# Double use of energy ? possibility of free energy ?

1. Jun 11, 2014

### stevan

I build simple electronic like this picture..

the capacitor I use is super capacitor...and is not leak
when I close the the switch, the lamp will light up...until the capacitor is full (energy out from the battery is equal energy in to the capacitor, right? because the capacitor is not leak)

then, I remove the battery, replace it with a simple conductor...once again the lamp will light up...this time, until the capacitor is empty

Is it mean I take energy from the battery once, and I can light up the lamp twice ?

maybe anyone know how to use energy more than twice like this ?

maybe someone think I'm searching for the possibility of free energy ? of course :D

2. Jun 11, 2014

### 256bits

This isn't the place to be looking for free energy, as this site deals with physics in the real world.

3. Jun 11, 2014

### derek10

When the capacitor is being charged, its resistance increases so the lamp current decreases until it's almost 0. The lamp won't get all the current provided from the battery. So no "free" energy here :-)
Also good look finding that. It doesn't exist.

4. Jun 11, 2014

### stevan

I give example in real world...and I see that look like overunity where the ouput is twice of input...

still, it's about physics..and it's in the real world...

If my analysis is wrong, then correct me

5. Jun 11, 2014

### stevan

yes sir, I know the lamp will dimmer and dimmer till it's completely off when the capacitor is full.
but, at that time, the energy (current) that come out from the battery also smaller and smaller till it's completely stop.

What I see is : energy that come out from the battery is entered the capacitor, and then the full capacitor once again can be used to light up the lamp

from this I think the energy from the battery is converted to light&heat energy in the lamp twice...

6. Jun 11, 2014

### MisterX

Assume the lamp has resistance $R$ and the battery is ideal with voltage $V_0$. In the first phase of operation, the current is given by $I = \frac{V_0}{R} e^{-t/(RC)}$. The power given by the battery is $V_0 I = \frac{V_0^2}{R} e^{-t/(RC)}$. The power given to the lamp is $I^2R = \frac{V_0^2}{R} e^{-2t/(RC)}$.

Let's assume for simplicity we charge and discharge for an infinite amount of time. The total energy given by the battery is

$E_{battery} = \int_0^\infty = \frac{V_0^2}{R} e^{-t/(RC)} = V_0^2C$

The energy given to the Lamp in the first phase is
$E_{lamp, 1} = \frac{V_0^2}{R} \int_0^\infty e^{-2t/(RC)} = \frac{1}{2}V_0^2C$

The 2nd phase is an RC discharge starting from $V_0$

$I = V_0e^{-t/(RC)}$

The energy given to the Lamp in second phase is
$E_{lamp, 2} = \frac{V_0^2}{R} \int_0^\infty e^{-t/(RC)} = \frac{1}{2}V_0^2C$

So you see,

$E_{battery} =E_{lamp, 1} + E_{lamp, 2}$.

So, this is not a free energy scheme. The energy taken from the battery exactly equals the total energy absorbed by the lamp.

7. Jun 11, 2014

### derek10

Yes, the battery current drops in a exponential way while the cap charges but still the capacitor gets energy from the battery. That's why the capacitor resistance increases towards infinity causing the lamp to dim

8. Jun 11, 2014

### stevan

ugh, it's difficult for me analyse this from the formula...

what I see is electron come out from battery terminal (-) to the one plate of the capacitor,
that make electron from other plate of the capacitor pushed away and enter the battery terminal (+)

when the voltage between battery and capacitor is same, the electron is stop moving...Is it mean no more energy transfered ?

for lamp, is it just only provide resistance ? Is it only make the electron movement slower and make capacitor need more time to get fully charged?

I only see the electron move through the lamp...no electron lost and changed to heat and light in the lamp ?

Is there any process that I missed ?

maybe something wrong when I think electricity like "water" ?

Last edited: Jun 11, 2014
9. Jun 11, 2014

### Staff: Mentor

You're taking energy from the battery once, but you aren't using the same energy twice.

While the capacitor is charging, some of the battery's energy is heating the filament of the lamp and making light that is radiated away. Some of the battery's energy is turned into potential energy stored by the capacitor. When we replace the battery with a conductor the capacitor discharges and the potential energy it had been storing lights the lamp.

The confusing thing about this particular setup is that although we aren't using the same energy twice, the same charge does flow through the lamp twice, once in each direction. I could do the same thing with water: pump it uphill, then let it flow back downhill, with a paddlewheel being turned by the water as it moves on either direction.

10. Jun 11, 2014

### stevan

that's what make me confused sir..
I think the battery's energy form is the moving electron, and lamp only slow down the electron movement because the filament is very thin.

What if I don't use battery? Instead, I use fully charged super capacitor with same capacity with capacitor that I used in the circuit.

What in my mind is charged capacitor have surplus electron in one plate, and deficit electron in other plate..
when I put it as battery in the circuit, of course that 2 capacitor will share electron until they have same potential..and that process make the electron move..and lamp only slow down the process...but as long as the potential is not same, the electron will keep moving.

example:
at first the full charge capacitor 1 (as battery replacement) have 400 electron...
when I put it into circuit as battery, and close the switch, the capacitor 1 (as baterry) will send 200 electron to the capacitor 2 (empty capacitor) to make the potential equal..
with or without lamp in the circuit, the capacitor 1 send 200 electron..
the lamp only make the "send" process slower...but dont reduce the amount of the electron sent, right ?
Is it mean the lamp only use electron's speed to light up?
Then, is it mean the capacitor 2 still receive 200 electron (but the process slower than without lamp)?
Then, is it mean capacitor 2 can light up the lamp again with it's 200 electron ?

or maybe something wrong in this analysis ?

Last edited: Jun 11, 2014
11. Jun 11, 2014

### Staff: Mentor

The problem is that you don't understand what electrical energy is. You are describing charge movement (albeit incorrectly), but energy is charge times voltage. The same charge is moved, but the voltage across the elements is not the same.

12. Jun 12, 2014

### stevan

I don't understand why the voltage is not the same sir..

maybe easier when I assume the battery voltage is stable at 12 volt..
when I put on the battery and close the switch, the voltage of the capacitor start from 0 climbing to 12 volt...and the difference voltage between battery and capacitor start from 12 down to 0 volt
then when I replace the battery with conductor, voltage of the capacitor start from 12, down to 0 volt...

maybe someone can give plain explanation for this ?

Last edited: Jun 12, 2014
13. Jun 12, 2014

### Staff: Mentor

14. Jun 12, 2014

### willem2

The total loss of energy is constant, because the battery delivers an amount of energy equal to CV^2 and (1/2)CV^2 ends up on the capacitor. Of course if you charge it faster, you will lose energy faster.

15. Jun 12, 2014

### stevan

If I use battery 12 volt to charge a capacitor.... in this formula CV^2 , what voltage is V ?
is it the battery voltage ( constant 12 volt), or difference voltage between battery and capacitor (at first 12 volt and gradually down to 0 volt) ?

then, if I add one more battery 6 volt at the bottom of the first 12 volt battery...in backward position...now what voltage is V in the formula?

is the capacitor just like this 6 volt battery (the difference is the voltage is grow from 0 to 12 because of charging)?

Last edited: Jun 12, 2014
16. Jun 12, 2014

### willem2

V is the battery voltage. The capacitor voltage will be equal to it when it is charged. The energy of a charged capacitor is (1/2)CV^2.

17. Jun 12, 2014

### stevan

then, if I add one more battery 6 volt at the bottom of the first 12 volt battery...in backward position...now what voltage is V in the formula?

is the capacitor just like this 6 volt battery (the difference is the voltage is grow from 0 to 12 because of charging)?

18. Jun 12, 2014

### Staff: Mentor

V is the final voltage across the capacitor after it is charged. Let's not add upside down batteries to our circuit since we wouldn't do this in real life. Keep it simple.

19. Jun 12, 2014

### 256bits

The battery has to supply the energy that is heating up the resistor ( yes, resistors do heat up when electrons flow through them ) and charging up the capacitor (stored as an electric field ).
You have to admit this is true - it certainly is different than just a resistor connected to a battery or just a capacitor connected to a battery.

So why does the energy delivered to a resistor from the charged capacitor not equal to, but less the, the energy the battery had to supply when connected to a resistor to charge up the capacitor?

Both scenarios, it would seem at first glance, since we are starting at 12 volts, should be equal.
In fact, as you have pointed out, the same electrons have to pass by the resistor, first from the battery to the capacitor, and then from the capacitor through the resistor with the battery disconnected. And in both cases the voltage through the resistor drops from 12 volts to 0 volts, so the resistor should heat up by the same amount, which it does.

So what is different?
Well it must have something to do with the electric field stored in the capacitor. and how that changes when electrons are removed, and how electrons in the battery get their 12 v potential.

For the capacitor, as we remove electrons, the voltage drops - the electric field is not as strong as it once was. At the point where the last electron is removed and passes through the resistor, there is zero charge on the capacitor which then has zero volts and zero electric field. We have used it all up, or rather the resistor has converted all that energy into heat. Note that all electrons are NOT at 12 volts but the ones coming after the each one previous has a lower potential ranging from 12v to 0v.

That is similar to when the capacitor is charging up, only in the reverse direction.

So what about the battery.
If it is a good battery, and we do not drain it by charging the capacitor, then the electrons are all coming out are at 12 volts. No electron has any potential lower then that - Not 11 v, Not 10 v, Not 0 v. All are at 12 v. And therein lies the difference between the battery and the capacitor.

How do they get to 12 volts? The chemical energy of the battery boosts them up from 0v to 12 volts. If you send one out to the resistor and capacitor, a new electron has to take its place to be ready for sendoff. Expenditure of chemical energy within the battery increases all those electrons
waiting to be part of the circuit all the way up from 0 v to 12 v.

So, you can see then, that when charging the capacitor, these electrons from the battery boosted up from 0v to 12 volts from the expenditure of chemical energy, will expend some of their energy in the resistor, or contribute to the electric field of the capacitor, depending upon how fully charged the capacitor is. When discharging the capacitor, these electrons can now expend the rest of their energy saved in the electric field of the capacitor by flowing through the resistor.

In essence, there is no dilemma, and no energy imbalance.

20. Jun 12, 2014

### stevan

:rofl:

No sir, I don't want to make this complicated..

My point is: is the capacitor do the same act as the inverted battery? the difference is the voltage of the capacitor is not constant 6 volt but growing from 0 to 12 volt

If so, then formula energy come from battery CV^2 is not right because the voltage is not constant (from 12 to 0)?

21. Jun 12, 2014

### CWatters

Instead of replacing the battery with a simple conductor why not replace it with another capacitor? If your theory is correct you should be able to repeat the process forever with energy just flowing back and forth.

It obviously won't do that but you might learn something in the process.

22. Jun 12, 2014

### Staff: Mentor

We know that the energy stored in the capacitor is one-half that which was given up by the battery. So if the energy in the capacitor is 1/2CV2, then obviously CV2 is the energy lost from the battery.

For a proper explanation of the math, visit the page I linked. It should explain it pretty well.

23. Jun 12, 2014

### stevan

I know this wont work, because each time replacing with empty capacitor, the charge split up between 2 capacitor and getting smaller and smaller

24. Jun 12, 2014

### Staff: Mentor

The light and capacitor are in series. At any time, the voltage across them sums to 12 (the voltage of the battery), with the capacitor going from 0-12 and the light going from 12-0. So each gets half of the energy produced by the battery.

It sorta seems like you don't think a light bulb uses energy, because you seem to be ignoring its energy use.

Last edited: Jun 12, 2014
25. Jun 12, 2014

### Staff: Mentor

Closed pending moderation