The easiest way to find out any parameter in a/this situation, is to remove one of the other factors influencing the over all values. [Algebra]
When you drop the temperature of the two caps to zero, or where they may act as super conductors, all resistance is gone.
When resistance is gone the two caps will equalize when connected, and no loss will be had in the transition. A cap's value is equal to a Charge. [Coulombs]
Because the magnetic properties are also affected by this temperature, no loss through magnetic effects will occur either.
You will simply empty one half of one bucket into the other bucket, and two halves still equal the same sum as you started with.
Therefore,...Any and all losses occur because of resistance.
All losses due magnetic effects and resistance have their own [reasons and ] formulae to find where the loss occurred, and are relevant directly to those properties and can be charted according to ohms law.
On another note, If you could set up a ripple in a superconductor ring, it would act like a “tankâ€
Switching energy from one cap to another
Hi Chris,
I think that if you connected two super conductor caps where one was
charged and the other wasnt (as original problem) there would
still be a loss somewhere because that would (supposedly) develop
an infinite current through the (superconducting) switch.
What i think would REALLY happen is once the switch was about to
be closed, when it gets down to the microinches from touching
the contact, a spark would occur and that would be across a
non-superconducting path and that would eat up some of the
energy, probably back to 50 percent again.
Make sense?
Thanks for your other input too...
Take care,
Al
I think that if you connected two super conductor caps where one was
charged and the other wasnt (as original problem) there would
still be a loss somewhere because that would (supposedly) develop
an infinite current through the (superconducting) switch.
What i think would REALLY happen is once the switch was about to
be closed, when it gets down to the microinches from touching
the contact, a spark would occur and that would be across a
non-superconducting path and that would eat up some of the
energy, probably back to 50 percent again.
Make sense?
Thanks for your other input too...
Take care,
Al
LEDs vs Bulbs, LEDs are winning.
-
Chris Smith
- Posts: 4325
- Joined: Tue Dec 04, 2001 1:01 am
- Location: Bieber Ca.
AL
We have to make one of two assumptions here for the argument sake to work.
ONE:
Either we have the absolute temperature or super conductor.
or
Two, we do add in a fact that we don’t or cant achieve this status perfectly?
Proposition one, the perfect scenario says we will have zero loss, no loss even in a arc because the temperature will affect air down to zero, or any type of gas surrounding the point contacts. This would still keep the loss due to heat it in the realm of 99.99% or better? [nothing is actually perfect?]
Proposition two is major loss included, but I doubt at near zero temperatures a loss of 50% would be realistic?
My guess here is still less than a few percent loss.
To visualize a perfect situation, take two tanks of the same volume and property.
One is filled with liquid oxygen in the pressure range of say 2000 PSI [or where ever it forms a liquid] and a tank size of one cubic foot of liquid.
The second tank is empty of liquid, but its still at 2000 psi. [or 1999 psi for argument sake, any point just below the formation of that liquid?]
Tap the two tanks together and open the valves.
What happens to the two tanks is they equalize perfectly because with out a leak, there is no loss of that liquid.
In the end you will have two tanks, each with one half cubic foot of liquid, and the pressure for argument sake will still be 2000 psi [or 1999.5 psi. ]
We have to make one of two assumptions here for the argument sake to work.
ONE:
Either we have the absolute temperature or super conductor.
or
Two, we do add in a fact that we don’t or cant achieve this status perfectly?
Proposition one, the perfect scenario says we will have zero loss, no loss even in a arc because the temperature will affect air down to zero, or any type of gas surrounding the point contacts. This would still keep the loss due to heat it in the realm of 99.99% or better? [nothing is actually perfect?]
Proposition two is major loss included, but I doubt at near zero temperatures a loss of 50% would be realistic?
My guess here is still less than a few percent loss.
To visualize a perfect situation, take two tanks of the same volume and property.
One is filled with liquid oxygen in the pressure range of say 2000 PSI [or where ever it forms a liquid] and a tank size of one cubic foot of liquid.
The second tank is empty of liquid, but its still at 2000 psi. [or 1999 psi for argument sake, any point just below the formation of that liquid?]
Tap the two tanks together and open the valves.
What happens to the two tanks is they equalize perfectly because with out a leak, there is no loss of that liquid.
In the end you will have two tanks, each with one half cubic foot of liquid, and the pressure for argument sake will still be 2000 psi [or 1999.5 psi. ]
Hi Chris,
Interesting ideas!
"Proposition one, the perfect scenario says we will have zero loss, no loss even in a arc because the temperature will affect air down to zero, or any type of gas surrounding the point contacts. This would still keep the loss due to heat it in the realm of 99.99% or better? [nothing is actually perfect?] "
Well, either we try to find every loss or we dont.
Also, how do you explain the current which would have to be
infinite? Ok, so then we would still have inductance, which by
the second circuit i presented would make a good oscillator,
and as already found out, we'd have to open the switch at
the perfect time in order to get full energy transfer. But, as
i also found out, if that inductance has any loss associated with it
(resistance or radiation or core losses) that brings us back to
square one again, where we lose energy...is it 50 percent?...
depending on how the core works we could lose more..what
if the inductor is a perfect antenna?..we lose everything
I know it seems hard to believe, but we have to keep in mind
that the current goes up the closer we approach zero resistance,
and if there is no resistance then the inductance limits the current,
and then we have to figure out how the inductor loses energy...
if the core isnt perfect than again we loose energy. Maybe less
if the core loss is low and radiation is low? But, if any resistance
in that inductor we loose 50 percent of the energy.
"Proposition two is major loss included, but I doubt at near zero temperatures a loss of 50% would be realistic? "
Well, if you draw up the circuit and place a small resistance in between
the two caps, no matter how low you make that resistor we still
end up with 50 percent loss. It looks like the reason for this is
because as the resistance goes down, the current pulse height
increases, so the loss stays the same (even though the time
goes down too, ie shorter pulse).
If you have a circuit analysis program you can easily try this by
connecting two caps by a resistor and grounding the other two
cap leads, and also setting the first cap's initial conditions to be
some voltage, like: ic=10v, or something like that. Then, close
the switch and watch the response of both caps, then compute
the energy once everything settles down. Try this with 0.1 ohms,
then with 0.01 ohms, then with 0.001 ohms, etc. Watch as
the resistance goes down the current pulse goes up and the
time to settle down gets shorter, but the energy in the two caps
once it settles is still 50 percent of what was started out with.
I guess if the inductance limits the current than if we open the
switch after one cycle a small resistance would lead to small loss.
The circuit damps out to 50 percent energy if allowed to run
until it settles naturally (ie no switch opening).
So for the circuit with an inductor we have to divide the cases in two:
case 1 where the circuit is allowed to damp out naturally, and
case 2 where we open the switch after only one cycle.
We then have to figure the energy for each case.
Make sense?
Interesting ideas!
"Proposition one, the perfect scenario says we will have zero loss, no loss even in a arc because the temperature will affect air down to zero, or any type of gas surrounding the point contacts. This would still keep the loss due to heat it in the realm of 99.99% or better? [nothing is actually perfect?] "
Well, either we try to find every loss or we dont.
Also, how do you explain the current which would have to be
infinite? Ok, so then we would still have inductance, which by
the second circuit i presented would make a good oscillator,
and as already found out, we'd have to open the switch at
the perfect time in order to get full energy transfer. But, as
i also found out, if that inductance has any loss associated with it
(resistance or radiation or core losses) that brings us back to
square one again, where we lose energy...is it 50 percent?...
depending on how the core works we could lose more..what
if the inductor is a perfect antenna?..we lose everything
I know it seems hard to believe, but we have to keep in mind
that the current goes up the closer we approach zero resistance,
and if there is no resistance then the inductance limits the current,
and then we have to figure out how the inductor loses energy...
if the core isnt perfect than again we loose energy. Maybe less
if the core loss is low and radiation is low? But, if any resistance
in that inductor we loose 50 percent of the energy.
"Proposition two is major loss included, but I doubt at near zero temperatures a loss of 50% would be realistic? "
Well, if you draw up the circuit and place a small resistance in between
the two caps, no matter how low you make that resistor we still
end up with 50 percent loss. It looks like the reason for this is
because as the resistance goes down, the current pulse height
increases, so the loss stays the same (even though the time
goes down too, ie shorter pulse).
If you have a circuit analysis program you can easily try this by
connecting two caps by a resistor and grounding the other two
cap leads, and also setting the first cap's initial conditions to be
some voltage, like: ic=10v, or something like that. Then, close
the switch and watch the response of both caps, then compute
the energy once everything settles down. Try this with 0.1 ohms,
then with 0.01 ohms, then with 0.001 ohms, etc. Watch as
the resistance goes down the current pulse goes up and the
time to settle down gets shorter, but the energy in the two caps
once it settles is still 50 percent of what was started out with.
I guess if the inductance limits the current than if we open the
switch after one cycle a small resistance would lead to small loss.
The circuit damps out to 50 percent energy if allowed to run
until it settles naturally (ie no switch opening).
So for the circuit with an inductor we have to divide the cases in two:
case 1 where the circuit is allowed to damp out naturally, and
case 2 where we open the switch after only one cycle.
We then have to figure the energy for each case.
Make sense?
LEDs vs Bulbs, LEDs are winning.
-
Robert Reed
- Posts: 2277
- Joined: Wed Nov 24, 2004 1:01 am
- Location: ASHTABULA,OHIO
- Contact:
I have been following this post with some interest and every body is making good replys here. Basically it seems like we are talking energy loss due to the transfer of that energy. In a parrallel resonant circuit, energy is alternately transferred back and forth from capacitor charge to inductor field. When I view a tank circuit ( with an o-scope probe 10 meg Z) that has had one single pulse injected into it, I don't see the 50% loss with each excursion. This would show up as each succeeding cycle being 0.707V amplitude of the former. What I do see is much less loss per cycle for the duration of its ringing period, and this is at room temp. and with tank circuits having a `Q' of around 50. A lot less loss would occur with higher `Q' circuits. If all this supposed energy loss should be taking place, why don't I see it in this application? Also as to radiation in your hypothetical circuits, this would depend a lot on frquency and actual conductor length as the conductor lenth versus frequency would determine radiation loss due to antenna effect. Also has any one considered VSWR at higher transfer rates as this may limit power transfer.
No arguments here, just curiosity.
No arguments here, just curiosity.
-
Chris Smith
- Posts: 4325
- Joined: Tue Dec 04, 2001 1:01 am
- Location: Bieber Ca.
Hello Chris,
We must be talking about different things here.
QUOTE
“Also, how do you explain the current which would have to be infinite? “
Actually, you cant have a infinite current.
END QUOTE
That's what i was trying to say, but you insisted there was nothing
to limit the current flow from being infinite, so i asked you how
you could explain it.
We must be talking about different things here.
QUOTE
“Also, how do you explain the current which would have to be infinite? “
Actually, you cant have a infinite current.
END QUOTE
That's what i was trying to say, but you insisted there was nothing
to limit the current flow from being infinite, so i asked you how
you could explain it.
LEDs vs Bulbs, LEDs are winning.
-
Chris Smith
- Posts: 4325
- Joined: Tue Dec 04, 2001 1:01 am
- Location: Bieber Ca.
AL, some one else stated the word infinate?
Cato said...
I suspect the instantaneous and infinite current flow causes a magnetic field so strong that there is a rip in the space time continuum and changes the cosmological constant accelerating the expansion of the universe to the point where the weak force is weakened so that every atom in the experimenters body is ripped apart by the electrical attraction of protons and electrons.
Current always has a limit.
X = electrons,.... in any given force or mass.
Cato said...
I suspect the instantaneous and infinite current flow causes a magnetic field so strong that there is a rip in the space time continuum and changes the cosmological constant accelerating the expansion of the universe to the point where the weak force is weakened so that every atom in the experimenters body is ripped apart by the electrical attraction of protons and electrons.
Current always has a limit.
X = electrons,.... in any given force or mass.
Hi again,
Perhaps we should limit our discussion, or divide it into two parts:
Part 1: Modern Theoretical Physics, that we cant confirm ourselves.
Part 2: Easily Reproduceable Circuit Theory, that we can experiment with directly in our own homes.
I like Part 2 best because that means we talk about stuff that can
be easily proved or disproved.
When i started this discussion i had in mind part 2 because i dont expect
anyone to have the lab equipment to test down near to abs zero, and
I'm pretty sure NOBODY here can go down to a perfect abs zero
to test any ideas or whims out in real life
Perhaps we should limit our discussion, or divide it into two parts:
Part 1: Modern Theoretical Physics, that we cant confirm ourselves.
Part 2: Easily Reproduceable Circuit Theory, that we can experiment with directly in our own homes.
I like Part 2 best because that means we talk about stuff that can
be easily proved or disproved.
When i started this discussion i had in mind part 2 because i dont expect
anyone to have the lab equipment to test down near to abs zero, and
I'm pretty sure NOBODY here can go down to a perfect abs zero
to test any ideas or whims out in real life
LEDs vs Bulbs, LEDs are winning.
-
Chris Smith
- Posts: 4325
- Joined: Tue Dec 04, 2001 1:01 am
- Location: Bieber Ca.
Al I agree
But keep in mind part one explains all the reasons part two exist.
Resistance is everything.
But this theoretical physics can be met half way in your basement and its fun.
You can purchase liquid gasses from the welding/ gas supplier and do some half way experiments on your own and then compare them to the same experiment at room temperature.
Did you know that the lowly 250 MW TO93 type transistor is theoretically possible of handling 250 watts of power with the proper cooling?
So If you really want to preform some experiments in the work shop, do them at room temperatures and then get a small bottle [ about 4 bucks last time I rented one] worth of liquid Co2 used for beverage dispensers and repeat the experiment and compare your results. [minus 134F?]
Liquid oxygen and liquid nitrogen can also be used, but you need a permit.
But keep in mind part one explains all the reasons part two exist.
Resistance is everything.
But this theoretical physics can be met half way in your basement and its fun.
You can purchase liquid gasses from the welding/ gas supplier and do some half way experiments on your own and then compare them to the same experiment at room temperature.
Did you know that the lowly 250 MW TO93 type transistor is theoretically possible of handling 250 watts of power with the proper cooling?
So If you really want to preform some experiments in the work shop, do them at room temperatures and then get a small bottle [ about 4 bucks last time I rented one] worth of liquid Co2 used for beverage dispensers and repeat the experiment and compare your results. [minus 134F?]
Liquid oxygen and liquid nitrogen can also be used, but you need a permit.
Hi again,
"But keep in mind part one explains all the reasons part two exist. "
The reason i mentioned two parts is to avoid anything like this, ie
the crossover between the two groups, when there isnt a separation
any more.
The experiment sounds like fun, but maybe we also need another
catagory:
Part 3: Harder (but possible) to reproduce experiments
"But keep in mind part one explains all the reasons part two exist. "
The reason i mentioned two parts is to avoid anything like this, ie
the crossover between the two groups, when there isnt a separation
any more.
The experiment sounds like fun, but maybe we also need another
catagory:
Part 3: Harder (but possible) to reproduce experiments
LEDs vs Bulbs, LEDs are winning.
-
Chris Smith
- Posts: 4325
- Joined: Tue Dec 04, 2001 1:01 am
- Location: Bieber Ca.
Al,
With all experiments especially those relating to the basics of physics, you must always set the parameters first so that the experiment has meaning.
Its like saying I have a box?
What is it made of, how big is it, how much does it weigh and what are its dimensions?
Having a box explains nothing, knowing everything about the box allows you to decide what to do with it.
Part three is not a super conductor or even close but resistance is fairly linear so the colder you go the more results you can chart.
The more data you have in the chart, the more conclusions you can infer or dispel.
Think of it like a Led, the less resistance on the potentiometer feeding the power, the brighter the LED becomes.
Chart all these positions of the pot on a scale, and you can conclude the brightness out put of the led based on the current, resistance, or angle of the pot knob and other values fairly accurately.
With all experiments especially those relating to the basics of physics, you must always set the parameters first so that the experiment has meaning.
Its like saying I have a box?
What is it made of, how big is it, how much does it weigh and what are its dimensions?
Having a box explains nothing, knowing everything about the box allows you to decide what to do with it.
Part three is not a super conductor or even close but resistance is fairly linear so the colder you go the more results you can chart.
The more data you have in the chart, the more conclusions you can infer or dispel.
Think of it like a Led, the less resistance on the potentiometer feeding the power, the brighter the LED becomes.
Chart all these positions of the pot on a scale, and you can conclude the brightness out put of the led based on the current, resistance, or angle of the pot knob and other values fairly accurately.
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