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But, I have problem with the concept that less energy is required to place charge on a capacitor at the beginning of charging process. The article shows the charging process as a polynomial integral for a graph of (voltage vs. time). But, uses an intergration of (voltage vs. charge).

I think a graph of (voltage vs. charge) for a capacitor would be more linear with constant slope. This is reflected in the result to the intergration of (voltage vs. charge), 1/2(QV). The the energy to place charge on the capacitor is the same for nearer the beginning or more toward the end.

Inside the capacitor, I think, the first charges have little resistance going on the plate (low voltage) and have more energy to push off charges on the opposite plate (high current). As more charges are added to the plate, more energy is required to place the charges on the plate (higher voltage), and less energy to push charges off the opposite plate (lower current).

Voltage and current change during the charging, but the energy stored is restricted to (1/2)QV for each charge placed on the capacitor. Something specific about the capacitor is limiting the energy stored. I disagree with the article (well, for now). Perhaps, charges placed on a capacitor have to come in with current and are in motion, but then come to rest on the plates coupled to opposite charges, resulting in an unavoidable, but constant energy loss.

Energy on a capacitor is proportional to the square of the voltage so obviously is takes more energy to add charge to the capacitor at the end of the charge then at the beginning. This is a simple result of the capacitor voltage increasing as charge is added.

The motion of the charges has nothing to do with the energy. There is no energy loss in an ideal capacitor.

 

I like your post, your explanation is clear. The inductor can transform Q into V, but QV remains the same and is equal to the power source QV (ideally). The capacitor is at a higher voltage, but did the energy loss still occur? If the total QV from the power source were determined, would the capacitor discharge more than 1/2(QV)?

If an ideal voltage source is connected in series with an ideal inductor, ideal capacitor and an ideal diode, the capacitor will end up charged up to twice the voltage of the source with no loss; the energy ultimately stored in the capacitor will be equal to the energy supplied by the source. Ideal means no resistive loss in any of the components.

But, if there is no inductance and there is some resistance, simply connecting a voltage source to a capacitor will charge the capacitor to the same voltage as the source, but with only half the energy supplied by the source left in the capacitor.

In the real world using an inductor and a suitably timed switch can transfer energy to the capacitor with much less than a 50% loss of energy. There will be some small loss, of course; that's how the real world works.

 

Energy on a capacitor is proportional to the square of the voltage so obviously is takes more energy to add charge to the capacitor at the end of the charge then at the beginning. This is a simple result of the capacitor voltage increasing as charge is added.

The motion of the charges has nothing to do with the energy. There is no energy loss in an ideal capacitor.

The square of the voltage is an intermediate term. C=Q/V, Q=VC, and (Q^2/2C) is result of intergration.
(Q^2/2C) = ((VC)^2 /2C) = (V^2C/2) = (V^2)(Q/V)/2 = (QV/2) = 1/2(QV).

I agree, for an ideal capacitor, the motion of the charges has nothing to do with the energy. There is no energy loss in an ideal capacitor.

I would say (for now) that the energy required to place a charge on a capacitor is matched by the energy required to displace a charge on the opposite plate. The capacitor can only store half the energy, the other half is required to displace charge out of the capacitor. If QV goes into the capacitor then 1/2(QV) is used to place charge on the plate and 1/2(QV) is used to displace charge off of the opposite plate and out the capacitor.

The energy used to displace charge out of an ideal capacitor is not lost. The energy is used to continue the voltage and current in the circuit. For a charging capacitor, the voltage opposes the source and the current is in line with the source.

 

If an ideal voltage source is connected in series with an ideal inductor, ideal capacitor and an ideal diode, the capacitor will end up charged up to twice the voltage of the source with no loss; the energy ultimately stored in the capacitor will be equal to the energy supplied by the source. Ideal means no resistive loss in any of the components.

But, if there is no inductance and there is some resistance, simply connecting a voltage source to a capacitor will charge the capacitor to the same voltage as the source, but with only half the energy supplied by the source left in the capacitor.

In the real world using an inductor and a suitably timed switch can transfer energy to the capacitor with much less than a 50% loss of energy. There will be some small loss, of course; that's how the real world works.

I see, energy does go out of the capacitor while charging to continue the circuit, but that energy can still be used to further charge the capacitor with a proper inductor circuit. As in,

In theory, yes. But it's not as simple as that.

What you have to do is throw the switch to connect the source, inductor, and capacitor in series. Then at just the right moment when the inductor current reaches a maximum, you have to throw the switch to put the inductor and the capacitor in series (or parallel -- same thing with only two components) and then, just as the current dies in the inductor, open the switch to prevent the capacitor from discharging back through the inductor. When you throw the switch to remove the source you will have an inductive kick that will cause arcing across the switch contacts unless you snub them out with a diode or capacitor across the contacts and there will be losses associated with that.

 

Hello there,

With any circuit that has a resistor that has a voltage across it there will always be energy loss. Thus any circuit on earth that has a resistor will loose some energy because the resistor would be doing nothing if it did not have voltage across it at least for part of the time.

A capacitor being charged with a voltage source with at least a tiny resistance always looses 1/2 of the energy taken from the source. With an added series inductance the circuit COULD change quite a bit because what we might end up with is a second order system that makes up a temporary oscillator. This is only one of the possibilities however, and the other possibility is that we still end up with only 1/2 the energy taken from the source in the capacitor. This means that there is a special condition necessary just to get it to work as a temporary oscillator, and once we get that, then we can move on to the second necessary condition and also get more detailed about the first condition and how the degree of compliance with the first condition affects the operation too.

The first necessary condition is that the circuit be under damped. This means the resistor value has to be below a certain value relative to the inductor and capacitor. This is found by a quick analysis that reveals what maximum resistor value is required to get an under damped circuit.
The second necessary condition is that we can open circuit the inductor when the inductor current goes through zero.

To get more detailed on the first condition first, we want R as small as possible. When R is near max value in order to satisfy this condition at all we wont see much more energy transfer than without the inductor even after we invoke the second condition. With R near zero we will see max transfer without any other need for switching.

To get more detailed about the second condition, we want to catch the zero crossing of the inductor current at the first zero crossing. Waiting for the second crsosing for example means part of the energy in the cap had time to transfer some of it's energy back into the source, which is not what we want to happen.

A nice fairly straightforward way to analyze this is to use two capacitors instead of one cap and one source. This way we can easily watch the energy level in both caps as the circuit response settle over time. The first cap C1 is given some initial energy, and the second cap is given none. The circuit is then connected at t=0. Some of the energy from C1 transfers to C2, then given enough time, some from C2 transfers back to C1, and this repeats until they both have the same energy (for two equal valued caps which is also an easier situation to look at).
The crossover resistance value for two caps of the same value and one inductor is:
R=sqrt(8*L/C)
which means the resistor value must be less than that or the inductor does nothing for us, and the lower R is the more energy is transferred at the point where the inductor goes though zero for the first time.

Another interesting fact is if we wait for t near infinity we dont get any benefit at all from the inductor, and this is almost equivalent to waiting too long to open circuit the inductor.

Of course for all of the above the effects of radiation is ignored and it's assumed that is true in this whole thread, as obviously radiation means more loss.

 

I would say (for now) that the energy required to place a charge on a capacitor is matched by the energy required to displace a charge on the opposite plate.

If that were the case, then no net energy would be stored since, as you claim, they would be matched and cancel each other out.

You seem to be stuck with this notion that you put charge onto one side of a capacitor and, as a result, charge is then displaces from the other plate. That's not a good way to look at it. Think of the process of a battery (or other source) connected to capacitor (you can ignore the other components, be it resistor or inductor or whatever, for now)

A battery uses chemical energy to move an electron from the positive terminal, through the battery, to the negative terminal. The creates a voltage difference between the two terminals. If these terminals are connected to a capacitor, then there is a voltage that results in a current flowing in the capacitor with an electron flowing from the negative terminal of the battery to the negative terminal of the capacitor and an electron flowing from the positive terminal of the capacitor to the positive terminal of the battery. Both sides of this process are effectively equal and it makes no sense to talk about it being easy to change the charge on one side and hard to change the charge on the other. Both take the same effort and both happen simultaneously.

The end result is that one electron worth of charge flowed through the battery and had an amount of work done on it the magnitude of which is |q·Vb| where q is the charge on the electron Vb is the battery voltage. The energy stored on the capacitor is equal to the work required to move an electron from one side to the other and the magnitude of that is |q·Vc| where Vc is the voltage across the capacitor. If Vc is not equal to Vb, then there is an energy imbalance and that energy has to be accounted for. If there is a resistor in series, then that energy is dissipated as energy in the resistor, which can't in any realistic way be recovered and converted back to electrical energy. If there is an inductor, then that energy is dumped into the magnetic field of the inductor and that CAN be recovered and converted back into electrical energy. How to do that in such a way that that energy ultimately ends up in the capacitor is the point of the discussion (or at least it is for most of us).

 

If that were the case, then no net energy would be stored since, as you claim, they would be matched and cancel each other out.

You are correct, a method for increasing the energy stored on a capacitor is the goal. I am stuck on understanding the mechanism upon which a capacitor, realistic or ideal, is limited to storing 1/2(QC) upon input from a source. Understanding why the capacitor does this, with resistance (realistic) or without resistance (ideal), can help in designing the inductor circuit.

The capacitor does store a net charge. Half is stored, and the other half goes through. If the capacitor stored all the energy inputted from the source then the capacitor would be a black box and nothing would be returned to the circuit. When the capacitor charges, the opposite plate discharges. The process is symmetric. Half the energy is stored and half the energy is returned to the circuit.

If the energy that is discharged from the opposite plate is captured by an inductor (the battery now removed) and used to further charge the capacitor with a pulse then half that energy would be stored and half would be returned to the circuit. The next pulse would continue the process, half stored and half returned. More energy is stored on the capacitor, but each pulse reduces the amount being stored by half.

Using an inductor can eliminate the problem of half the energy being lost in resistance when charging a capacitor. A switch in series to disconnect at the right time, when the inductor current is zero, is key. A diode can do this.

Also see here:

http://forum.allaboutcircuits.com/threads/lc-circuit-with-diode.43848/#post-283597

The Electrician has a good post. But, observe plot three. The final energy on the capacitor depends upon the number of pulses from the inductor. There is current (energy) leaving the opposite plate of the capacitor as the inductor inputs current into the capacitor. If this is an initial pulse from the inductor then the inductor has only changed the energy from the source to a different, but equivalent QV. The same energy could have been supplied by a source with twice the voltage and no inductor. The energy leaving the opposite plate of the capacitor is (1/2)QV, and the energy stored on the capacitor is (1/2)QV. If the inductor captures the energy leaving the capacitor and inputs that energy in a second pulse then the energy stored on the capacitor is now greater than (1/2)QV.

Correction: The same energy could have been supplied by a source with twice the voltage and half the current and no inductor.

 

You are correct, a method for increasing the energy stored on a capacitor is the goal. I am stuck on understanding the mechanism upon which a capacitor, realistic or ideal, is limited to storing 1/2(QC) upon input from a source. Understanding why the capacitor does this, with resistance (realistic) or without resistance (ideal), can help in designing the inductor circuit.

The capacitor does store a net charge. Half is stored, and the other half goes through. If the capacitor stored all the energy inputted from the source then the capacitor would be a black box and nothing would be returned to the circuit. When the capacitor charges, the opposite plate discharges. The process is symmetric. Half the energy is stored and half the energy is returned to the circuit.

You are using 'charge' and 'energy' interchangeably. They are NOT the same thing! You can have a lot of current flowing through zero voltage and have zero power or you can have a lot of voltage but zero current flowing between them and have zero power. Power requires a non-zero current flowing through a non-zero voltage difference.

Also, you need to be very careful about what you mean by 'net charge'. A charged capacitor has ZERO net charge -- it have a charge of +Q on one plate and -Q on the other plate, but overall the capacitor itself has not net charge at all. Relatively few things every develop very much of a net charge because the resulting voltages that develop quickly grow so large that it becomes impossible to further charge the object or you get a breakdown and discharge through the medium to something else.

 

You are using 'charge' and 'energy' interchangeably. They are NOT the same thing! You can have a lot of current flowing through zero voltage and have zero power or you can have a lot of voltage but zero current flowing between them and have zero power. Power requires a non-zero current flowing through a non-zero voltage difference.

Also, you need to be very careful about what you mean by 'net charge'. A charged capacitor has ZERO net charge -- it have a charge of +Q on one plate and -Q on the other plate, but overall the capacitor itself has not net charge at all. Relatively few things every develop very much of a net charge because the resulting voltages that develop quickly grow so large that it becomes impossible to further charge the object or you get a breakdown and discharge through the medium to something else.

Yes, I see. Net energy, not net charge.

 

You don't need a voltage difference for power. Only charge flow or movement. A voltage difference is just one way to get the charge to move. There are others. A voltage difference is not required.

 

So now hopefully you can see that talking about either in terms of all of something entering the capacitor and half of it being stored and the other half going through makes no sense.

It takes work to store an additional incremental amount of charge, dq, into a capacitor that has a voltage Vc across it (noting that this is accomplished by moving a charge of dq from the positive plate to the negative plate. Thus the work performed is Vc·dq. This work is converted from the electrical energy needed to move the charges around into electrical potential energy in the electric field between the capacitor plates. This increases the voltage by an amount dVc=dq/C so that the next packet of charge requires more energy to store it and stores more energy in the capacitor. This is why the first packet of charge stores no energy into the cap while the last packet stores the most. However, if you are using a constant voltage DC source then each packet of charge draws the same amount of energy from the source, namely an amount equal to what will eventually be stored in the cap by the last packet of charge.

If you instead use a variable voltage source then you can ramp the source voltage up linearly and, as you do, you will transfer charge from the source to the capacitor without any energy losses (assuming an ideal source and an ideal capacitor).

 

You don't need a voltage difference for power. Only charge flow or movement. A voltage difference is just one way to get the charge to move. There are others. A voltage difference is not required.

So how much power is there in 100 A of current flowing through a voltage difference of 0V? Let's say that it is flowing in a loop of wire (of zero resistance). Where is it coming from? Where is it going.

 

That's easy......what is the voltage on the wire?

 

That's easy......what is the voltage on the wire?

Not so easy -- voltage, by definition, only has meaning relative to a reference.

The wire in question is electrically isolated; the voltage could be 0V, 1000V, -1000V, or whatever else you want it to be. Nothing changes as a result.

 

ok...what is causing the flow?

 

Nothing -- the current is flowing around the resistanceless loop of wire and will continue to do so indefinitely.

 

What was the voltage that induced the flow, even if it was taken away? The power will be product of that voltage and the current. When you take the voltage off.....it did not go away. If no resistance.....the voltage will stay.

I guess it's definitions....every bit of charge has voltage. So you will always have a voltage. I never said other wise.
To me a voltage difference is two or more charge accumulations. Or two areas of different magnitudes.

An isolated particle does not have a voltage difference with itself........but it has energy and power.

Of course, an isolated charge, in the middle of nothing, in the strictest sense, that is a voltage difference.

But only because you are using space as the other pole(reference).............and it is mathematically.......but not physically.

A space pole is not physically accumulated and localize like an electric pole. Therefore a space pole has no direction.
Also a space pole does not have reactance.

If you take the isolated particle out of the space pole and put it some other medium.......it still has it's power.....it didn't need space or the absents of voltage close by to have and maintain it's power.

The pwr in the wire will be stored in the electric and magnetic fields of the wire ring.

Wait a sec.....pardon me....this is electron flow......a lot of the energy will be in the momentum of the current.

 

Wait a sec.....pardon me....this is electron flow......a lot of the energy will be in the momentum of the current.

I think a review of superconductor conduction is needed.
http://www.doitpoms.ac.uk/tlplib/superconductivity/index.php

 

Many, many theories need to be reviewed.

 

What was the voltage that induced the flow, even if it was taken away? The power will be product of that voltage and the current. When you take the voltage off.....it did not go away. If no resistance.....the voltage will stay.

Okay, let's say that the wire loop was charged at 2V. So you are saying that the power in the wire is 200 W? So how much energy has this power produced after, say, six months?