Energy Loss

[SY]

Do not confuse losses and work.

[ChocoHolic]

Originally posted by SY
A resonant circuit is not an apt analogy because you're exchanging energy back and forth in a closed system. This is purely a matter of doing work...

I do disagree again.
It would be a matter of work, if we would have losses and would generate heat (lowest form of work). But in an ideal resonant circuit... there is no work done! If we look to p(t) in each component it has equal areas in negative and positive direction. Resulting work is zero.

[ChocoHolic]

Originally posted by SY
Do not confuse losses and work.

You are right. Losses and heat are not work.

For work we would need to introduce an ideal DC motor...., the consumed energy and done work would look for the resonant circuit like lost energy.

[ChocoHolic]

For the resonant circuit ist does not change anything if we would have work or heat. As soon as there is real power, not only appearant power, then we would shift energy of the choke-cap system into heat or work. But a cap and a choke do not make work, nor heat.

[SY]

That's correct. And that's exactly why it's a poor analogy. There's no mass and springs here, only springs. Consider a massless, lossless ideal spring, contrained on one end. Compress it a distance X. What is its potential energy? 1/2 k x². Now butt up an identical uncompressed spring. Release the first one so that in equilibrium, each spring is now compressed 1/2 x. The energy of each spring is then 1/2 k (x/2)², or 1/8 k x². Two springs, so the total energy is 1/4 k x².

What happened to the energy? Work was done compressing the second spring even though there's no mass or friction. And that work is subtracted from the initial energy.

[ChocoHolic]

Did we loose our original topic?
Losses in a zero Ohms connection due to indefinite high current?

[ChocoHolic]

Originally posted by SY
That's correct. And that's exactly why it's a poor analogy. There's no mass and springs here, only springs. Consider a massless, lossless ideal spring, contrained on one end. Compress it a distance X. What is its potential energy? 1/2 k x². Now butt up an identical uncompressed spring. Release the first one so that in equilibrium, each spring is now compressed 1/2 x. The energy of each spring is then 1/2 k (x/2)², or 1/8 k x². Two springs, so the total energy is 1/4 k x².

What happened to the energy? Work was done compressing the second spring even though there's no mass or friction. And that work is subtracted from the initial energy.

Your model without mass would ring with indefinite high frequency.

Normally the resonant circuit is a nice analogy to a spring-mass system.
A ideal spring mass system does not make work, if we calculate the full period.... But I agree, within one half period there is positve work and in the other one is negative work.

Analogue we could consider the same in the resonant circuit. Means
moving charge up or down , positive area of p(t) and negative area of p(t). I agree to this way of modelling. But still no losses.

[ChocoHolic]

...hm, would your system ring with indefinite frequency?...
No, I think it would not really ring, but it would balance in indefinite short time, with indefinite high velocities and start and stop with indefinite high accellerations...
Hey that's a cool analogy to our original system !

[SY]

This is a problem in statics. Release it as slowly as you want.

edit: for you p-chem veterans, think of an isothermal expansion of an ideal gas with volume increasing from V to 2V.

[ChocoHolic]

... release it slowly, means keeping it in hand until balanced situation?
With this we would not loose energy, but would get well defined mechanical work out of it by integral F(s) ds .
That's boring. Would be a analogy to a mechanically loaded DC motor between the caps.

Interesting is the situation of simply releasing it and let it move freely to balanced situation. There you would loose the energy and would get similar indefinite values like in our cap-cap circuit.