# Magic capacitors!

Discussion in 'Electronic Design' started by George Jefferson, Jul 21, 2010.

1. ### George JeffersonGuest

Suppose you have two capacitors connected as

--*--
| |
C1 C2
| |
-----

where * is a switch.

What is the total energy before and after the switch is closed(in general).

If you want to make it easier assume C2 is initially discharged.

Is the energy before and after the same? If not explain why and why it is
not a violation of the conservation of energy law.

2. ### Michael FörtschGuest

The total electrical energy after the charge has been transfered is 50%
of the electrical energy that was stored in C1. The residual 50% had
been handed to the government as a charge transfer tax.

3. ### George JeffersonGuest

You didn't answer the question. I assume this because you don't know.

4. ### George JeffersonGuest

Um you don't get it. Your ignorance in basic electronics amazes me. Michael
got it(although he didn't explain where the energy went but I think gets
it).

Assume the second cap is initially "uncharged" and has the same capacitance
as the first.

Then the initial energy is

Wi = 1/2*C*V^2
Wf = 2*1/2*C*(V/2)^2 = 1/4*C*V^2 = 1/2*Wi

Hence the final energy of the system 1/2 what we started with.

I'd really like to hear your explanation but I know thats impossible(as
you'll steal someone elses). After all your the one that believes charge
isn't conserved... heres your change to *prove* it.

5. ### Tim WilliamsGuest

Wrong -- the inductance and resistance of the circuit is unspecified, so 1., there is no correct answer, and 2. it's a nonphysical circuit, so not only is it unspecified, it's meaningless.

Tim

6. ### George JeffersonGuest

WRONG. That has nothing to do with it. We can "specify" that the inductance
and resistance is 0. Just because this is physically not possible does not
mean we cannot create such a hypothetical mathematical model.

7. ### Tim WilliamsGuest

Since we're taking the limit of two variables (R --> 0 and L --> 0), this is a multivariate calculus problem, and therefore it matters which variables we take to zero first, or in what proportion.

It is only necessary to find two different answers to prove the limit does not exist, so we shall take simple directions (R and L axes, respectively).

1. IFF the limits are equal, then the limit exists.
2. In the L = 0 case, half the energy disappears (E = 1/2 Eo); in the R = 0 case, the energy remains (E = Eo).
3. Because the answer is not equal under different conditions, the limit is undefined. QED.

Tim

8. ### Tim WilliamsGuest

The presumption is radiation is able to leave. Expressing radiation as a lumped constant equivalent series resistance, assume zero ESR. In other words, inside a superconducting box, so there is no way to measure the resonance (it's a theoretical problem, that's okay) and it resonates forever.

Tim

9. ### George JeffersonGuest

1/0?

I don't see what this has to do with the problem though... I'm sure you
do... of course this assumes you are right.

10. ### George JeffersonGuest

You have no clue. The "lost" energy must go somewhere. It goes into moving
the charge from one plate to another. If there is no resistive force then no
energy is lost in the resistance. If there were a resistance then it would
reduce the overall voltage and that loss would be converted into heat in the
resistance.

When the switch is closed the negative charge must move from one plate to
another. Hence a force exists and because of the finite length between the
plates work is done.

Regardless of how the caps are connected the there will be work done in
moving the charge. The link provided by phantom shows that in the ideal case
the energy is radiated away. This should make absolute sense as the medium
surrounding the wire must absorb the energy(hence it is radiated).

Your "proof" is just pure nonsense. The idea is that you must put in half
the energy to move half the charge(which is 1/4 the original energy) and
regardless of how the energy is dissipated it must be done in some way. For
an ideal wire it obviously can't be done through heat.

11. ### George JeffersonGuest

I'm glad my post got what it was suppose to get out. I kinda feel like
Breitbart.

12. ### Guest

Which is why I didn't say "model R => 0".

13. ### Guest

Nope. Tautology is not a mathematical model.
Model infinite current and zero time. The equations blow up (Larkin's
"singularity").

14. ### Martin RiddleGuest

So, is this like "Wait 30 days and get a different answer?" , I play
this all the time with those management types.

Cheers

15. ### markpGuest

The simple answer is that for zero resistance and zero inductance the
voltage and current would have to be in phase, so as you switch there's an
instantaneous drop in voltage across a zero ohm resitor and hence infinite
current for an infinitesimily small time. There is no solution in this case.

Mark.

16. ### Dirk Bruere at NeoPaxGuest

And do we also assume it does not oscillate nor radiate?

17. ### Don LancasterGuest

If you have a resistor between the capacitors,half the energy is
dissipated. The math does NOT depend on the resistance value.

In the case of zero resistance, you get infinite current and a huge
spark that radiates the equivalent energy.

However, you can charge the second capacitor more efficiently with an
inductor in a resonant circuit.

--
Many thanks,

Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
rss: http://www.tinaja.com/whtnu.xml email:

Please visit my GURU's LAIR web site at http://www.tinaja.com

18. ### JosephKKGuest

Sorry Don. that does not stand up to the physics or the arithmetic.
Energy may be radiated, but not by the switch. Nor is there a
requirement for a spark.
Not part of the question.

19. ### markpGuest

<snip>

A 'spark' needs some medium to form, in air this spark is caused by electron
avalanche. Since it involves electrons moving and being accelerated it will
take time. The question didn't say how fast the switch was closed, it could
be much faster than than the time for the avalanche to happen. Also there's
no requirement for any air at all, a vacuum is a pretty good insulator. With
relatively low voltages and an absense of any free electrons (like no sharp
points like dust) there wouldn't be a spark in a vacuum:
http://www.newton.dep.anl.gov/newton/askasci/1993/physics/PHY102.HTM

Mark.

20. ### markpGuest

Yes, Q=CV equation is somewhat misleading in this context. A capacitor
doesn't store electrical charge, it stores energy. This is a very common
misconception, when we say 'charge a capacitor' we don't mean put electrical
charge into it, we mean put energy into it. The plates are equal and
opposite in electrical charge due to an abundance of electrons on one plate
and an equal and opposite charge on the other. The total stored electrical
charge in a capacitor is zero, and the Q=CV equation relates to how much
charge flowed *in and out* of the capacitor (in fact since electrons can't
cross the barrier between the plates, it actually describes the *modulus* of
the abundance of charge on each plate, one abundance is positive and the
other is negative).

Mark.

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