QuoteCharging a capacitor means storing energy in to it.
No energy went around the capacitor but in to capacitor.
So what caused the resistor to heat up? If it had been an LED it would emit light. If it were a motor it would turn. You cannot charge that capacitor without those things happening and yet, according to you, there was no energy available to them. They would be perpetual motion machines running on nothing!
Lets do a simple question.
the circuit here
https://en.wikipedia.org/wiki/Capacitive_power_supply
If a DC voltage ( 20 V) is switched into X1 and X1 and left connected indefinitely will there be any current through R4 and R5 temporarily?
Simple yes /no question,
Well I used it cause the simple example seems to be mixed up with a lot of phoohey.
So we know ... and you agreed .. that some energy made it through the circuit to R4 and R5 with the application of a dc source to the circuit input.
Do you dispute this ?
This energy seems to have "gotten by" the capacitor. It is a series element in the circuit.
So what do you say about this energy. We know that if an AC source is connected the circuit functions as a power supply allowing energy to be extracted from X3 and X4. A suddenly applied DC signal has some AC in it.
If you still say that NO energy is transferred by the capacitor where did the energy for R4 and R5 come from?
Where does it come from if AC is applied to X3 and X4.
How come that after capacitor is charged no more current is flowing ?
The answer should be obvious and that is energy went in to capacitor not trough.
The cap is fully charged so cannot be charged more.
The CHARGE went into the capacitor. The energy went around the circuit to allow charging. Once the capacitor is charged there can be no more circulating power, so no more energy. But if there is charge going INTO the capacitor then there is power going AROUND the circuit. If that wasn't so you wouldn't NEED a circuit!
This charging energy is a result of electrons going into one terminal of the capacitor and out the other.
So we have electrons entering and leaving the terminals of the capacitor that go on to impart some energy in other things in the "return path"
And if we keep switching the polarity of the input we can keep getting power out in the return path.
This power will not flow without a route into and out of the capacitor terminals. the capacitor in integral to the transfer of power to the load. All this is very simple.
Can you understand the relation and also the difference ?
QuoteCan you understand the relation and also the difference ?
Can you not understand that to charge the capacitor you need a completed circuit for SOMETHING to go around. If you leave off the return path then nothing will happen. Therefore something must be going down that return path.
Are you saying that this is wrong? That you don't need a completed circuit for the capacitor to charge?
so current will flow from the source through the wires in the energy storage device (capacitor)
Quoteso current will flow from the source through the wires in the energy storage device (capacitor)
And the same current will flow back to the source from the other side on your energy storage device. If not, explain WTF is going through that resistor, and why you need a return path.
Quoteso current will flow from the source through the wires in the energy storage device (capacitor)
And the same current will flow back to the source from the other side on your energy storage device. If not, explain WTF is going through that resistor, and why you need a return path.
See the problem I asked snarkysparky to solve in post 784. Result form that will blow your minds.
I asked to reverse the polarity of the voltage source after the capacitor has charged so circuit is in steady state.
I can provide the answer but likely you won't believe if you do not do that yourself.
I am wrong there - got carried away. But there is A current going through that resistor isn't there? Go on, you have to admit to that. Let me remind you with the simulation:
But the new question is what happens if after you charged the capacitor so no current but capacitor contains those 72mJ and you reverse the polarity on the source.
No, you're jumping ahead and going off on a diversion again. We have yet to determine exactly what is going through that resistor since you insist it can't be energy.
Are you really arguing that work is not required to store energy in capacitor (or in this case a cars battery)?
Are we in Humpty Dumpty land again, where words only mean what you say the do?
If transfer could be done at 100% efficiency then yes no work will be done to transfer energy from source to capacitor.
In the particular example efficiency was just 50% so half of the energy ended as heat during transportation.
You can use a DC-DC converter and transport the energy with about 90% efficiency so that only 10% is lost for moving the charge from a source to a capacitor (energy storage device).
What part exactly is not clear.
It is clear that you are completely wrong.
When there is 0V across a 1F capacitor it takes a very tiny bit of work to remove a charge from the - plate, and add on to the + one. It doesn't matter if it is the same electron or not - it's a little bit of work to pull the electron out of one plate, leaving it slightly positively charged, and a little bit of work to push it into the other plate, which was quite happily neutral and now won't be.
But for the next time there is 1/(6.24 x 10^18) of a V between the plates, so that requires more work. If you do the math it will be 4x that of the first electron.
But for the next time there is 2/(6.24 x 10^18) of a V between the plates, so that requires even more work than the previous one. Once again, if you do the math it will be 9x that of the first electron.
For the (6.24 x 10^18)th electron there is now (6.24 x 10^18)/(6.24 x 10^18) = 1V between the plates. It is now taking a whopping (6.24 x 10^18)^2 times the energy it took to move the first electron.
In fact if you sum up all the work done on each charge it will equal 0.5 * C * Vfinal^2.
And all work being performed by your source capacitor.
The inductor in the DC-DC configuration when the switch is closed it is taking lower current at the higher voltage to make establish a magnetic field, and then when the switch opens it the collapsing magnetic field moves higher current at a lower voltage, at just slightly higher than the voltage of the uncharged capacitor.
It is clear that you are completely wrong.
When there is 0V across a 1F capacitor it takes a very tiny bit of work to remove a charge from the - plate, and add on to the + one. It doesn't matter if it is the same electron or not - it's a little bit of work to pull the electron out of one plate, leaving it slightly positively charged, and a little bit of work to push it into the other plate, which was quite happily neutral and now won't be.
But for the next time there is 1/(6.24 x 10^18) of a V between the plates, so that requires more work. If you do the math it will be 4x that of the first electron.
But for the next time there is 2/(6.24 x 10^18) of a V between the plates, so that requires even more work than the previous one. Once again, if you do the math it will be 9x that of the first electron.
For the (6.24 x 10^18)th electron there is now (6.24 x 10^18)/(6.24 x 10^18) = 1V between the plates. It is now taking a whopping (6.24 x 10^18)^2 times the energy it took to move the first electron.
In fact if you sum up all the work done on each charge it will equal 0.5 * C * Vfinal^2.
And all work being performed by your source capacitor.
The inductor in the DC-DC configuration when the switch is closed it is taking lower current at the higher voltage to make establish a magnetic field, and then when the switch opens it the collapsing magnetic field moves higher current at a lower voltage, at just slightly higher than the voltage of the uncharged capacitor.
You forget one super important thing.
That is reversible so energy storage and not work.
The work will be done when you discharge that capacitor.
Oh come on, you know the formula: work = force x distance,
If you or I push down on a lever, lifting a weight, it is still work, even though the weight on the far end can move the lever back to the original position, undoing our efforts.
Here is the mechanical analogy to the two capacitor problem that actually fits.
Springs can store energy, Stored energy is proportional to the 'stretch' squared - just like a how capacitors store energy. In this analogy, the spring is our capacitor.
Flywheels can store kinetic energy. They also have very good bearings that cause minimal energy loss. This is our deliberately introduced inductance.
Ropes can transfer the forces (admittedly only when it tension, but we will make sure they stay in tension). Here I am making no statement if these are the wires, or the electric field. These are just a mechanical way to transfer the forces. When the rope is moving it is assumed to have minimal momentum. Consider the momentum of the rope as self-inductance - if the flywheel is very light enough (or things are arranged so no flywheel is needed), this might becomes important in in how the system behaves. But with any sensible flywheel it becomes unimportant.
The flywheel has a clamp on it. When the clamp is on the rope can't move, so no force can be transferred through a clamped flywheel. This is our switch.