Capacitors Paradox: Can Energy Transfer Require Less?

[TAAREK]

if two equal capacitor one of then is charged and the other is not.then the switch is closed, half of the stored energy of the 1st capacitor will be lost.

if there is a way to move electrons from one of the positive plates to the other positive of the other capacitor,will it require less energy than half energy stored in the beginning ?

 

[mfb]

Probably not like that, but there are ways to do the transfer more efficiently.

 

[technician]

with 2 identical capacitors as you have described you will always find that 1/2 of the energy is 'lost'.
There will be lots of discussion about how it is lost but you cannot escape it. It is in the mathematics.
You cannot make it 'more efficient'

 

[mfb]

You cannot make it 'more efficient'

With a different setup, it is possible.
If you just connect the two capacitors, you have to waste 50%, but that is not the only option.

 

[technician]

one capacitor: energy stored = 0.5QV
2 identical capacitors in parallel Charge still = Q; V now V/2: energy stored = 0.5Q x 0.5V !
what options are there?

 

[cabraham]

If the cap originally charged is large, then the switch closes allowing it to discharge into a smaller cap, the efficiency > 50%. I had it computed but I don't know where I put it. When I find it I'll post it. In other words, if the 1st cap is 1.0uf, and the 2nd cap is also 1.0uf, then the transfer efficiency is 1/(1+1)=0.50. But if the 1st cap is 2.2uf, 2nd cap is 1.0uf, efficiency is 2.2(1+2.2)= 0.688. If 1st cap is 4.7uf, 2nd cap is 1.0uf, efficiency is 4.7/(1+4.7)=0.825. Did I help?

Claude

 

[technician]

the original post specified 2 equal capacitors and you have correctly calculated 50%
It is a very straightforward capacitor calculation ! turns up time and time again.

 

[technician]

If the cap originally charged is large, then the switch closes allowing it to discharge into a smaller cap, the efficiency > 50%. I had it computed but I don't know where I put it. When I find it I'll post it. In other words, if the 1st cap is 1.0uf, and the 2nd cap is also 1.0uf, then the transfer efficiency is 1/(1+1)=0.50. But if the 1st cap is 2.2uf, 2nd cap is 1.0uf, efficiency is 2.2(1+2.2)= 0.688. If 1st cap is 4.7uf, 2nd cap is 1.0uf, efficiency is 4.7/(1+4.7)=0.825. Did I help?

Claude

effectively you are calculating the voltage across the parallel combination of these capacitors as a fraction of the original voltage across the single capacitor.
One fact is undeniable: energy is lost when capacitors are connected together.
It is 'over egging' it to call it 'transfer efficiency'. I have never met this term in the analysis of capacitor circuits

 

[cabraham]

Well I just introduced you to this term. BTW, what is "over egging"? I have never met this term in the analysis of cap circuits, nor elsewhere. Thanks in advance.

Claude

 

[technician]

putting more eggs in a cake than are recommended in the recipe

 

[mfb]

one capacitor: energy stored = 0.5QV
2 identical capacitors in parallel Charge still = Q; V now V/2: energy stored = 0.5Q x 0.5V !
what options are there?

You can increase Q.

I am sure there are better setups, but in doubt, use a DC-DC-converter. This will increase the charging efficiency, as charging current in the (initially) low-voltage capacitor starts larger than discharge current in the (initially) high-voltage capacitor.
A coil and a quick switch could be even better.

 

[technician]

You can increase Q.

I am sure there are better setups, but in doubt, use a DC-DC-converter. This will increase the charging efficiency, as charging current in the (initially) low-voltage capacitor starts larger than discharge current in the (initially) high-voltage capacitor.

You cannot increase the charge (Q) if all you have is 2capacitors.
What do you mean by 'better set ups'?
What do you mean by ' increase the charging efficiency'
What you have written is nonsense.

 

[mfb]

You cannot increase the charge (Q) if all you have is 2capacitors.

I never questioned that.

What do you mean by 'better set ups'?

Setups with additional components.

What do you mean by ' increase the charging efficiency'

See my previous posts.

What you have written is nonsense.

It is not.

 

[technician]

'Additional components'. ,!??
Nonsense
Where is the physics explanation?

 

[technician]

Probably not like that, but there are ways to do the transfer more efficiently.

How?

 

[TAAREK]

as you can see in the attached picture.how can you calculate the energy required to move half the electrons from the 1st positive plate to the 2nd positive plate,if we will assume the distance between the two positive plates is d and the starting voltage on the 1st capacitor is V

 

[mfb]

'Additional components'. ,!??
Nonsense
Where is the physics explanation?

How?

Read my posts, please, before you call them "Nonsense".
I proposed two methods, a DC-DC-converter and a coil with a switch.

@TAAREK: Multiply voltage difference with the charge you want to transfer.

 

[TurtleMeister]

mfb is correct. The simplest way to increase efficiency is to use an inductor.
http://www.smpstech.com/charge.htm

 

[The Electrician]

This has been dealt with many times before. For example, have a look at these (and the references therein):

http://www.hep.princeton.edu/~mcdonald/examples/twocaps.pdf

http://puhep1.princeton.edu/~mcdonald/examples/seriesrlc.pdf

And, in this thread, some of the issues are discussed:

https://www.physicsforums.com/showthread.php?t=292838

 

[technician]

Electrician is correct. This question is standard textbook physics.
It is easy to calculate the energy 'lost' when capacitors are connected together. The references discussing circuits with inductors do not change the physics.

 

[TAAREK]

energy to move charge Q from the positive plate to the 2nd positive plate = Q*V
but if V is less that he voltage which was applied on the 1st capacitor then the power dissipated will be less than the half.

so what is the potential defrance between the two positive plates ?
the applied voltage on the 2nd plate equals the 1st plate

 

[Windadct]

It seems there are two discussions here - 1) The OP was about the Paradox, and 2) How to make this more efficient.

As for the paradox it appears the energy is irradiated - in an ideal system, the current flow, would be extremely large - you would hear the "click" on a radio for example. An apparent mathematical proof is here : http://puhep1.princeton.edu/~kirkmcd/examples/twocaps.pdf

The fact this was still being done in 2002 - seems remarkable to me.

As for the 2nd "discussion" -- what is the actual question you are trying to settle?

 

[technician]

It seems there are two discussions here - 1) The OP was about the Paradox, and 2) How to make this more efficient.

As for the paradox it appears the energy is irradiated - in an ideal system, the current flow, would be extremely large - you would hear the "click" on a radio for example. An apparent mathematical proof is here : http://puhep1.princeton.edu/~kirkmcd...es/twocaps.pdf


You are correct. This is standard A level textbook stuff. If the 2 capacitors are equal then 1/2 the energy is 'lost' and in post #6 cabraham effectively gave more examples of energy lost for a range of different capacitors connected in parallel.
The energy is lost by any resistance in the circuit, any sparking at the switch and ultimately by electromagnetic radiation caused by the changing current.

 

[DrZoidberg]

Electrician is correct. This question is standard textbook physics.
It is easy to calculate the energy 'lost' when capacitors are connected together. The references discussing circuits with inductors do not change the physics.

Maybe you simply misunderstood mfb? He was talking about an LC circuit with two capacitors in it. You do know how oscillations in an LC circuit work, do you? The energy is being transferred from C to L and then back again. Depending on the circumstances the energy lost in each cycle can be quite small. There is no law of physics that would stop you from transferring nearly 100% of the energy contained in a capacitor into another one.

 

[Dadface]

This thread is about the situation described in post one where two equal capacitors, one charged and one uncharged, are connected by a switch. In this case half of the charge will flow from the charged capacitor and to the second capacitor. Since energy stored equals Q squared/2C the final energy stored will be half of the initial energy stored.
There is no paradox since the "lost" energy can be accounted for in terms of heating of the connecting wires, sparking and radiative losses.

 

[technician]

Maybe you simply misunderstood mfb? He was talking about an LC circuit with two capacitors in it. You do know how oscillations in an LC circuit work, do you? The energy is being transferred from C to L and then back again. Depending on the circumstances the energy lost in each cycle can be quite small. There is no law of physics that would stop you from transferring nearly 100% of the energy contained in a capacitor into another one.

I did not misunderstand the original post.
Yes I do know about AC C and L circuits. I have made contributions to questions regarding these in the past.
I think everything relating to the original post is now clear.

 

[Windadct]

It is a commonly referred to Paradox ... and the losses are not due to heating in the wires, they are assumed ideal - they are however circuit elements (ideal conductors) and subject to Maxwells - the current creates a magnetic field, that expands and collapses - generating classic EMP - To the circuit the energy is lost - but in physics energy is not "lost", that is why this is an interesting case.

 

[The Electrician]

if two equal capacitor one of then is charged and the other is not.then the switch is closed, half of the stored energy of the 1st capacitor will be lost.

if there is a way to move electrons from one of the positive plates to the other positive of the other capacitor,will it require less energy than half energy stored in the beginning ?

If you simply connect the two capacitors with a switch, half the energy in the system will be lost.

However, if an inductor and diode are used to transfer the energy and disconnect at the right time, much less energy will be lost.

Here's a schematic of a circuit I wired up. The two capacitors are 470 μF each and the inductor is an "air core" 500 μH unit. C1 is initially charged to 30 volts, and C2 is initially uncharged.

When the switch is closed, the capacitors and inductor form a resonant loop and a half cycle "ring" of the circuit begins. The current in the inductor builds up and then decays. When that current reaches zero and attempts to reverse (as it would without the diode), the diode "disconnects" the inductor, and most of the energy has been transferred to C2.

Here's a scope capture of the voltages across C1 and C2, and the inductor current. The voltage across C1 is yellow, across C2 is green, and the inductor current is red. The current was measured with a clamp-on current probe and the scale is 5 amps/cm, reaching a peak of about 17 amps.

C1 starts out at 30 volts and ends up at about 5 volts. C2 starts out at zero volts and ends up at about 25 volts. C2 doesn't charge up to the full 30 volts as it would if everything were lossless, but at 25 volts it contains much more energy than it would if the two capacitors were simply connected with a switch.

So, it is possible to transfer energy from one capacitor to the other without losing a full 50% of the energy.

 

[technician]

@electrician...this is brilliant! I would love to check your numbers...what is the time scale on the plot?
What do you get without the diode? Do you get an oscillation.
I have no means of reproducing your measurements but I am impressed by this...one of the best things I have seen on PF and in the true spirit of physics...experimental evidence to discus.
Congratulations

 

[TurtleMeister]

I agree, that's a nice illustration by The Electrician. You have a polarity indicated on the inductor. Would that reverse after the peak current was achieved?

 

[jim hardy]

Great post indeed !

I think the 'paradox' arises from the connection of two unequally charged ideal capacitors through ideal wires. That causes division by zero, I = ΔVolts/zeroOhms , so anything goes.

Adding an ideal inductor prevents division by zero. Further, an ideal inductor can dissipate no energy so the energy should all show up in the capacitors afterward.
Ahh if only you had an ideal diode...

It's not unlike the thermo question - a gas is confined to one half of a chamber, the other half is evacuated. What happens if the wall separating the chamber halves is suddenly removed?

 

[The Electrician]

With no diode the result is:

Note that the ground reference for the two voltage traces (yellow and green) is at the bottom 1 cm; you can see a ground symbol down there at the far left. The final result is 15 volts on both caps.

Here I have inserted a 10 ohm resistor in series. Now we're overdamped.

Still 15 volts on both caps at the end.

Here's a picture of the experimental setup:

 

[The Electrician]

@electrician...this is brilliant! I would love to check your numbers...what is the time scale on the plot?
What do you get without the diode? Do you get an oscillation.
I have no means of reproducing your measurements but I am impressed by this...one of the best things I have seen on PF and in the true spirit of physics...experimental evidence to discus.
Congratulations

You can see the horizontal sweep speed at the top of the image--500 μs/cm for the first image. Later images are 2 ms/cm. Channel 1 and 2 are 5 V/cm, and the current scale is 5 A/cm.

 

[jim hardy]

Those are very nice traces. What machine do you use to make them? I'm still on a analog HP180...

 

[The Electrician]

Those are very nice traces. What machine do you use to make them? I'm still on a analog HP180...

It's one of these:

http://www.home.agilent.com/en/pd-6...hannels?nid=-536902766.536908096&cc=US&lc=eng

For a while recently it looked like they were going to discontinue it, but apparently they decided to raise the price and continue making it!

 

[Dadface]

I like this. It's interesting to note that if the diode is not used both capacitors eventually reach 15V with each capacitor having the same charge and the energy loss being 50 percent.
If the circuit is switched off before charging/discharging is complete then the energy loss is reduced. The same applies when the capacitors are connected by wires. The energy loss is 50 percent only when charging/discharging is complete.

 

[jim hardy]

Interesting that the energy loss is same for the real case of resistance in wires and theoretical one of zero resistance in wires.

There must be a relation between capacitor ratio and energy lost.
That'd be an interesting piece of algebra, and might uncover one of those "little gems of wisdom" a fellow likes to keep in his bag of tricks.

I'd just embarrass myself with clumsy Latex if I tried...

old jim

 

[TurtleMeister]

Yes, this is a good learning exercise. If the circuit is allowed to oscillate (without the diode) then energy is lost mostly through the resistive component of the inductor. So in that sense it is similar to an RC circuit except that it oscillates.

Since no one has volunteered to answer my question in post #30, I will attempt to answer it myself. The question is: With the diode in the circuit will the polarity on the inductor reverse after the peak current is reached?

The answer is yes it will. This may seem non intuitive at first because the direction of current flow through the inductor does not change. During the first quarter cycle energy is being stored in the magnetic field of the inductor with C1 being the source. At the peak of current flow the voltage on C1 and C2 are equal (15 volts) and the voltage on the inductor is zero (voltage and current on the inductor is 90 degrees out of phase). During the second quarter cycle the magnetic field of the inductor is collapsing with the inductor being the source. The energy stored in the inductor reduces the voltage on C1 to it's final 5 volt value and increases the voltage on C2 to it's final 25 volt value. The diode prevents oscillation. The circuit cannot go for more than one half cycle, giving a maximum transfer efficiency. You can see the one half cycle of current in the oscilloscope trace. Note: because of losses, voltages may be slightly less than those stated.

 

[The Electrician]

Since no one has volunteered to answer my question in post #30, I will attempt to answer it myself. The question is: With the diode in the circuit will the polarity on the inductor reverse after the peak current is reached?

I thought your question was rhetorical. The polarity indication on the inductor is an artifact of the spice program I used to produce the schematic. Spice indicates polarity on inductors simply as a way of declaring the reference direction for inductor voltage when plotting waveforms. It was happenstance that the polarity indication matches the initial polarity in the behavior of the circuit.

Your detailed explanation is right on the money. The person who hasn't seen this type of circuit in action before could persuade his intuition about the inductor's voltage polarity in two ways. Remembering that for an inductor e=L*di/dt, we see that the slope of the inductor current versus time changes sign at the peak of the inductor current, hence the voltage polarity changes at the peak; after the first quarter cycle, as you point out.

Also, because the inductor is connected between the two capacitors, the voltage across the inductor is the difference of the capacitor voltages. The capacitor voltages cross over at 15 volts at the same time as the inductor current reaches its peak, and that's when the inductor voltage polarity changes.

This circuit is a simple demonstration of how "resonant switching" works in switching power supplies.

Thanks for your detailed further explanation of the circuit's operation.