My idea is:

If two charged capacitors are connected in parallel so that their positive plates are connected together and their negative plates are connected together. Then, if the capacity of one of them changes, a significant current is created in the circuit. If we put a consumer (such as a lamp) in the path of this current, this energy can be used to turn on the lamp.

changing the capacity of one of the capacitors must be done continuously and current must be maintained in the circuit By doing this alternately, it can be seen that the current in the circuit is established, but its direction changes alternately. But the total electric charge difference of two capacitors is constant. That is, if we apply the same initial capacity to the capacitor whose capacity we have continuously changed, we will see that the total energy in the two capacitors has not changed in the ratio of the first state.

My question is:

• 1.Can the energy produced in this circuit be enough to compensate for the energy lost (for example, heat due to the change of capacitor capacity)?

• 2.If we can compensate for the energy loss, can such a system be used as a sustainable energy source?

3.This assembly consumes mechanical energy to change the capacitance of the capacitor. and produces electrical energy. If possible, specify with a theoretical or experimental example that the mechanical energy consumed is greater or the electrical energy produced

I am looking to understand what is the power dissipation in variable capacitors and whether this idea is practically or theoretically feasible. Any theoretical or practical views on this matter would be very helpful.

 

No. The answer to all of the above is no.

 

The concept of a variable capacitor is valid for regenerative energy systems, in theory. Increasing the value of the capacitance can store electrical energy while maintaining a constant voltage. Reducing the capacitance will allow the energy to be delivered to the load while maintaining a constant voltage. The value of the capacitance has to be of significant size in order to make this a viable scheme.

In order to change the value of the capacitance, additional energy must be supplied to the system. In other words, you always have to input more energy into the system than what you get out of it.

The bottom line is you cannot violate the law of conservation of energy, i.e. the First Law of Thermodynamics.

 

The bottom line is you cannot violate the law of conservation of energy, i.e. the First Law of Thermodynamics.

This, precisely this.

 

This assembly consumes mechanical energy to change the capacitance of the capacitor. and produces electrical energy...

We're not considering violation of the laws of thermodynamics here, guys.

If possible, specify with a theoretical or experimental example that the mechanical energy consumed is greater or the electrical energy produced.

This is not a schoolwork exercise, by any chance?

 

This is not a schoolwork exercise, by any chance?
[/QUOTE]
No, it's not school practice.

 

My idea is:

If two charged capacitors are connected in parallel so that their positive plates are connected together and their negative plates are connected together. Then, if the capacity of one of them changes, a significant current is created in the circuit. If we put a consumer (such as a lamp) in the path of this current, this energy can be used to turn on the lamp.

changing the capacity of one of the capacitors must be done continuously and current must be maintained in the circuit By doing this alternately, it can be seen that the current in the circuit is established, but its direction changes alternately. But the total electric charge difference of two capacitors is constant. That is, if we apply the same initial capacity to the capacitor whose capacity we have continuously changed, we will see that the total energy in the two capacitors has not changed in the ratio of the first state.

My question is:

• 1.Can the energy produced in this circuit be enough to compensate for the energy lost (for example, heat due to the change of capacitor capacity)?

• 2.If we can compensate for the energy loss, can such a system be used as a sustainable energy source?

3.This assembly consumes mechanical energy to change the capacitance of the capacitor. and produces electrical energy. If possible, specify with a theoretical or experimental example that the mechanical energy consumed is greater or the electrical energy produced

I am looking to understand what is the power dissipation in variable capacitors and whether this idea is practically or theoretically feasible. Any theoretical or practical views on this matter would be very helpful.

Here I gave a theoretical example using a virtual laboratory








Test description: This test has seven stages. At the end of the test, the capacitors have not changed compared to the first stage. Of course, this is a simple test and does not take into account the energy loss in the capacitors. The important point is how much the energy loss in the capacitor is normally. If the value is small, the system or the circuit is good. If it is too much, this system is not good. With a detailed example, the energy consumed in this circuit should be compared with the energy produced
If possible, guide me with a theoretical or theoretical example and compare the energy consumption and production of this circuit

 

The ESR of the caps, the resistance of the wires, and the resistance of the load all turn some of the energy to heat. This is lost to the system. Eventually, it will consume all of the energy that can be consumed by the movement of the charge.

The system you describe can be simplified to a single capacitor and a lamp. That simplified system will be more efficient than your complicated arrangement.

Simply: any time you do work, power changes to heat (or light) and is lost to the system. The inevitable losses in any system guarantees it will become as energetic as everything around it and inert.

And note that you are not producing any energy by moving charge around—you will always have a net loss.

 

I have sent some pictures in my answer to show the desired circuit well

 

In this electric circuit, pay attention to the electric charge difference in the two capacitors, which remains constant at the end. I put the photos of my test in the comments

 

You are ignoring two things.

1. In order to reduce the capacity, you have to introduce additional energy to the system.
2. As long as current is flowing, you are losing energy.

The bottom line is, you are suggesting an over-unity situation, discussion of which is banned on AAC forums.

 

My question is:
• 1.Can the energy produced in this circuit be enough to compensate for the energy lost (for example, heat due to the change of capacitor capacity)?

Nu.
There are two plates charged with electric charge

If you want to change the capacity of a capacitor, you have to move the plates further apart, or change the common area.

Coulomb force appears between two electrically charged plates. That is, the force of attraction as the plates of a capacitor are charged with charges of the opposite sign.

To move the capacitor plates away from each other or to bring them closer
You have to overcome the Coulomb force. That is, to act on them with a greater force

Move the plate a distance with a force = mechanical work = energy
W=F*d

The greater the electric charge of the plate, the greater the current obtained, but the Coulomb force is greater.
That means higher mechanical energy.

This is how the electric voltage is defined as the mechanical work per electric charge
U=W/q

I-Q/dt

U=V2-V1
V=F*q Electric potential energy

C=Q/U Capacitor capacity
If C increases at the same Q, U decreases

If I move the boards apart from each other by x meters

 

Or change the area like the variable capacitor from old AM radios

 

The electrical energy obtained will be::

But,


W is the mechanical energy consumed to obtain the electrical energy E.

In ideal conditions they are the same.
This is if the capacitor plates were superconductors and if there were no friction when we turn the capacitor knob.

 

Of course, the capacity of the capacitor can be changed by removing or placing the dielectric between the two plates

 

Nu.
There are two plates charged with electric charge

If you want to change the capacity of a capacitor, you have to move the plates further apart, or change the common area.

Coulomb force appears between two electrically charged plates. That is, the force of attraction as the plates of a capacitor are charged with charges of the opposite sign.

Of course, the capacity of the capacitor can be changed by removing or placing the dielectric between the two plates

 

consider more things to get brownie points.

see what is the capacity (and physical size) of the largest variable capacitor. compare that to a single super cap of comparable size.

check its voltage rating. then calculate energy it can store at its maximum capacity. then extrapolate how much volume would be needed to store energy of a single AA battery.

 

Of course, the capacity of the capacitor can be changed by removing or placing the dielectric between the two plates

It's the same thing.
The dielectric consists of polar molecules.
That is, a molecule in which the nucleus of an atom is larger with more protons and attracts the cloud of electrons towards it.
Towards it, the molecule becomes a little negative and on the opposite side a little positive.

This "+ -" is oriented in the electric field between the armatures of the capacitor and the story is repeated.

Due to the Coulomb forces, placing and removing it between the capacitor armatures causes mechanical work theoretically equal to the electrical energy obtained.

 

Of course, the capacity of the capacitor can be changed by removing or placing the dielectric between the two plates

You seem to be assuming that you can change the capacitor plate's location and/or the dielectric and not use any energy to do that. You have to realize that there are forces internal to the capacitor that require overcoming in order to do any of that, and that takes additional energy.

In one of your drawings you show one of the capacitors with a certain plate separation distance, then with a different plate separation distance, yet you do not show how that was accomplished. Once you show that, you will see more energy enter the system.

Now the only question left is, is the required added energy higher than, equal to, or lower than the energy that the light bulb gets. You have to answer that question in order to figure this out for sure. That's if we ignore the conservation of energy for the moment so we do not have to assume anything at all.

For problems like this you have to be able to prove it works, and that means considering everything possible, without ignoring anything. We can't know for sure if things like this can work sometimes until we try them, because we can be overlooking something in ANY assumption, true or false.
An example of this kind of error due to an assumption is an air conditioner. If we do not analyze it correctly, it looks like a device that can put out more energy than it takes in, yet a more concise analysis shows us that we are not doing that.

So far, no person or thing has been able to violate the conservation of energy. Can it be done somehow with the help of quantum physics, that's probably the only possibility left, and there is a good chance that may not work either. Should that stop us from trying? Maybe not, because we learn from the experience how hard and fast that law of nature is, and if there is a bypass somehow we may someday discover it. An example here is fusion power, which does not violate that law, but we can bypass it for the time being by generating power with energy that is very densely packed.

 

Again, discussion of over-unity systems is prohibited on AAC forums.

Amir, you are the only member in this thread making this proposal, while all others have stated why this is not possible.
For your benefit and your benefit only, this discussion has been allowed to continued so that you can grasp why this is called over-unity and defies the 1st Law of Thermodynamics.

This thread will be closed in due course.