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Magic capacitors!

G

George Jefferson

Jan 1, 1970
0
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.
 
M

Michael Förtsch

Jan 1, 1970
0
George said:
[rather old riddle]

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.

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.
 
G

George Jefferson

Jan 1, 1970
0
John Larkin said:
Energy is conserved, so it's the same, if you account for all the
manifestations of energy.

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

George Jefferson

Jan 1, 1970
0
John Larkin said:
State the question unambiguously and I will.

As I said, the puzzle is both ancient and trivial, so probably JT
invented it. There are web sites and even academic papers devoted to
it. Given all that, how could I not understand it?

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.
 
T

Tim Williams

Jan 1, 1970
0
Michael Förtsch said:
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.

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
 
G

George Jefferson

Jan 1, 1970
0
Tim Williams said:
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.

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.
 
T

Tim Williams

Jan 1, 1970
0
Ok, give us a mathematical model of "divide by zero".

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
 
T

Tim Williams

Jan 1, 1970
0
The Phantom said:
In the R = 0 case, L is not zero and the energy is lost by radiation. See:

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
 
G

George Jefferson

Jan 1, 1970
0
Ok, give us a mathematical model of "divide by zero".

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.
 
G

George Jefferson

Jan 1, 1970
0
Tim Williams said:
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.

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.
 
G

George Jefferson

Jan 1, 1970
0
Jim Thompson said:
Let the hedging begin...

In Message-ID: <[email protected]>

You said:

"Right. If you dump all the energy from one charged cap into another,
discharged, cap of a different value, and do it efficiently, charge is
not conserved."

Note the NOT CONSERVED.

Now you say, "...the C*V (charge) on the first cap obviously becomes a
different C*V on the second one".

Where did the charge come from/go to?

John "The Bloviator" Larkin is totally incapable of admitting error.

I truly suspect you're too ignorant to understand :-(

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

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

Martin Riddle

Jan 1, 1970
0
Jim Thompson said:
Let the hedging begin...

In Message-ID: <[email protected]>

You said:

"Right. If you dump all the energy from one charged cap into another,
discharged, cap of a different value, and do it efficiently, charge is
not conserved."

Note the NOT CONSERVED.

Now you say, "...the C*V (charge) on the first cap obviously becomes a
different C*V on the second one".

Where did the charge come from/go to?

John "The Bloviator" Larkin is totally incapable of admitting error.

I truly suspect you're too ignorant to understand :-(

...Jim Thompson

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

Cheers
 
M

markp

Jan 1, 1970
0
George Jefferson said:
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.

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.
 
D

Dirk Bruere at NeoPax

Jan 1, 1970
0
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.

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

Don Lancaster

Jan 1, 1970
0
Let the hedging begin...

In Message-ID:<[email protected]>

You said:

"Right. If you dump all the energy from one charged cap into another,
discharged, cap of a different value, and do it efficiently, charge is
not conserved."

Note the NOT CONSERVED.

Now you say, "...the C*V (charge) on the first cap obviously becomes a
different C*V on the second one".

Where did the charge come from/go to?

John "The Bloviator" Larkin is totally incapable of admitting error.

I truly suspect you're too ignorant to understand :-(

...Jim Thompson


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: [email protected]

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

JosephKK

Jan 1, 1970
0
If you have a resistor between the capacitors,half the energy is
dissipated. The math does NOT depend on the resistance value.

Sorry Don. that does not stand up to the physics or the arithmetic.
In the case of zero resistance, you get infinite current and a huge
spark that radiates the equivalent energy.

Energy may be radiated, but not by the switch. Nor is there a
requirement for a spark.
However, you can charge the second capacitor more efficiently with an
inductor in a resonant circuit.

Not part of the question.
 
M

markp

Jan 1, 1970
0
You are an idiot. If the two contacts are to be closed, and the
resistance is zero, there WILL be a spark BEFORE closure
occurs. And in ANY case of ANY spark, there WILL be radiation.
<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.
 
M

markp

Jan 1, 1970
0
John Larkin said:
That's funny. But people can choose to be amazed in all sorts of ways.


Michael

Miraculous calculation. Yours and about 300 web sites that admire this
puzzle.

You didn't wxplain where the energy went - see those 300 web sites -
but you are assuming losses. Another solution is that no energy is
lost, and it rings forever, in which case the final state that you
cite never happens. The exact waveforms are actually interesting.


Check my previous posts. I noted the exact waveform across a resistive
switch, for any values of C1 and C2, and an independent way to compute
the energy lost in that switch.

Given an inductor, one can move all the energy from one charged cap to
another, uncharged one. If the C values are unequal, the C*V (charge)
on the first cap obviously becomes a different C*V on the second one.
I noted that here some weeks ago, too.

This is all EE101 stuff.

John

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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