Capacitor charging loss (Not the two capacitor issue.)

[The Electrician]

I see his latest response posted an hour or so ago.

He says: "Because it is irrespective of the element value it cannot be calculated, it is just there. So also with a simulation program it cannot be calculated, when the loss is independent of the element value!"

You probably aren't going to get very far with someone who believes this effect is so magical that it can't be calculated, and it can't be simulated.

I wonder what he would say if offered experimental evidence? It wouldn't be hard to solder up an RLC circuit and capture some current and voltage waveforms.

 

[spaceball3000]

I see his latest response posted an hour or so ago.

He says: "Because it is irrespective of the element value it cannot be calculated, it is just there. So also with a simulation program it cannot be calculated, when the loss is independent of the element value!"

You probably aren't going to get very far with someone who believes this effect is so magical that it can't be calculated, and it can't be simulated.

I wonder what he would say if offered experimental evidence? It wouldn't be hard to solder up an RLC circuit and capture some current and voltage waveforms.

Sigh, your likely right.
I thought of doing that, but my thought he would just say my testing equipment is inaccurate when testing my RLC circuit.

 

[The Electrician]

My father used to say:

"There are none so blind as those who will not see."

 

[The Electrician]

It occurred to me to wonder why, if he really believes that "Because it is irrespective of the element value it cannot be calculated, it is just there. So also with a simulation program it cannot be calculated, when the loss is independent of the element value!", why would he provide calculations attempting to prove his own point of view is correct?

Wouldn't his own calculations be incorrect, because "...it cannot be calculated..."?

 

[spaceball3000]

It occurred to me to wonder why, if he really believes that "Because it is irrespective of the element value it cannot be calculated, it is just there. So also with a simulation program it cannot be calculated, when the loss is independent of the element value!", why would he provide calculations attempting to prove his own point of view is correct?

Wouldn't his own calculations be incorrect, because "...it cannot be calculated..."?

Thats a good point.

 

[The Electrician]

It would be interesting if you would press him on this point.

By adopting the notion that "...it cannot be calculated...", he puts himself in a position where there is nothing more to say to him.

But, if his theory is to be considered a good theory, it must be falsifiable:
http://en.wikipedia.org/wiki/Falsify

Ask him what would be a simple test to show the truth of his assertion that charging a capacitor loses 50% of the energy, no matter what method is used. It should be a test that he admits will show a failure of his theory if the theory is wrong.

Ask if a simple series circuit of diode, resistor (small one), inductor and capacitor will exhibit the 50% loss.

If he says yes, then ask what measurements should be made in performing the experiment.

And, if he says something like "I don't need to prove anything; I know it works", then we see that he is being just as close-minded as the orthodox scientists he disparages.

 

[spaceball3000]

It would be interesting if you would press him on this point.

By adopting the notion that "...it cannot be calculated...", he puts himself in a position where there is nothing more to say to him.

But, if his theory is to be considered a good theory, it must be falsifiable:
http://en.wikipedia.org/wiki/Falsify

Ask him what would be a simple test to show the truth of his assertion that charging a capacitor loses 50% of the energy, no matter what method is used. It should be a test that he admits will show a failure of his theory if the theory is wrong.

Ask if a simple series circuit of diode, resistor (small one), inductor and capacitor will exhibit the 50% loss.

If he says yes, then ask what measurements should be made in performing the experiment.

And, if he says something like "I don't need to prove anything; I know it works", then we see that he is being just as close-minded as the orthodox scientists he disparages.

I like your approach, but given his previous responses and equations I honestly expect he will not give useful information back. Though I'll ask him about this later on if were still at a standstill.

 

[spaceball3000]

Kirk T. McDonald, who wrote the paper "A Capacitor Paradox" referenced on my http://www.hep.princeton.edu/~mcdona...es/twocaps.pdf

Excerpt

...if the battery is disconnected after 1/2 cycle, the stored energy can be large compared to the energy lost to heat.

I'll post this information tonight, I wonder what he will say.

 

[The Electrician]

Check for an email on your hotmail account.

 

[spaceball3000]

Check for an email on your hotmail account.

Thanks got it, I'm sure it will be useful.

 

[uart]

Good luck with this guy Spaceball, but I don't think you'll get anywhere with him. I've tried to argue with people like this, fixed ideas but very little understanding of the fundamentals, before and it's usually very frustrating.

How on Earth this Rolf character could have any kind of grasp of quantum physics when he can't handle simple circuit analysis or even a very straight forward linear differential equation is totally beyond me. It certainly seems that you were correct to question his credentials.

 

[spaceball3000]

Uart, I sure your right that I'll get no where.

Well I just did LTSpice simulations (below), It calculates efficiently just fine, I'm going to post this new info all your guys new info too, and I think that would do it for me.

If he was honest and fixed his mistake early on, I was going to forward his info to a few important people that would have invested in him, barring the device actually did live up to the capacitance and voltage numbers claimed.

------------
I ran some tests with LTSpice and had it spit out all the joule numbers.

Here is an RC circuit - See attached file RC.JPG
5v power supply (constant voltage), 0.08 ohm resistor, 1F cap
Power supplied 25J
R loss to heat 12.504J
C Energy stored 12.496J

Energy Lost to heat 50.1%
Efficiently 49.9%

Adding an L to create an RLC circuit - See attached file RLC.JPG
5v power supply (constant voltage), 0.08 ohm resistor, 1F cap, 1H, Inductor
Power supplied 24.704J
R loss to heat 1.4894J
L loss to heat 11.011J
C Energy stored 12.204J

Energy Lost to heat 50.6%
Efficiently 49.4%

Now same RLC circuit, but added a diode, and pulsed power supply. -See attched file RLC_D_Pulse.JPG
150v power supply (Pulse), 0.08 ohm resistor, 1F cap, 1H inductor, HV diode.
V 150v pulse for 0.1 second = 33.658J
R loss to heat 3.3397J
L loss to heat 38.191mJ = 0.038191J
D loss to heat 6.1062J
C Energy stored 24.174J

Energy Lost to heat 28.2%
Efficiently 71.8%

In my first two LTSpice examples, both the RC and RLC circuit have at least 50% loss as expected. But this minimum 50% loss isn't guaranteed in all cases, just look at my last LTSpice example. It's similar to the idea of disconnecting the battery after 1/2 cycle as described by the Kirk T. McDonald "Charging a Capacitor via a Transient RLC Circuit" paper.

 

[spaceball3000]

I think I have another person who thinks there is always at 50% energy loss when charging an capacitor. Unless I misinterpreted their post (below), though they did mention (I^2*R*T), so with an very low I and R that should reduce the loss a lot.. but then they later mention "the total loss is only half if I*R*T = .5*C*V^2" which the .5*C*V^2 brings back in the 1/2 energy lost during an capacitor charge :(

Any ideas what I should say? I've tried my way (see LTSpice example above) and I rather try another approach, anyone got an good way to explain there is other ways to charge an capacitor without the 1/2 energy loss that normally occurs in a static RC, RLC circuit.

A capacitor can only be charged without loss if:
It is supplied from a voltage ramp from 0 to the fully charged voltage, and if there is zero internal resistance.

As soon as there is series resistance there is lost energy basicly I^2*R*T. Easy to see with a linear ramp, given T, constant charge current is Q/T. With resistance, the source voltage must be higher for the same charging conditions, and the resistor energy increases. The total loss is only half if I*R*T = .5*C*V^2. I constant if start V = I*R and finish at end T is end Vc + I*R. The loss is variable vs R.

 

[uart]

Yeah I read that reply and had difficulty understanding exactly what he was trying to say. It seems like a pretty useless/incoherence post to me but what can you say except maybe ask him to clarify his position.

 

[uart]

BTW Spaceball, just in case you wanted a mechanical analogy to use at any time, this one is almost exactly analogous.

When you drop a stationary parcel onto a conveyer belt moving at constant speed, the parcel will always slide a little as it comes up to speed but eventually the two (conveyer and parcel) will be moving at the same speed. If the coefficient of friction is high then the parcel only slips for a very short time, but during that time the frictional force and hence losses are large but over a short time period.

If on the other hand the coefficient of friction is low then the parcel slips for rather a long time before reaching full conveyer speed. During that time the frictional force is much lower though acting over a much longer time and distance. In fact when you do the calculations in detail you find that if the conveyer speed is "v" then you get precisely 1/2mv^2 of energy lost during the slip period regardless of how much friction is present (and even independent of whether it is linear or nonlinear). That is one half the input energy from the conveyer goes into increasing the parcel's KE and the other half is lost in the slipping process. This is basically an exact analogy of the capacitor charging from a constant supply voltage via a resistor alone.

The introduction of the inductor in the resonant charging scheme is equivalent to having a conveyer running at half speed and with very low friction surface that carries a spring to latch onto the parcel as it's dropped onto the conveyer. Now friction can't bring the parcel quickly up to speed but the spring can do so. Initially the parcel lags and the spring stretches as the parcel builds up speed. Eventually the spring reaches a maximum extension at which time the parcel is running the same speed as the conveyer. But the parcel will still increase in speed now as the stretched spring pulls back toward it's equilibrium position, and by the time the spring is back to this equilibrium position the parcel is traveling forward at speed 2v (that is at a forward speed of v relative to the conveyer). If we do nothing then this thing will just oscillate, but we can in theory use a "kicker" to kick the parcel off the convey, at that exact instant, and onto parallel conveyer running at double the speed of the original. If we do this we've actually managed to losslessly bring the parcel up to speed on this second conveyer.

Ok that was a bit long winded but it really is an exact mechanical analogy. :)

 

[spaceball3000]

BTW Spaceball, just in case you wanted a mechanical analogy to use at any time, this one is almost exactly analogous.

When you drop a stationary parcel onto a conveyer belt moving at constant speed, the parcel will always slide a little as it comes up to speed but eventually the two (conveyer and parcel) will be moving at the same speed. If the coefficient of friction is high then the parcel only slips for a very short time, but during that time the frictional force and hence losses are large but over a short time period.

If on the other hand the coefficient of friction is low then the parcel slips for rather a long time before reaching full conveyer speed. During that time the frictional force is much lower though acting over a much longer time and distance. In fact when you do the calculations in detail you find that if the conveyer speed is "v" then you get precisely 1/2mv^2 of energy lost during the slip period regardless of how much friction is present (and even independent of whether it is linear or nonlinear). That is one half the input energy from the conveyer goes into increasing the parcel's KE and the other half is lost in the slipping process. This is basically an exact analogy of the capacitor charging from a constant supply voltage via a resistor alone.

The introduction of the inductor in the resonant charging scheme is equivalent to having a conveyer running at half speed and with very low friction surface that carries a spring to latch onto the parcel as it's dropped onto the conveyer. Now friction can't bring the parcel quickly up to speed but the spring can do so. Initially the parcel lags and the spring stretches as the parcel builds up speed. Eventually the spring reaches a maximum extension at which time the parcel is running the same speed as the conveyer. But the parcel will still increase in speed now as the stretched spring pulls back toward it's equilibrium position, and by the time the spring is back to this equilibrium position the parcel is traveling forward at speed 2v (that is at a forward speed of v relative to the conveyer). If we do nothing then this thing will just oscillate, but we can in theory use a "kicker" to kick the parcel off the convey, at that exact instant, and onto parallel conveyer running at double the speed of the original. If we do this we've actually managed to losslessly bring the parcel up to speed on this second conveyer.

Ok that was a bit long winded but it really is an exact mechanical analogy. :)

lol a little long winded, but I get it!

Just confirming, the springs are attached to the slower conveyor belt, and attached automatically to the package as they are dropped, when they slingshot forward at 2x speed the spring detaches, and package lands on 2x speed conveyor belt. Since the package is already is at 2x speed, and the 2x speed conveyor belt are the same the package just drop on top and doesn't slip at all.

 

[uart]

lol a little long winded, but I get it!

Just confirming, the springs are attached to the slower conveyor belt, and attached automatically to the package as they are dropped, when they slingshot forward at 2x speed the spring detaches, and package lands on 2x speed conveyor belt. Since the package is already is at 2x speed, and the 2x speed conveyor belt are the same the package just drop on top and doesn't slip at all.

Yeah that's it. I was envisaging the spring as pulling the parcel up to speed but you could have a setup with the spring pushing instead (in this case the spring would compress instead of stretch as it got the parcel up to speed and wouldn't need a latching mechanism, though it might need lateral stabilization). In either case the principle is the same.

A friend of mine is a Mech Eng who specializes in conveyer systems and he mentioned this problem to me a few months ago. As soon as he said it I realized it was analogous to the capacitor charging problem, the differential equations that describe it are identical in nature.

BTW. In this problem it's very easy to understand exactly why half the energy is lost when friction alone brings the parcel up to speed. It's just Newton and the good old equal and opposite reaction thing. Whatever force the belt applies to the parcel the parcel applies an equal and opposite force to the belt, trying to slow it down. Since the force is the same and the relative distance slipped is the same it's clear the energy lost to friction is identical to the energy gained by the parcel, not just at the end but at every instant during the acceleration.