where did the energy go? / General Science and Electronics / Forums

Forums

4hv.org :: Forums :: General Science and Electronics

« Previous topic | Next topic »

where did the energy go?

1 2 next

Move Thread

LAN_403

IamSmooth

Sun Jan 24 2010, 04:56AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

See this for two capacitors, C1 and C2, where C1 = C2.

If C1 is charged, and a switch is thrown to dump half the charge into C2, one can see that half of the energy is gone. The explanation is parasistic resistance, but in the example, this seems to happen if there is no resistance in the circuit (ideal). Can someone explain this if there is an ideal circuit with no resistive element.

radiotech

Sun Jan 24 2010, 08:31AM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546

When the switch is closed, each capacitor will have an equal charge.
charge is in Coulombs.

A first look: ...premise..half the energy is gone.. but energy is constant.

C1 = 1F charged to 1 V Joules = 0.5

C2 = 1F discharged 0V Joules = 0

after some event

C1 + C2 = 2 F Joules = 0.5

by conservation of charge

Charge in C1 = Charge in C2

Voltage level C1 = voltage level C2

by parallel circuit, constant voltage
energy is conserved

Joules = 1/2 C V^2

0.5 = 1/2 * (2F * V^2)

the voltage across C1 + C2 will be 0.707 Volts.

in experimental work seperating the plates
increases the voltage, bringing them together
reduces it. ( when the second teathered satelite mission failed as the kevlar reinforced wire broke, Considerable excitement resulted as he parting ends tried to maintain the current furnished by the ion sources on the shuttle.

No switch can open instantly .It rises from closed resistance to an acceptable value but passes through a point where half the load voltage is across the contacts. At that point, half the power dissipated in the load is also dissipated in the switch. But since the load resistance seen by the source has doubled, the system will only be dissipating 25 % of closed switch level. Adding to the problems is the negative resistance tendancy of high current density arcs.

Why not look at this as a problem of a given space losing energy to another space which when added to the first space, the average energy density reduces. Is this not the electrical case of entropy?

~~

Sulaiman

Sun Jan 24 2010, 08:45AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

Energy loss in arcing is rubbish .. the switch is closing not opening.

IF the capacitors and wires are 'ideal' (no resistance or inductance) then the instant the switch closes an infinite current pulse would result.

Resistance could be minimised to a negligible value (superconductors etc.), but the inductance cannot,
so we are left with a resonant L.C circuit which will radiate electromagnetic waves/energy.
In a real life situation the resistance(s) cannot easily be eliminated so the 'lost' energy will be shared between resistive heating and electromagnetic radiation.

Mattski

Sun Jan 24 2010, 10:52AM

Registered Member #1792 Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527

I did the math several years ago (integrate I^2*R) from which I concluded, for any nonzero resistance between the two capacitors, the energy loss in the resistor would account for all of the cap PE loss.

I showed this to the physics prof who first posed me this two capacitor problem, and he said that it also occurs in circuits built out of superconductors, and that in that case all of the lost power goes to radiation. Now the way to look at this case is that as Sulaiman mentions, you have an LC resonator. We know that energy is going to slosh back and forth between the two caps, and if there were no losses at all, it would continue doing so without end, and we would never reach the end state we already know we must reach. So therefore if there is any loss at all due to radiation, even an amount that would normally be negligible, then the circuit will keep oscillating until it has lost half of its energy. Now if you could every make a truly lossless system, it would simply oscillate forever.

An interesting analogy which is almost identical, is consider two columns of water, one with a height h1 of water, the other empty. The potential energy is integrate(m*g*h) from 0 to h1 over h = 1/2*m*g*h1^2. Now connect the two columns and allow to equalize. The height must now be equal between the two, otherwise water would flow. The math, which is essentially identical, shows that your potential energy is now int(m*g*h,h,0,1/2*h1) = 1/4*m*g*h^2. The lost energy was probably mainly consumed doing work on the air in the columns, which if the columns were sealed would mostly be turned to heat, or radiated into the surrounding air if they are open. And again, if the system could be made lossless, you can probably see that the water would just oscillate between columns forever.

There is probably a nice thermodynamics explanation for this, but basically, you have a lot of potential energy, and you set something in motion. At the end of it, you have turned some of that into entropy, cuz entropy's a bitch that way, so you now have less potential energy in the form that you started with.

radiotech

Sun Jan 24 2010, 04:59PM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546

When a switch closes, the same thing happens as when it opens.
I guess you could say, a river cant be crossed without crossing
the middle. One of the issues with the early solid state electronics
involving switches, because things were faster, we had to debounce contacts. Also two metal surfaces touching do not equal a single piece of metal easily.

Faraday has written much on capacitors and dielectrics and strangely not much has changed since.

IamSmooth

Sun Jan 24 2010, 06:16PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

In the water column example, potential energy is M*g*h.

What is the integral of the potential energy?

WaveRider

Sun Jan 24 2010, 06:51PM

Registered Member #29 Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500

Any loop containing the two capacitors would have a finite inductance and resistance, hence:

Closing the switch would close the loop of an "inductor" that, with the caps, would form an LCR circuit.
This circuit would ring (oscillate)
These oscillations will decay away as a result of resistance in the
circuit as well as radiation of energy from the "loop". (The loop is an antenna as well as an "inductor")

Hence, there is an apparent energy loss when all the charges equalise... It is interesting to note that this type of problem highlights the limitations of circuit theory. The redistribution of charge happens on such short time scales that you cannot neglect "field theory" effects like radiation...

radiotech

Sun Jan 24 2010, 07:47PM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546

If a charged capacitor charges another, both plates are involved, the current in the bottom connecting wire equals the current in the top connecting wire, which means that no magnetic field is radiated if the field from one wire cancels the field from the other. Now its a shape problem.

The only reason a capacitor works in the first place, is we have nullified the forces (Kqq/r^2) that want to push on the plates. We dont let hem move usually, electrostatic voltmeters and loudspeakers aside.

WaveRider

Sun Jan 24 2010, 08:58PM

Registered Member #29 Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500

Can you explain exactly what the forces on the plates have to do with the loss mechanisms at work and why this would somehow negate all the physics of the last 150 years since Maxwell proposed his famous theory of electromagnetism?

radiotech

Sun Jan 24 2010, 09:17PM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546

My position is that the energy is conserved and the charge never diminishes. I believe the 'marx generators' just re-connect capacitors to achieve a higher voltage, by just moving the plates of a 'unit capacitor' further apart and the number of joules remains the same.

1 2 next

Moderator(s): Chris Russell, Noelle, Alex, Tesladownunder, Dave Marshall, Dave Billington, Bjørn, Steve Conner, Wolfram, Kizmo, Mads Barnkob