Testing the TK Tar Baby

[Magluvin]

Yes i can. I see why now also.  The source trace is longer.

Also with the cap in series with the load, we have a voltage division of the source.

Sooo, with a 5v source, as the cap charges, the source -cap=vr

So the load only gets half of what the source lost, and the cap gets the other. Then the caps "half" delivers the other half to the load.

Yup. Looked good in the brain. ;] 

Soo is that why we seem to lose 50% of energy when we discharge a cap into another cap till they even out. Voltage division. Not a loss in heat.  No?

Mags

[MileHigh]

TK:

Are we permitted a little diversion while the tar is brewing?

Scenario 1:  So, you have a 1 kg mass sliding happily along on a magic frictionless plane at 1 m/s.

Scenario 2:  The 1 kg mass sliding happily along at 1 m/s on a frictionless plane hits a stationary mass of 1 kg.  When they hit they stick together and keep sliding happily along.

Scenario 3:  The 1 kg mass sliding happily along at 1 m/s on a frictionless plane hits a stationary mass of 0.5 kg.  When they hit they stick together and keep sliding happily along.

So the question is, what are the energies in scenario 1, scenario 2, and scenario 3?

MileHigh

[Magluvin]

Here is one more using the same to charge the cap through the load, but instead of just discharging the cap to the load, we put the source and cap in series(switches).

The first one we see half of the source dissipated into the load and the other half into the cap.
But when we add the cap in series with the source, we have  source+cap=vr.  The energy from the source is equal to the load overall in the series switch mode.

But we still lost during the charging of the cap.

Just fiddling

Mags

[TinselKoala]

Yes i can. I see why now also.  The source trace is longer.

Also with the cap in series with the load, we have a voltage division of the source.

Sooo, with a 5v source, as the cap charges, the source -cap=vr

So the load only gets half of what the source lost, and the cap gets the other. Then the caps "half" delivers the other half to the load.

Yup. Looked good in the brain. ;] 

Soo is that why we seem to lose 50% of energy when we discharge a cap into another cap till they even out. Voltage division. Not a loss in heat.  No?

Mags

No, I don't think so. The losses will always wind up as heat, or maybe RF radiation, which is the same thing just lower frequency. The cap has an "ESR" or equivalent series resistance which is dissipative, and also it does take work to stretch lattices and move electrolytes around and jiggle ions and stuff like that there. But that's not what's causing the voltage drop equalization.
The Energy in Joules on a capacitor goes as the square of the voltage and linearly with the capacitance: E = (C)(V^2)/2. If the caps were perfectly lossless you wouldn't lose energy by the voltage division, just voltage. Energy is the conserved quantity. So you have a cap with known capacitance in Farads and you charge it to a certain Voltage. This gives you a certain amount of Energy in the cap. Then you discharge into another uncharged cap. The voltage will equalize. Now you have apportioned the original energy between the two, in ratio determined by the ratio of the capacitances, at the new equilibrium voltage. Minus some losses from heating and RF and such. If you know the second capacitance you can calculate the equilibrium voltage, and vice versa.

The hot trick, for me, is to charge caps at high voltage in series, then discharge them at lower voltage in parallel. This is how to extract energy from the Earth's electric field and "down-convert" it to useful power.

[TinselKoala]

TK:

Are we permitted a little diversion while the tar is brewing?

Scenario 1:  So, you have a 1 kg mass sliding happily along on a magic frictionless plane at 1 m/s.

Scenario 2:  The 1 kg mass sliding happily along at 1 m/s on a frictionless plane hits a stationary mass of 1 kg.  When they hit they stick together and keep sliding happily along.

Scenario 3:  The 1 kg mass sliding happily along at 1 m/s on a frictionless plane hits a stationary mass of 0.5 kg.  When they hit they stick together and keep sliding happily along.

So the question is, what are the energies in scenario 1, scenario 2, and scenario 3?

MileHigh

The energy is of course conserved, if you have a frictionless plane and so on. And the Kinetic energy of a moving mass, not accelerating, is given by E=(m)(v^2)/2, a formula with a familiar form. This is what you start with in all three cases, and this is what you wind up with in all three cases. But the energy is apportioned into the sliders according to their masses. Momentum is conserved too, and momentum is just mv. So you know you must have E initial = E final, and also you must have mv initial = mv final. Velocity is a vector quantity, so it has direction (or sign) as well as magnitude.
So to calculate the resultant velocity after the collision you just substitute in the masses and Einit=Efinal; you know the masses and v init, so it's easy to solve for v final, and get the apportioned energies which go as the square of the velocity.
If all you need is the final velocity just use CofM. Mv init = (M+m)v final.

Unit dimensions are important here. Energy can be in Joules, Dynes, Ergs, even Electron Volts. Mass can be in grams, kilograms, bushels and pecks, and velocity is of course measured in miles per hour or kiloparsecs per generation. This is the SR (systeme Rosemarique).

But it might be simpler to use cgs or SI units like kilograms, meters, and seconds.

(Was this a trick question?)

((Choice of reference frame is important too. Since motion is relative, so is kinetic energy.))