MH's ideal coil and voltage question

[partzman]

You can do many things that bring things back into equal,, but that is adding in the mechanism that the formulas "as used" with these ideals do not have.

Say if you use pneumatic rams,, then the rams connect the air tank to the flywheel and you have the mechanism,, same thing if you use springs,, or if you use charge separation within the ideal voltage supply <= but it does not have that since it is ideal.

Now what if you are using rams,, what would happen to the pressure inside the ram that is going to stop the flywheel?

Well that depends doesn't it,, an infinite size air tank would not see a pressure change and there you go,, however, there is the adding in of the rams and pipes. <= a mechanism

If you have a limited air tank size then there will be a decrease in pressure as you are spinning up the flywheel and an increase in pressure when slowing it back down.

So using the ideals then we need a mechanism in place that itself does not consume energy but makes the transfer of the potentials equal,, same quantity transferred as stored and then as returned.

Without it the ideal voltage source is only blowing air,, 14.4J worth to spin it up and then another 14.4J worth to stop it,, and like I have said,,, even in this ideal space that can not happen.

Webby1,

Do I understand you correctly that you are saying 14.4J is consumed from the positive 4v supply during T0 to T3 to produce 2.4 amps in the 5H inductor and then as we apply a negative 4v supply during T5 to T8 to reduce the 2.4amps in the 5h inductor to zero, we consume another 14.4J?

partzman

[myenergetic]

High there

Can't say it better. hope the attached helps.

Anybody that cares to read this thread from the beginning will see how truly crazy and nonsensical it could get at times. 

[partzman]

Sort of,

The system starts at zero stored

We add 14.4J and it is stored

We then use another 14.4J to extract the 14.4J we added

If you will
14.4+14.4-14.4=14.4

With a resistor you have
+14.4-14.4=0 dissipated as heat

This should be the outcome for either method but using the formulas and ideals it does not.

If you apply the formula to the negative part of the cycle to get the change in current, that cost is the same as what was stored,, so then you would need to reverse the current value from the source which would be a full stop of current instantly.

If you ramp up the change then it will slowly slow down and stop the current.

The energy differences would be moved into and out of the source, so a sink of ~10.15J from the inductor for a cost of ~4.25J then the last of the extraction of ~4.25J would cost ~10.15J

So it cost me nothing to kill the energy that was stored,, what was taken out equals what was put in to kill it,, but I had to put it in to start with.

This must be wrong,, or incomplete.

Worse is the flywheel analogy using air,, makes an easy think since you are reversing the direction of the input influence to stop the flywheel, so you now have to spend the energy to spin it up,, you also then have to spend the energy to stop it since the air is the brake.

OK, let's take a closer look at the symmetrically equal charge and discharge cycles of the 5H inductor with equal and opposite +4/-4 voltage sources over equal time periods.  I've attached a schematic with Circuit A which is an equivalent to the previous sim, and Circuit B with a re-positioned ground connection but otherwise identical in operation to Circuit A.

I will attempt to point out that during the ramp down phase, the energy stored in the inductor is returned to the -4 supply so theoretically no energy is lost with ideal components.

Looking at Circuit B, when switch W1 is closed, L1 begins to receive a positive current flow out of V1 entering the dot end of the coil and exiting the non-dot end. When W1 opens and W2 closes, the current stored in L1 still flows in a positive direction out of the non-dot end of L1 into the positive terminal of V2 and by the time the current in L1 has reached zero, the energy taken from V1 is returned to V2 due to equal supply voltages and ideal components. The voltage across L1 changes polarity at the switching W1 and W2 but the current begins a gradual decline. These actions are able to be seen in the simulation.

The same energy action occurs in Circuit A but is harder to "see" because of the layout but may become apparent after careful study.  The nomenclature below each circuit calls out the voltage polarities across L1 for the various switch positions. Note that they are identical.

partzman

[verpies]

I will attempt to point out that during the ramp down phase, the energy stored in the inductor is returned to the -4 supply so theoretically no energy is lost with ideal components.

Yes but with the compressed air and flywheel analogy the air is not pushed back where it came from when the wheel is braked so the analogy fails.

With this analogy energy has to be actually expended to create the reverse torque and brake the flywheel, because the air leaks out after it hits the flywheel.  In ideal electronic circuit there is no such leakage.

[verpies]

Since there is no resistance within this circuit there is no loss from the stored energy,, so to MAKE it loose the stored energy you need to supply more energy.

No, it is enough introduce any series resistance into the L circuit in order to make that circuit perform work on that resistance and convert the stored energy into heat ...or introduce an empty capacitor into the L circuit in order to make that circuit perform work on that capacitance and charge it up.