Voltage, current, energy and charge

cala

Moderators love me. PLEASE, DON'T CLOSE THIS TOPIC BEFORE I OBTAIN SOME ANSWERS!

My last post was closed. It's something i can't understand, i'm censored, and if you want to discuss with me, you're censored also.

Alexander, you help me a lot. I see you really understand about all physics phenomena, you have physics quite clear in your mind, so i'm glad to see you take the time to answer me.

You said:

"There is complete balance of energy in both processes (charging 2C capacitor and discharging C/2 capacitor back to the battery)..."

"Parallel transfer of energy: CV^2 (battery) = CV^2/4 (resistance) + 3CV^2/4 (capacitors)" (the 3CV^2/4 on capacitors add to the initial CV^2/4 pre-charge, then capacitors are charged at CV^2)

"Series transfer of energy:CV^2 (capacitors) = CV^2/4 (resistance) + CV^2/2 (battery)"

"So, no net gain/loss of any energy is detected".

You see that the battery gaves CV^2 energy on parallel proccess and takes CV^2/2 on series proccess, But what this means?.

The charges on the battery are the same at starting and ending points of the proccess. Will the battery discharge by doing less energy on it?

Energy is related to V and I, but finally all this terms are related to charge.

Voltage is related to a concentration of charges from one spatial point to another, and current is related with how fast this charges move from one point to another in time.

In the parallel proccess, Q charges of the battery goes to the capacitors, because there is a related V from the battery to the capacitors, and the speed of the charges "diffusion" proccess gives the I, so the energy is CV^2.

In series proccess, the V is the same than before (from the capacitors to the battery), but the "diffusion" proccess of the Q charges to the battery takes another time or speed to happen, thus giving a different current, so energy is less than before on the battery: CV^2/2.

As you said, there is no gain/loss of energy, but the resistor disipated CV^2/4 on both proccess, and also you've got the same Q charges that run out the battery coming in again.

Energy on battery is different, but the charges that "produced" this energy are the same as before. Then the battery doesn't discharge, so you can play with the same charges for a long time doing this cycle without waste their concentration (their V), and then creating energy by the charges transfers without finally spending the source of charges.

If you have a concentration of charges, and you move them to another place at certain speed, you've got some energy. Now, if you can move them back to the first place without loosing their concentration at another speed, making no work to force them, you obtain another energy, but you're also able to repeat the cycle the times you want, obtaining energy without loose the conditions that lets the energy appear.

There is no physical regret to use the same charges once and again to obtain energy, you only need a method to maintain the charges ordered into a certain V at initial and final steps, and the current is given by the proccess you do.

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chroot

Staff Emeritus
Gold Member
The reasons your thread was closed were:

a) What you propose does not actually work.
b) It was explained to you several times why it does not work.
c) You're not really interested in a discussion. You don't take the time to really understand the explanations provided to you, and instead just say the same (incorrect) things over and over.

Here are some of (many) examples of your misunderstanding of physics:
You see that the battery gaves CV^2 energy
Capacitance times voltage squared does not have dimensions of energy.
Will the battery discharge by doing less energy on it?
As has been said, you can shuffle charge back and forth as long as you want. Assuming superconductors, no work is being done. This is not free energy.
and then creating energy by the charges transfers without finally spending the source of charges.
You don't create energy by moving charges around.
If you have a concentration of charges, and you move them to another place at certain speed, you've got some energy.
Once again, no.

The bottom line is this: you can slosh charge back and forth forever between your battery and capacitors. Assuming ideal (resistanceless) conductors, you can repeat such a cycle indefinitely. Once you introduce real (resistive) wires, or some kind of dissipative device like a resistor or a lamp or a motor, you'll find that sloshing the charges back and forth is not as successful.

Moderators, I suggest this thread be closed as well.

- Warren

Alexander

Well, go easy on Cala, you may scare him away from physics or engineering, but may be he is next Edison or Tesla - who knows today?

So while in general yes, you can't beat energy conservation, I would go a step further and am willing to explain him details of his perp mobile (#5?), for learning purpose - after all this is why Cala and most of us are here.

Cala, there is a current involved charging/discharging capacitors and battery in you capacitor device. This current dissipates energy on resistor which shall be taken into account. And this is where eventually the difference in what battery gave (CV^2) and received (CV^2/2) went into. If you calculate how much is energy dissipated on the resistor E = Integral(I^2R)dt then during first half cycle when current is I(t) = (V/2R)exp(-t/2RC) you will get after integration (from t=0 to t=infinity) that E = CV^2/4.

Similar in second half of the cycle, when current I=(V/R)exp(-2t/RC) is flowing from capacitors (now in series) back into battery, then the energy dissipated on the resistor R is again Int(I^2R)dt = CV^2/4

So, CV^2/4+CV^2/4 = CV^2/2 - net energy dissipated in the resistor R in entire cycle, the battery gave up CV^2 in first half cycle and received back CV^2/2 during second, CV^2 = CV^2/2 + CV^2/2 = CV^2, full balance. At the end of the day you have half energy of battery back, half dissipated in heat into resistor. No net loss/gain - full balance.

Alexander

Originally posted by chroot

Here are some of (many) examples of your misunderstanding of physics:

Capacitance times voltage squared does not have dimensions of energy.

- Warren [/B]
How come? capacitance times voltage squared does indeed have energy dimensions: 1F x (1V)^2 = 1 J

Will

How come? capacitance times voltage squared does indeed have energy dimensions: 1F x (1V)^2 = 1 J

Yup. Looks that way to me capacitance =(coulumbs/volt) and volts = (joules/coulumb) so cv^2 = (c/v)v^2 = cv = c(j/c) = j

cala

When I used CV^2 as energy measuring, i was taking it from Alexander, but also i understand it's a valid measure of energy.

That is what happen when i open my mouth, some people think i'm wrong not for what i say, only because is me who is saying that thing. Alexander, I finally understand what you are talking to me:

If we use capacitors, when we connect it in series, the positive plate of one cancels the charge of the negative one of the other, so we loose some "concentration" of the charges, so the battery will receive half the charge that it gave.

If you use batteries instead capacitors, the same thing happens. To make an electron go into one of the parallel batteries, you waste 2 electrons in the series batteries, so finally, to charge the parallel batteries you waste 4 times the same charge, not double. There is a problem of "disorder" of charges with the series connection of capacitors or batteries.

But what about something like spheric conductors charged with excess electrons? In this case, the charges don't "disorder". There is only a kind of charges, not negative on one side and possitive on the other... But is there a way to connect this kind of devices in series, making their inner voltage add? (I think maybe it can be done reducing the surface of the devices).

Do we need to expend power to reduce the surface of a charged conductor, and then make this charged conductor take another V value?

Imagine a charged metallic ball rolling in contact outside a charged metallic box. Do this ball need an external force to introduce itself inside the box if it's always in contact with the box?

cala

I post an idea of how such a system could be:

www.geocities.com/k_pullo/cacharro.htm

The system is made with two charged conductors with same kind of charges.

Some MOSFET creates or collapse a channel for the charges, so the topology of the conductors change, and then, the surface of the conductors "grow" or "shrink", so the charges stored keeps the same, but it's density change, and so the voltage. Then we connect the load, and then the charges tend to rise a stable density state, but before it happens, we disconnect and change the conductors topology, so the surface change, thus the charge density, and so the voltage. Now, if we connect the load, the charges will want to rise the stable state, but on the other direction.

This system has no charge looses by "cancelling" with opposite ones, it's based only on one kind of charges and potential differences induced by changes on the surface of conductors.

Alexander

Cala, let me give you more simple example of two capacitor system which seems to not comply with energy conservation. (For the sake of you to learn better chargeing-dischargin process).

Charge two similar capacitors C to some voltage V (say, connecting them in parallel and applying voltage V to them). They got CV^2 joules of energy. Now disconnect them from voltage sorce, and then re-connect capacitors in series (so + of one is connected to - of another and vice versa). Charges cancel each other and net voltage on each capacitor vanishes.

Now energy in each capacitor is 0.

Where did CV^2 joules go?

cala

Yes Alexander, you're right, the energy is lost because the charges "disorder" or cancel negative with possitive ones. They are there, but they loose their "order", and that is why it seems to be a loose of charge, but in fact there is a loose of "order" of charges.

Now, what do you think about my last system? If using only one kind of excess charges (not "ordered" negative and possitive charges) the "disorder" can't happen.

Do you know any method to connect such devices in series (to add their voltage), or to change the potential or charge density without spending work or little work?

Alexander

Hold on. Energy is NOT lost in my last example of connecting charged capacitors by opposite poles. Not a tiny bit of it.

Let's consider even more simple example: let's charge a single capacitor C to some voltage V and then disconnect it from voltage source and short circuit capacitor's terminals. Positive charge from one plate will reunite with exactly same amount of negative charge on opposite plate, and the capacitor will be discharged. No more voltage between plates. But what happened to CV^2/2 joules of energy? Where exactly did they go in this example?

cala

Well, i see it that way:

Initially you've got CV^2, because you ordered the charges on the capacitor (by the battery voltage).

Now, as you connect the terminals, the charges re-order (or disorder), so the voltage goes decreasing, so the energy goes decreasing also.

Meanwhile the charges distribute, there is a current that runs through the inner ressistance of the capacitor and wire, so this energy is converted into heat i suppose.

But energy is not what i'm looking for, i wanted to know what happened with the charges and the charges order.

In the case of the capacitor, once you discharge it, you have to apply more energy to re-order the charges again, to separate possitive ones from negative ones, and create the voltage difference again.

If we were able to make the charges run through the circuits changing potential conditions, but maintaining them ordered (not mixing with the opposite charges), we could make these charges go on and on just changing the potential conditions.

What i'm looking for is a way to change potential conditions of a fixed quantity of charges, changing geometry or other methods.

I'm sorry because my last partial explanation. I only talked about the part i was interested in.

I hope my answer be correct this time. If not, keep going, i'm a bit "heavy-head"

Alexander

Yes, energy here (in the example with single capacitor) is of course not lost but is dissipated mostly into Joule heating of wires (and some on e/m radiation). Because wires have quite low resistance, usuallyat the moment when the voltage is zero, current is not zero yet and the capacitor is recharging to opposite polarity. Then process repeats in opposite direction until capacitor is charged again almost to full initial charge, then again swing to opposite polarity and so on, cycle by cycle loosing energy on Joule heating. Basicly you have RLC circuit here (yes, wires and inner conductors of any capacitor indeed have inductance), thus damped oscillations.

Indeed, connect a capacitor to oscilloscope and observe voltage on it while short circuiting it - you'll see damped oscillations.

But oscillations are of not primary importance here. What is important is that if you integrate power dissipated on a resistor (I2R) over sufficiently long time (till oscillations die out), you get complete balance of energy: CV2/2=int(I2R)dt. So, losses on resistance shall always be included into energy balance. (In case of really small resistance oscillations may proceed for very long time, and then prime channel of energy loss will be conversion into e/m radiation - but this is another story and not related much to your example).

In you example with two capacitors C (and a resistor R) being charged by a battery V when in parallel and discharged back to same battery when reconnected in series energy goes from battery into capacitors and then back into battery, but only 50% returns to the battery - another 50% of energy goes into heat in the resistor R. Indeed, in the first half of cycle battery loses CV2 Joules out of which CV2/4 goes into heating of resistor R, and 3CV2/4 goes into capacitors. During second half (when you reconnect capacitors and let them charge battery back) the amount of CV2/2 returns to the battery but the rest CV2/4 is again heating resistor R during this process.

Full balance thus: out of 100% lost by battery in first half cycle exactly 50% returns back into battery, and 25%+25% = 50% turns into heat in the resistor R.

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