Thanks for showing the courage to participate.
For you, since you concluded 1 & 2 have the same answer, question four would be "If you took two 100,000 uF capacitors each charged to 100 volts and placed them in series would the effective capacitance now be 50,000 uF and at 200 volts?"
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Here is my take (I see the bait only).
Using the textbook formula W=1/2*C*V^2 expressing the energy stored in the electric field within the capacitor:
The answer to 1 is W=500Joule. Charge=10Coulomb
The answer to 2 is W=1000Joule. Charge=10Coulomb
The answer to 3 is 50000uF
Question 4 I think implies this: Energy Stored on a Capacitor(which has to do with questions 1,2)
The answer to 5 is from the charger (say battery)
Where is the hook?
George
Energy = 1/2 C V^2, so the second one has twice the energy of the first one. If you connect two of the first one in series then you have twice the energy, exactly as expected. The energy is exactly the same however you prepare the system; it comes from whatever produced the charging current(s).simon7000 said:
Why are you posing trivial problems? This is not the same as the old charged caps in parallel 'puzzle'. It is not the inverse of that, either.
Two equal caps C in series, one charged to V, the other not: Energy = 0.5 C V^2.
Two equal caps C in series, each charged to V/2: Energy = 2 x (0.5 C (V/2)^2) = 0.25 V^2. Same as one cap of C/2, charged to V.
These two situations cannot be compared. The first one really does have more stored energy, which you can extract by making a connection to the join. Perhaps another way of looking at it is that the first arrangement is more ordered so less entropy. Given an opportunity it would like to gain entropy and lose energy and attain the balanced condition, but if no current flows it can't do this.
People often think they can fool physics, but the universe is smarter than us.
Two equal caps C in series, each charged to V/2: Energy = 2 x (0.5 C (V/2)^2) = 0.25 V^2. Same as one cap of C/2, charged to V.
These two situations cannot be compared. The first one really does have more stored energy, which you can extract by making a connection to the join. Perhaps another way of looking at it is that the first arrangement is more ordered so less entropy. Given an opportunity it would like to gain entropy and lose energy and attain the balanced condition, but if no current flows it can't do this.
People often think they can fool physics, but the universe is smarter than us.
The simple physics answer to all this class of problems moving charge across a potential requires work. Setup the contour integrals and let any energy dissipating mechanism go to a limit at infinity (i.e. if you assume any R in series with the capacitors take the limit as R goes to zero and that will account for all the energy "lost"). You need a contour because energy can be radiated, AM radio makes click when you short a capacitor out).
EDIT - I guess that's the caps in parallel, but same issue - look for what is doing work on the charge or the work the charge is giving back.
DOUBLE PLUS EDIT - No ED's last post is the caps in parallel problem. When the caps are paralleled the charge gives work back to the system as radiation or heat. That is saying the cap has zero series resistance does not help, then the current is infinity the integral limits take care of the singularity. It is trivial to set it up.
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Scott,
Yes you understood the issue and gave the right answer by proper application of theory. Now where does the energy go in practice when you have a charged capacitor and connect it to an uncharged one?
Hint it's a fun thing to do, IN SOME CASES. (Other cases might kill you!)
ES
EDIT for Scott's edit. Yes when you series an uncharged capacitor with a charged one it is the same as the parallel version as soon as any current (charge) can flow (transfer).
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He already told you. Why do people keep dragging this issue out? The solution is well known. There is no mystery. Energy apparently disappears; where does it go? Somewhere else, either as heat or radiation. In case of doubt, this is what happens in theory and in practice.simon7000 said:
Ed's probably alluding to exploding capacitors. We should ask jn to compute how much second harmonic is radiated away if it is shorted with a super-conducting inductor.
It's interesting that a client right now is puzzled about the power consumed by an LED backlight circuit and has somehow come to believe that another product has some far lower power. The engineers at the company are suddenly concerned, and are making a number of conceptual errors along the way.
The key problem is the use of a charge pump to generate the voltage ahead of the LED and dropping resistor. This was already there and needed to take the nominal 3V battery and boost it to 5V to run the LCD display. As long as the LCD current is very low it's not a significant power consumption. However, charge pumps work best when their outputs are near-integer multiples of the input voltage. Going from 3 to 5 is not a very good approximation to a factor of two. And the there is the substantial power wasted in the dropping resistor for a ~3V forward drop "white" LED.
But, having developed a very efficient boost regulator that regulates current using a rather small current sampling R, I decided that, since the LED use is discretionary, it was going to complicate things too much to proceed with it. The product is late, but it's conceivable that future products could use the circuit. Or, perhaps by then, something comparable will be produced as an IC, requiring minimal external components. Indeed it is surprising that none such exists, for these very low powers.
The key problem is the use of a charge pump to generate the voltage ahead of the LED and dropping resistor. This was already there and needed to take the nominal 3V battery and boost it to 5V to run the LCD display. As long as the LCD current is very low it's not a significant power consumption. However, charge pumps work best when their outputs are near-integer multiples of the input voltage. Going from 3 to 5 is not a very good approximation to a factor of two. And the there is the substantial power wasted in the dropping resistor for a ~3V forward drop "white" LED.
But, having developed a very efficient boost regulator that regulates current using a rather small current sampling R, I decided that, since the LED use is discretionary, it was going to complicate things too much to proceed with it. The product is late, but it's conceivable that future products could use the circuit. Or, perhaps by then, something comparable will be produced as an IC, requiring minimal external components. Indeed it is surprising that none such exists, for these very low powers.
Ed's probably alluding to exploding capacitors. We should ask jn to compute how much second harmonic is radiated away if it is shorted with a super-conducting inductor.
Yes Scott,
The practical answer is a loud noise and a flash, if you are lucky. A large enough charge on the capacitor and it might actually injure you.
If we use the example of 100,000 uF at 100 volts (or 50,000 across both rails) which is a value actually used in some audio amplifiers (and close to it in many) and connected another equal capacitor we would expect a 250J discharge. You were worried about putting a 1.5 watt laser into hobbiest hands!
So where I saw one fellow post the best way to test for adequate power supply capacitor size was to load the amplifier with an 8 ohm load and listen with headphones as you switch in capacitance to the power supply until you cannot hear a difference... It is probably a valid low equipment method. However I think you just might want to use a resistor across your capacitor A/B switch.
ES
Still am, highly attenuated even glancing reflection would burn a hole in your retina almost instantly or anyone around for that matter (depends on wavelength, there are some almost 100% absorption regions of the aqueous humor).
So this silly game about series capacitors was just a long-winded way of making a safety announcement about parallel capacitors? Wouldn't it have been more helpful to just say "don't connect big caps in parallel without equalising voltage first"?simon7000 said:
Or was it that the series capacitor game didn't actually develop in the way you expected? No energy disappeared, so nothing to talk about so change the subject to parallel capacitors?
No, it was a tutorial on energy storage in capacitors. Almost everyone got it.
However the point that got lost was that the Voltage = Charge x Capacitance and the Energy = 1/2 x Capacitance x Voltage Squared. This was shown by all the combinations that conserved energy when recombined. As SY and Scott understood the kicker was what happens when charge moves energy appears to be lost from the simplified math. Scott chipped in the proper approach a bit too soon.
The theory explains what happens, the practice is quite important.
When I ask such basic questions, things get awfully quiet, as many of the regulars don't want to risk a wrong answer and don't know the basics. They do get the fun of learning a bit without the usual hostile noises.
So tell us some more about series capacitors. In the series case you originally proposed no energy appears to disappear, so the 'puzzle' is not even a puzzle.
Why ask a question about series caps then claim it was all to make a safety point about parallel caps?
For parallel caps the 'simplified math' is correct: the final state has less electrical energy than the initial state - the energy really has been lost; it has been lost to the environment. Doing the full math doesn't change this, but it does give a clue where the energy has gone.
For series caps no energy is lost, as I showed. If there apears to be more that is because for some setups there is more.
Why ask a question about series caps then claim it was all to make a safety point about parallel caps?
For parallel caps the 'simplified math' is correct: the final state has less electrical energy than the initial state - the energy really has been lost; it has been lost to the environment. Doing the full math doesn't change this, but it does give a clue where the energy has gone.
For series caps no energy is lost, as I showed. If there apears to be more that is because for some setups there is more.
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