Brad:

Still hung up on this nonsense are you?

Here is the question:

You have an ideal voltage source and an ideal coil of 5 Henrys.  At time t=0 seconds the coil connects to the ideal voltage source.  For three seconds the voltage is 4 volts.  Then for the next two seconds the voltage is zero volts.  Then for two seconds the voltage is negative three volts, and then for the next six seconds the voltage is 0.5 volts.  Then after that the voltage is zero volts.

The question is what happens starting at t = 0

Let's change it up and make it more difficult, and revamp the question:

You have an ideal voltage source and an ideal coil of 5 Henrys.  At time t=0 seconds the coil connects to the ideal voltage source.  The voltage source waveform is 20*t^2.  So as the time t increases, the voltage increases proportional to the square of the time.

The question is what happens starting at t = 0

The answer:

The current through the ideal coil starts from zero at time t = 0 and then increases with this formula:  i = 1.33*t^3.

Time..........Voltage.........Current
0...............0.................0
1...............20...............1.33
5...............500.............166.67
10.............2000............1333.33
20.............8000............10666.67
50.............50000..........166666.7

There you go, harder question answered.

So why don't you go reassemble your device for John and even take a shot at the original question.

MileHigh

And there you go.
I knew you could not answer the question correctly--you are way off.

Care to have another go MH?.

Some hints for you.
1-An ideal voltage is one that dose not change in selected value.
2-An ideal inductor has no resistance nor capacitance.
3-What is the current value of 1 volt across a resister with a value of 0 ohms ?.


Brad

I am not sure what to say, I am not sure you correctly processed what I posted.  Perhaps try again tomorrow when the neurons will be firing differently?

<<< 1-An ideal voltage is one that dose not change in selected value. >>>

After six years you are lucky that you still have people that want to work with you and make attempts to help you.

Magluvin:

There is an answer to the conundrum with ideal electrical components.  We do have to add the ideal diode, but what you missed was the impossibly infinitely fast switching function.  I will explain but first let's go back to the air tanks.

In the air tank example, and in fact using an unreal model where we ignore temperature for a second and have idealized components, when one tank discharges into the other tank, that spins up an ideal flywheel pump.   So you can set it up such that when the first tank drops to 70.7 psi, you then stop the air flow out of the first tank, and then the ideal flywheel pump takes over and pumps extra air into the second tank so that it also reaches 70.7 psi and therefore no energy is lost.  The net result is no energy was lost and extra air was pumped into the second tank by the ideal flywheel pump.

So, let's do that with ideal capacitors and ideal diodes and an ideal switch.

As Poynt said, an ideal capacitor connecting to an ideal capacitor is a no-no.  You were absolutely right though about the conundrum of "missing" electrons to get to 7.07 volts in each cap.

If you have two ideal caps, and an ideal inductor connecting between them, that is a manageable situation with no ripping of space-time.  In this case, Cap A discharges into Cap B via the coil and the charge goes back and forth forever.  There are no extra electrons and there is never a condition with 7.07 volts in each cap.  The best you can get is this in terms of equal voltage something like 5 volts in each cap, and current flowing through the coil.  I am not saying it has to be 5 volts in each cap either, just an equal voltage in each cap.  The coil discharges into Cap B and stops discharging when Cap A has zero volts and Cap B has 10 volts.  Then the whole process reverses.  During that infinite back-and-forth cycle there will be an instant in time where there is an equal voltage in each cap and the "missing" energy is in the coil.

I will do another post to explain the two caps with 7.07 volts each.

MileHigh

Ok. Thought about what it is I need to say to show my reasoning..... At least you said " You were absolutely right though about the conundrum of "missing" electrons to get to 7.07 volts in each cap."


What I had gotten from this discussion way back was that when we discharge a full cap into an empty directly, we lose half the energy 'because' of resistance. I frown on that. I can agree that some heat will occur because of the resistance and very high current flow during the transfer, but I fail to see that the heat generated caused the loss.

Like the ideal scheme. If we had 2 ideal caps, super conductive to say, one at 10uf at 10v, and the other at 10uf 0v, and we do the dump, Im wondering why there would be an unfathomable explosion or what ever when we hit the superconducting switch. In fact, there should be no heat generated at all with ideal caps and switch because there is no resistance. So if it says so in the book, Id like to read that book.

Now furthermore, if we have 2 ideal caps and dump the 10v cap to the 0v cap, there still should still be 5v each, considering the electron count measurements I described earlier. It seems the 'loss due to resistance' is made up for unknown reasons and we are just suppose to agree.  Well I cannot. Unless, whom ever made that statement back then 'meant' the energy lost is equal to the heat generated. But then there is still the blame on the resistance for the loss. So what Im saying is, it isnt the 'resistance that causes the loss', it is the fact that we expanded the electrical pressure into a larger container haphazardly, with which in the end, we wind up with only half of the usable energy. The energy wasnt simply used up as heat. it was reduced by letting the pressure change value and containment size get larger, and heat was generated because of the resistance.

So the energy that we lost in the transfer and equalization was due to stupidity of doing so by not using the flow of pressure from one cap to the other. Dumb, stupid, ridiculous thing to do.    

This may be showing that conservation of energy may not be all its cracked up to be. None of us(I dont believe) have ideal caps or switches or even inductors to test this. We are just told it is so. So there is no real reason why it cannot be questioned in such that I have.

Sure we can say that if we use an ideal inductor and ideal diode that we could convert all the energy from the 10v ideal cap to another 0v ideal cap and say that it would be a 100%transfer. But we can just about do that now with regular components, and eliminating the diode and replacing it with timed switching like I did in sim. Even with all the crappy resistance in the caps, the wires, the inductor and switches, we can come damn close to full transfer. So it leaves me to think that if the superconductor components are so superior, then why might we only be gaining that little bit we lost with the regular components? Maybe it can do better than that?? How would we know without access to these components. The best thing Ive seen with super conductors is we can float them above a magnet as long as we keep it super chilled. Probably wasting more energy chilling the thing than it would take to levitate a normal object of the same weight.


So really what Im looking for is the answer to the conundrum that you actually agree with me on. If you and I can agree that we cannot produce 2 caps with 7.07v each from a 10v cap 'by direct dump', then where would we lose half of the source cap energy in an ideal scenario by doing a direct cap to cap equalization? No heat. Where did it go? Excluding using an inductor and diode/timed switching, as it is another subject I will bring up after this cap to cap deal.


Mags

 

@Mag

you know it's definitely interesting thinking about it, I just don't get how it could mathematically be worked out that there is some gain when placing a charged capacitor in parallel with a discharged capacitor. The air tank analogy is definitely similar, but I don't think we can say that a battery, or capacitor, or any electrical power 'source' is under the same kind of pressure. it's actually a good analogy for voltage I guess, like water pressure in pipes, but when the charged caps plates coalesce with the neutral cap isn't it just a balancing-out transference at electron speed, and not the result of some form of actual pressure like that of air? when it comes to counting electrons, isn't the 'empty' cap more like a container that already has electrons in it just like the empty tank has air in it before being fed pressurized air from another tank? In ideal terms where we have a 100% transference of electrical energy from one place to another then it seems like we answered our own question. If we are talking about a current flow happening between one cap to another, and assuming there is 0 resistance, then if we're going with the flow and also assuming current will actually pass over 0 resistance in the real world then I don't think a 10v cap would have to balance out to 5v to have 'equalized the pressure' so to speak.

" I just don't get how it could mathematically be worked out that there is some gain when placing a charged"

I havnt said there was gain. Yet.


"The air tank analogy is definitely similar, but I don't think we can say that a battery, or capacitor, or any electrical power 'source' is under the same kind of pressure. it's actually a good analogy for voltage I guess,"

A cap yes, battery no. If the cap remains in perfect shape, it should always give accurate calculable results.

If we had an air tank that was so small that we could only pump in say 100 max atoms of oxygen, then there must be a determined psi for each number atoms of oxygen added or subtracted from the tank. Like 1 atom would be say 1psi and 100 atoms be 100psi.  So if we have a very tiny cap, each plate should be, if plates are exactly perfect/ideal, identical for the example, and equal number of electrons per plate at 0v potential, then if we take 1 electron from the + plate and add an electron to the - plate, there should be a voltage potential on the cap and it is considered charged to a particular potential.  Now lets say that the 1 electron taken and 1 given produces .0001v between the 2 plates, and we then discharge the cap to 0 v and do it all over again, would we not each time we charged the cap in such a way, end up with .0001v, every time? If not, please explain why. So with the cap to cap idea, there should be a number of electrons taken from the + plate and electrons added to the - plate in order to get exactly 10v. So isnt it logical that if we had a 10uf 10v cap and we were able to count the electron differential between the + and - plates, that when we do the direct cap to cap transfer that if we end up with 5v in each when done that there should be half the number count of electrons missing from the positve plates, and half the number count of electrons in abundance on the negative plates, as compared to the electron count number of the original 10uf 10v cap? ?    If not, lol, then again, please explain why.


" If we are talking about a current flow happening between one cap to another, and assuming there is 0 resistance, then if we're going with the flow and also assuming current will actually pass over 0 resistance in the real world then I don't think a 10v cap would have to balance out to 5v to have 'equalized the pressure' so to speak."

Ok, then how do you come to that conclusion figuring an electron differential number count as described above? If in your example above you start with 10v and end up with an alternative voltage compared to 5v in each cap when done, how can you explain the electron differential number count that determines that voltage? And if Im incorrect, then please explain why. But if you cannot, not beeing snooty, then why do you believe what you do? Books? Pun intended.


Mags

@Mag

you know it's definitely interesting thinking about it, I just don't get how it could mathematically be worked out that there is some gain when placing a charged capacitor in parallel with a discharged capacitor. The air tank analogy is definitely similar, but I don't think we can say that a battery, or capacitor, or any electrical power 'source' is under the same kind of pressure. it's actually a good analogy for voltage I guess, like water pressure in pipes, but when the charged caps plates coalesce with the neutral cap isn't it just a balancing-out transference at electron speed, and not the result of some form of actual pressure like that of air? when it comes to counting electrons, isn't the 'empty' cap more like a container that already has electrons in it just like the empty tank has air in it before being fed pressurized air from another tank? In ideal terms where we have a 100% transference of electrical energy from one place to another then it seems like we answered our own question. If we are talking about a current flow happening between one cap to another, and assuming there is 0 resistance, then if we're going with the flow and also assuming current will actually pass over 0 resistance in the real world then I don't think a 10v cap would have to balance out to 5v to have 'equalized the pressure' so to speak.

Think. 

100psi to 50psi
10gal to 5 gal
20lb to 10lb
10v to 5v

Any recognizable likenesses?

How about.... 
10 billion to 5 billion electrons?

10billion electron differential in the source cap and 5 billion differential in each cap after the cap to cap deal is done. How could it be any different than 10billion divided by 2? 1 cap into 2 caps. 1 air tank into 2. If we could we count the number of oxygen atoms in the source tank at 100psi, then when we do a tank to tank transfer and equalization, would we have the total of the number of oxygen atoms in 2 tanks, half in one tank and half in the other? How could we ever expect the electron differential between cap plates to be any different?

10uf cap plates say has 10 billion excess electrons on the negative plate and 10 billion taken away from the positive plate, just as an example, then we do the direct cap to cap thing. When all is said and done, can you argue that those 10 billion excess and missing electrons of the source cap + and - plates would not end up being split up between the 2 caps once the caps equalize to 5v each? 5 billion differential for each cap would be the number, wouldnt it? If it is a different number, how did you determine that? Where did the extra electrons come from if the number is larger than 5bil for each cap? Or if your number is less than 5bil in each cap, where did some electrons disappear to?

Mags

C = Q/V

Q = CV

So if Q is conserved and you double the capacitance to 2*C, then for the equation to hold then V has to be halved to V/2.

However, you lose half the energy when this happens.  Some of the charge moves through the resistor and suffers a voltage drop and loses some of its "bang."

Ah yes. The voltage drop. I was just thinking about that. Are you reading my mind through my phone?  lol  None the less, its true that I was.

Its funny the voltage drop across any value resistor doesnt change the outcome. It just changes the time till equalization as you said earlier. But when you first consider it, it makes you think a bit, dont it? In this case, what I believe, is that the resistor does just that and isnt a loss in this cap to cap deal. But thats just me until convinced otherwise. As for the answers so far, Im still at point A.

Mags

I should be able to shed some light on this subject.  We often say "things go to infinity" whereas MarkE would use a more commonly used term in scientific parlance, "undefined."  I will make use of both terms here.

Simple thought experiment for the real-world shorting one cap to the other and losing half the energy.  If you use say a 100 kohm resistor say it takes five minutes for the two caps to be equal in voltage.  (There is semi-related conundrum about the two caps "never" attaining the exact same voltage.)  Then you switch to a 5 kohm resistor and it takes 30 seconds.  Then a 100 ohm resistor and it takes two seconds.

Here is the simple conclusion:  It doesn't matter what value of resistor you use you always lose half the energy.  Hence when you short two caps together with "no resistor" you still lose half the energy.

Now for the ideal caps.  You can't short one cap to the other because you get infinite current for zero seconds.  In other words it is undefined.  Ending up with the same voltage in each cap or 7.07 volts in each cap is a non-starter, because you can never get there.  So let's switch to plan B and put the ideal inductor between the two ideal caps.  Now the energy cycles back and forth between each cap forever.  As you lower the value of the ideal inductor in an attempt to simulate a short with a value for zero for the inductance what happens?  The cycling frequency gets higher and higher until it approaches an infinite cycling frequency as the value of the inductance goes to zero.  One more time infinity crops up, and hence you can say the solution for a value of zero for the inductance is undefined.

Going back to losing half the energy for a real-world shorting of two caps together, you are over analyzing the situation.  Voltage times current through the resistor equals heat power.  So you convert energy stored in the electric field in the capacitor into heat energy.  It's actually very mundane, nothing to do with expanding into a new capacitor or volume.

There is a good old visualization trick for this one.  When the 10-volt cap is shorted to the 0-volt cap you lose half the energy and both caps are at 5 volts.  You can visualize this like a totally inelastic collision.  That simply means that energy is burnt off when two things collide.  So imagine a stationary metal block being hit by a ball of putty with a velocity of 10 meters per second.  After they hit they move together at 5 meters per second.  Here is the thing:  The ball off putty deformed when it hit and stuck to the metal block.  The deformation process was resistive in nature, like bending a coat hanger, and therefore the metal block and the ball of putty heated up due to the collision.  There is the signature of the resistive losses for the capacitor example.

Here is another way to look at the same thing:  On an axle you have three things:  A flywheel, a remote controlled clutch, and then another flywheel.  The clutch is between the two flywheels.  You spin up the first flywheel to 100 RPM.  The other flywheel is not turning.  Then you press the button and the clutch engages and connects them together.  It takes one second for the clutch to fully engage.  The net result is that both flywheels are now spinning together along with the "weightless" clutch at 50 RPM.  When the clutch engaged there was friction between the clutch plates producing heat.

MileHigh

I actually understand these examples and it has helped me to visualize what is happening.  I really get the 2 fly wheels example...as soon as I saw where you were going with the clutch idea...I yelled at my monitor 50 RPM! 50 RPM!  Then I read on to the point where you said they were all turning at 50 RPM.  This makes total sense.

Thanks.

Bill

Mag thanks for further explanation, but at this point from what I gather it seems there's a confusion as to why there actually would be some 'loss' at all when charging one cap with another.

First, why is resistance not an option? we can call it a 'loss' due to heat, or a dissipation of heat energy, or whatever, but why would there be no heat generated in the transfer of current across a conductor? If we're talking about a 10v 10uF cap 'dumping' into another 10v uF cap, it seems the 'heat' would have been in the tiny spark you may see, because the ultra fast discharge of those amps would not persist long enough to generate a level of heat within the conductor you would notice any other way other than let's say dumping that same cap into an indicator filament. Everyone here has certainly welded the legs of a cap to some metal during a discharge before. Just saying, why is heat not really considered a factor?

Ideally, were we to create a '100% transfer', it would be more like instead of 2 caps, the one cap magically grew twice the size in an instant, thereby retaining it's charge but doubling in capacitance.. am I way off there?

Alternatively I imagine it like two cups, same height (voltage). One is exactly twice the volume of the other (capacity). What happens when we pour the same amount of liquid into each cup. Obviously the height of each cups liquid level will be different, analogous to the voltage level of the caps. So in that ideal scenario, a gain or loss in 'pressure' would also accordingly result in the loss or gain of sustainability/duration. same amount of water, like same amount of air. ultimately releasing either the air or the water from these varying containers may not net you the same pressure as before, but that just requires 'tuning' now because we still have the same total water and air.


the electrical representation of that tank with the venturi metaphor for me is like taking a large amount of energy to charge a cap up, then slowly dump the cap into another cap through a resistor, making up for any current dissipated as heat in the process by providing the remaining charge via a solar cell. So, to do that with our circuits, we just need to add a solar cell to grab some of that sun, or wind turbine to grab some air, etc..

OK, well, then what did we even do? other than just magically increase the cap size? we can't magically also double the power in this scenario so that just has to 'spread out' so to speak right?
why would the voltage remain the same if we didn't also magically double the energy in the cap?

If we are looking at pressurized air spinning a turbine, and opening a valve in the air line increases the torque on the turbine, then we have clearly found a way to increase the efficiency of our air pressure-to-turbine exercise where our goal is to torque the turbine. but does the venturi actually increase the air pressure through the pipe? If so, isn't that just like lowering the resistance in a circuit and cranking up the current, draining the source faster? if it does not increase the air pressure in the pipe, then does it not simply just create a better air flow for the fan right at the exit valve? I mean look no lie that's badass but isn't just sort of just like finding the right capacitor that made your Joule Thief 75% efficient instead of 40%?

 I don't see the real point of the cap argument because it's like saying we have to charge a cap initially.. but after that, we can facilitate this ideal transfer between that cap and another cap in some perfect resonance, where that same amount of energy just sloshes back and forth, yet in order to facilitate that process, we are introducing an outside energy that could also be used to do work, thereby ultimately recooping the energy it took us to get that initial cap charge. That's great but that energy came from somewhere else, not from within the system, so what's the difference between that and adding a solar cell? well, for one, I guess the solar cell will only work in the sun..  but to say the air flow and venturi analogy is more efficient because of that would be to ignore the amount of mechanical energy it took to fill that first air tank compared to the mechanical energy you would get out of it.