SY said:
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.
SY said:
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.
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.
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.
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.
Did we loose our original topic? 
Losses in a zero Ohms connection due to indefinite high current?
Losses in a zero Ohms connection due to indefinite high current?
SY said:
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.
...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 !
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 !
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.
edit: for you p-chem veterans, think of an isothermal expansion of an ideal gas with volume increasing from V to 2V.
... 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.
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.
Yes, but that wasn't the question. 😀
In a dynamical situation where you're interested in the time behavior, there's nothing to do but to set up the diffy qs and the boundary conditions and flail away. If you get singularities, it means you've picked an inappropriate model- massless, lossless springs, frictionless pulleys, and infinitely rigid rods seem to be out of stock at McMaster-Carr these days...
Yep, that's one way to think about it.
In a dynamical situation where you're interested in the time behavior, there's nothing to do but to set up the diffy qs and the boundary conditions and flail away. If you get singularities, it means you've picked an inappropriate model- massless, lossless springs, frictionless pulleys, and infinitely rigid rods seem to be out of stock at McMaster-Carr these days...
Yep, that's one way to think about it.
It is really true, if we release it slowly, then the spring system will deliver work to the external releasing mechanism. It will always put some force (unbalance of F1 and F2) to the system. Resulting in a function F(x). Force time distance is work. Or better if force is not constant: E = F(x) dx
Nothing lost.
We only get a strange thing if we simply release it.
Do singularities really always mean that the chosen model is not valid? Why?
I really think that your model is a perfect analogy to the original question of the two caps.
Energy in starting situation is clear, balanced end situation and energy in the balanced situation are are clear. But during the transition from unbalanced to balanced we loose some energy and do not really know where. Also both systems make somehow sense, because balanced static situation is usually the lower energy situation. (Different to the ideal resonant circuit which has no balanced static situation and the system keeps ringing).
Furtheron the double spring system and the C-C system show singularities if start to solve it with the classic diffy qs.
If you look to your spring-spring system, there is a force inbalance in the beginnig. The difference of the force will start to move the system. But with zero mass you will get indefinite accelarations and velocity.
I love it !!!
Nothing lost.
We only get a strange thing if we simply release it.
Do singularities really always mean that the chosen model is not valid? Why?
I really think that your model is a perfect analogy to the original question of the two caps.
Energy in starting situation is clear, balanced end situation and energy in the balanced situation are are clear. But during the transition from unbalanced to balanced we loose some energy and do not really know where. Also both systems make somehow sense, because balanced static situation is usually the lower energy situation. (Different to the ideal resonant circuit which has no balanced static situation and the system keeps ringing).
Furtheron the double spring system and the C-C system show singularities if start to solve it with the classic diffy qs.
If you look to your spring-spring system, there is a force inbalance in the beginnig. The difference of the force will start to move the system. But with zero mass you will get indefinite accelarations and velocity.
I love it !!!
Not always, but when you're trying to calculate something simple and everything blows up, it's probably a safe assumption that you're trying to calculate something too simple.
edit: STRONG book recommendation: Shive and Weber, "Similarities in Physics." Really excellent.
Why are people getting in a tizzy about this one and making incorrect assumptions? The charged capacitor has a voltage across it. The uncharged capacitor has no voltage across it. Therefore, when we connect the two together, a charge will flow from the charged capacitor into the uncharged capacitor until both voltages are equal. The charge lost by one capacitor must be equal to the charge gained by the other. The final voltage will be dependent on the relative capacitances. It is not necessary to invoke differential equations, Dirac pulses, EM radiation or even L and R. I would scribble this all down properly with equations but I no longer have paper or pen (moving house in nine days).
But that doesn't answer the original question: why is the ending energy different than the starting energy? That took some explaining.
EC8010 said:
As you say: The charge lost by the first must be equal to the charge gained by the other.
And we made it even easier. We picked two identical caps. This leads to halving the voltage, if we follow the charge approach.
But if we calculate the energy in the beginning and the energy in the end, we loose half of the energy.
If we would chose the energy approach then the resulting voltages would be 70.7% of the original and funny thing the charge would have increased to 141%. That's the discussion.
Two fundamental approaches, which both usually would be easily accepted by most people, but here they do not match together.
Guys,
This is WAY to much much fun, but, I actually have to work today.
Keep in mind, "moving charge through a field" can mean a resistor... that's where the "missing energy" is going. This doesn't apply to caps... the field is between the plates, and there ain't 'posed to be no 'lectrons there. "But, but, but what about?" they say with theit hands flying in the air. Consider a cathode and anode... charge moving through a field... when those electrons land with their mv^2/2 the plate gets hot... hmmm. "How hot they ask", [Va-k * I] says me.
It is clear that energy leaves the system, then in what form?
Place a resistor across the two caps; you will find the missing energy. Now, choose any value of resistance; the amount of energy will stay the same. You will have to settle for 99.9% something becuase in this case the caps will take forever to charge. Somebody that can still manage that l'Hopital junk can do something more elegant. They should be able to prove that as R approaches zero the energy lost will always be 1/2 of the initial.
Go back to the basic RC charge scenario... e^(-t/RC) etc... Charge the cap... wait for eternity. The energy in the cap is exactly half of the energy applied to the RC system... sound familiar?
After 25 years, I couldn't do a differential equation to save my butt. EE's are tought diff. eq. for punishment only... it's then on to LaPlace (which I can do)
The spring thing is a suitable analogy as well as buckets on the dock. In the course of letting the springs move to the relaxed state 1/2 the initial energy is available to do work outside the system.
No R... no L... no reality... no unreal supporting math.
🙂
This is WAY to much much fun, but, I actually have to work today.
Keep in mind, "moving charge through a field" can mean a resistor... that's where the "missing energy" is going. This doesn't apply to caps... the field is between the plates, and there ain't 'posed to be no 'lectrons there. "But, but, but what about?" they say with theit hands flying in the air. Consider a cathode and anode... charge moving through a field... when those electrons land with their mv^2/2 the plate gets hot... hmmm. "How hot they ask", [Va-k * I] says me.
It is clear that energy leaves the system, then in what form?
Place a resistor across the two caps; you will find the missing energy. Now, choose any value of resistance; the amount of energy will stay the same. You will have to settle for 99.9% something becuase in this case the caps will take forever to charge. Somebody that can still manage that l'Hopital junk can do something more elegant. They should be able to prove that as R approaches zero the energy lost will always be 1/2 of the initial.
Go back to the basic RC charge scenario... e^(-t/RC) etc... Charge the cap... wait for eternity. The energy in the cap is exactly half of the energy applied to the RC system... sound familiar?
After 25 years, I couldn't do a differential equation to save my butt. EE's are tought diff. eq. for punishment only... it's then on to LaPlace (which I can do)
The spring thing is a suitable analogy as well as buckets on the dock. In the course of letting the springs move to the relaxed state 1/2 the initial energy is available to do work outside the system.
No R... no L... no reality... no unreal supporting math.
🙂
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