. so long as those electrons are moving and there is a current, that voltage will not only drop due to resistance, but due to the 'pressure' equalization so to speak when the joules are spread out like jelly long the capacitance of this extra capacitor introduced, which is like another way of saying the electrons are rearranging yet the differential remains the same.

 it's like, yea. naturally I want to say it in essence we are basically saying, what happens when I cut this in half? does it form 2 equal halves? sure, why not? I'm just not necessarily saying the voltage unit would be cut in half imo. I'd assume the same total energy, or joules, or whatever you want to call it that was stored, was cut in half.  so basically I think it COULD end up as 5v/5v so long as stored current was adjusted. how would we measure that? electrons I assume. I just don't necessarily think it 10v has to end up as 5v/5v after the dump using equal caps because it seems to work out that way in our other analogies in nature. it most certainly could though and probably does.

You raise important details as I go on.......

"but maybe this is what you are saying by differential. the electron ratio to positive charge which produces a voltage."

Yes, I probably should use the word ratio instead of differential. 


"Well I'm saying in reality, we shouldn't lose any electrons like we shouldn't lose any air or water"

Water is not a good analogy for pressures like air and charge. Water is not compressible. Air and electrical charge are compressible.


"we shouldn't lose any electrons like we shouldn't lose any air or water."

Ok, here is where I have not went into detail......

If we have 2 caps, one 10uf and the other 100uf, the 10uf cap plates will have less total atoms than the 100uf.  With no charge, there are no excess electrons on the neg plate and no electrons taken from the pos plate.  When we add an electron to the neg plate and take an electron from the pos plate of each cap, the 10uf cap will have more voltage/pressure than the 100uf cap. And if it were a 1uf cap the voltage/pressure would be even higher, just with the 1 electron difference for each plate.  So if we had a theoretical pair of tiny caps and the smaller cap had say 100 atoms and the larger had 1000 atoms for their plates, when we take one electron from one plate and put it into the other plate, the smaller cap will have more pressure than the larger cap. Similar to a small air tank and a larger one, it would take less atoms of oxygen pumped into the smaller tank to reach say 100psi than it would to get the larger tank to 100psi. In both examples, the containers determine how much is needed to be pumped in to get to a particular pressure or voltage.

So when we take that ratio of electrons that are in the 10uf cap at 10v, we are letting that same ratio of the same number of electrons to be in a larger cap/container by doing the cap to cap connection. So now we have the same 'total' number of excess electrons on the neg plates and the same number of electrons missing from the pos plates, but they are divided between the to caps, which if they are now in parallel equals a larger capacitor/container. So pressure/voltage is reduced. It is not resistance that caused the loss, it was the fact that we put our initial pressure into a larger container, there by reducing the total pressure. The only loss is that we didnt use the action of the transfer that happened to do work.



" it's like, yea. naturally I want to say it in essence we are basically saying, what happens when I cut this in half?"

Ah. Thats where you are stuck.  We are not cutting the capacitor or air container in half. We are dividing that pressure into a larger container of twice the size. that is why we have reduced pressure/voltage. 10uf cap in parallel with a 10uf cap equals a 20uf cap. A 10lb nitrous oxide bottle and connected it to another empty bottle, there would be 5lb in each. But we lost total pressure, even though we still have our total amount of gas in weight. We lost the energy of pressure. Resistance had nothing to do with the loss. I cannot use the 10lb of nitous in my car if the pressure were 0psi. As a further example, if we were to have a large enough container that if we sucked out all gas to a perfect vacuum and put in that 10lb of nitrous till it was 16lb, equal to ambient sea level air pressure, the gas will not come out when we open the nozzle. That 10lb of nitrous(about $35) is not usable.



"Well I'm saying in reality, we shouldn't lose any electrons like we shouldn't lose any air or water"

This is the point Im making...  If we did the cap to cap transfer and we start with 10v in the first cap and 0v in the other, the only way we can end up with the total amount of initial energy in the 2 caps total would be to end up with 7.07v each.  So that would mean that we would need a different ratio/number of electrons on the cap plates.  We would have to introduce more electrons in the system, or pull more from the pos plates and put them into the neg plates through some mechanism of doing so. So the cap to cap dump can only end up as half, .5, of the total electron ratio once spread between a container of twice the original size. How could we ever see a .707 in each of 2 caps when we started with 10 doing a cap to cap equalization?

If we have a 10uf cap at 10v and another cap 10uf at 0v, and instead of a direct cap to cap dump, we use an inductor between the caps and watched it in slow motion, we would see the field build in the inductor as current flows. The inductor acts as a flywheel. It stores energy as the current increases. It will pump the remaining electron ratio from the source cap into the receiving cap till the source is 0v and the receiving cap is 10v.  So now if we do the same and we cut off the source cap at 7.07v, the inductor is still charged, and we switch the inductor to be across the receiving cap when we disconnected the source cap, the inductor will pump electrons from the pos plate of the receiving cap into the neg plate till the receiving cap is at 7.07v and we disconnect the inductor. We will now have the total amount of energy in the 2 caps as we did in the source cap to begin with. Thats because we used the energy of the transfer to do work, where the direct cap to cap we did nothing with that work potential. Resistance has nothing to do with the loss because there is no other outcome no mater the value of resistance introduced whether it be 1Mohm or 0ohm, we will still end up with half of the electron ratio in each cap. The total number electron imbalance will be the same, just divided between the 2 caps as reduced pressure.


Mags

What if the other cap is adjustable??

So you start it out at 0.0MFD,, connect the caps together and then adjust the cap slowly up to the same value as the other one,, while doing this the current flow could be used in some sort of inductive fashion so that when current flow stops the collapsing induced field returns the energy it has taken,, that would be the energy not used but would be wasted,, and tops up the adjustable cap.

What if that slow induced field had a secondary interaction with something that allowed for the collapsing field to have a faster rate of change in the flux density??

If the cap is adjustable say 1uf to 10uf, then when you charge it at 10uf to 10v, the voltage goes up when you adjust the cap to a lower value. the electron ratio and count number remains the same. We have just increased the pressure by changing the value of the cap. The energy would remain the same, only the voltage value and capacitance value would change. Good question, as it helps my explanations.

Mags

Here I have edited one of the previous pics to describe what Im saying....

The first pic is of 2 caps, same value, both at 0v.

The second pic is of the left cap charged to a particular voltage.

The third pic is after cap to cap connection.

This is to show that the 'total number of electrons in both caps' is the same throughout the process and just shows the ratio when charged and then cap to cap connection. We never lose or gain electrons in the system. They are only moved or say pumped from one plate to the other, but while charging the total remains equal. One in and one out at a time to say as an example. So if we do a ratio electron count, and we know the other variables as in the caps capacitance value and the electron ratio change in the left caps plates when charged, then we should be able to calculate the voltage in the left cap when charged. Every time. Then when we do the cap to cap deal, we again should be able to calculate the voltage in each when they are equalized. And each time we do the count, we should be able to do the math to find the voltage.

So in an ideal system, how could 0ohm vs resistance(any value) affect the electron count values changing the voltage outcome for each cap? Where did the extra electrons come from to have 7.07v in each cap from 10v initial charge in the ideal model vs 5v on each cap in the resistance energy losing model? Thats the big question.

If we get the same outcome, ideal or real world, how could we say we lost half the energy in heat, if there wasnt any heat in the ideal model??

Ok. Im done for now.  I cannot further explain it. Let me know where I am going wrong, if you can.

Mags

One more thing...

"If we get the same outcome, ideal or real world, how could we say we lost half the energy in heat, if there wasnt any heat in the ideal model?"

Soo, if the resistance is NOT the cause of the loss, did we get the heat if any for free? ??

Mags

What if the actual electron count 'equalization' in the ideal system were as I show and it did give us 7.07v in each cap from the precharged 10v cap. Then we would have to had lose some electrons somewhere to be at 5v on each cap in the real world system doing an electron ratio count. If so, where did they go? If so our total electron count would be depleted somehow, then how? Certainly they are not stored or trapped in the resistor/resistance or they would become negatively charged components in the system. Also if we are losing electrons from the system then our whole system would be positively charged due to missing electrons. 

It doesnt make sense. the only thing I can say we lost is pressure and resistance didnt cause the loss. I mean, that is the unit of storage in this case, pressure.
1MOHM resistance, 50% loss. 1 ohm resistance, 50% loss.  1 pico ohm resistance, 50% loss. .0000000001pico ohm, 50% loss. Always 5v in each cap from 10v. But ideal caps are said to be zero loss.


Mags

Mags, I think something is wrong here.

How many electrons are in a charged cap of 2.7 volts and 3,000 Farads?

How many electrons are in a charged cap of 2.7 volts and 100 pico farads? 

According to what I think you are saying, the amount of electrons would be the same which is not possible in my opinion.  I think the problem is that I was taught that you need to consider capacitance and not just volts or amps.  My point is that 2 caps of the same voltage could have a varied capacitance and hence a totally different amount of stored energy. 

If I am mistaken in your taking voltage as the measure for how many electrons are in a cap, then I am sorry and must have misunderstood.

Bill