capacitor math anomaly
capacitor math anomaly
(OP)
I may have a mathematical anomaly concerning the energy transfer between two capacitors.
Here is the experiment:
CAPACITOR SPECS COULOMBS JOULES
CAPACITOR 1 aka C1
C1 11200UF@19V
@ 19V 212.8m coulombs @ 2.02 joules
@17.48V 195.77m coulombs @ 1.71 joules
TOTAL discharge 17.03m coulombs @ .31joules
CAPACITOR 2 aka C2
C2 1000UF@100V
TOTAL CHARGE/ C2@ 17.48V 17.48m coulombs @ .16joules
Capacitor calculator: htt p://highfi elds-arc.6 te.net/con structors/ olcalcs/ca clchrg.htm
Note: All non-polarized capacitors (polypropylene dielectric- ESR rating of .1582ohms) used in test are of high quality and were well within their tolerances and should maintain at least 90 to 95% efficiency through out all testing cycles.
Wire leads used to connect components to circuit are the actual leads on the capacitors, installed by the manufacturer.
Test:
C1 (19v starting voltage) discharges directly into C2. Both capacitors equalize to an identical voltage of 17.48v within 1 second.
Note: According to the math, C1 discharged 17.03m coulombs @ .31 joules during this cycle directly into C2, which can only hold .16 joules @17.48v.
What happened to the remaining .15 joules during this energy transfer?
This one has me wondering!!!!
Thanks for any help guys!!!!
Brad
Here is the experiment:
CAPACITOR SPECS COULOMBS JOULES
CAPACITOR 1 aka C1
C1 11200UF@19V
@ 19V 212.8m coulombs @ 2.02 joules
@17.48V 195.77m coulombs @ 1.71 joules
TOTAL discharge 17.03m coulombs @ .31joules
CAPACITOR 2 aka C2
C2 1000UF@100V
TOTAL CHARGE/ C2@ 17.48V 17.48m coulombs @ .16joules
Capacitor calculator: htt
Note: All non-polarized capacitors (polypropylene dielectric- ESR rating of .1582ohms) used in test are of high quality and were well within their tolerances and should maintain at least 90 to 95% efficiency through out all testing cycles.
Wire leads used to connect components to circuit are the actual leads on the capacitors, installed by the manufacturer.
Test:
C1 (19v starting voltage) discharges directly into C2. Both capacitors equalize to an identical voltage of 17.48v within 1 second.
Note: According to the math, C1 discharged 17.03m coulombs @ .31 joules during this cycle directly into C2, which can only hold .16 joules @17.48v.
What happened to the remaining .15 joules during this energy transfer?
This one has me wondering!!!!
Thanks for any help guys!!!!
Brad






RE: capacitor math anomaly
Alan
"The engineer's first problem in any design situation is to discover what the problem really is." Unk.
RE: capacitor math anomaly
Alan
RE: capacitor math anomaly
No this is not homework-just working on an experiment...
RE: capacitor math anomaly
RE: capacitor math anomaly
The fact that energy in a capacitor equals 1/2*U^2*C says it all. Even if you use superconductors and air as a dielectricum, you can not avoid losing half of the energy at each transfer. If there are no resistive losses, half of the energy will be lost due to arc and radiation, EMP if no resistance at all.
That is how nature behaves.
Gunnar Englund
www.gke.org
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RE: capacitor math anomaly
That was my exact thought at first.
But from experiments, it appears that when discharging C1 form 19v to 17.48 volts though a load, it produces the
same work as when C2 discharges through the same load from 17.48v to 0v.
The problem is if there is a 50% loss during the energy transfer, as the math predicts, how can these two test produce almost identical work?
Thanks for any help.
Brad
RE: capacitor math anomaly
Ball 1: mass m, velocity V, kinetic energy E = 1/2 * m * V^2
Ball 2: mass m, velocity 0, kinetic energy 0
After the inelastic collision, both balls are moving at V/2 (by conservation of momentum)
Ball 1: mass m, velocity V/2 kinetic energy E = 1/8 * m *V^2
Ball 2: mass m, velocity V/2 kinetic energy E = 1/8 * m *V^2
Total KE = 1/4 * m * V^2, or half of the initial
Where did the other half of the energy go? Heat from the collision, acoustic energy, etc.
Curt Wilson
Delta Tau Data Systems
RE: capacitor math anomaly
If I understand correctly, this is a representation of what is happening to the electrons during transfer, showing 50% energy loss.
From my response to Skogsgurra, I still do not understand why my test results apparently do not go along with the math.
Any additional ideas are appreciated.
Thanks Brad
RE: capacitor math anomaly
Gunnar Englund
www.gke.org
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RE: capacitor math anomaly
"not be surprised if you see a 5 - 10 percent error."
If I understand correctly, are you saying that there should be a 5-10% difference between the total work produced by C1 compared to C2 ?
Thanks for the fast response,
Brad
RE: capacitor math anomaly
It is not correct in the general case to say that half the energy transfered is lost. Take the case of two equivalent capacitors (or my example of two balls of equal mass). Of the original one unit of energy stored in the first capacitor, 3/4 of a unit is transfered out, of which 1/4 of a unit ends up in the second capacitor, so 1/2 of a unit is lost. This means that (careful now!) 2/3 of the energy transfered out is lost.
In Brad's first case where the second capacitor is less than 10% as big as the first, it just so happens that about 1/2 of the energy transfered out is lost.
In Brad's second case of discharging the second capacitor (transfering to an "infinite" capacitor), virtually all of the energy transfered out is lost.
You should be able to derive a formula for percentage lost as a function of the ratio between C1 and C2.
As to where the lost energy goes, remember that a "Circuits 101" model cannot properly handle the sudden shorting of 2 different voltages together. Also, I strongly doubt that a capacitor's ESR rating covers the case of short-circuit discharges.
Curt Wilson
Delta Tau Data Systems
RE: capacitor math anomaly
For Curt. The 'half is mine' (uttered by Nature) is something you need to accept. It is a law of Nature. In other words, whenever charge is exchanged between two capacitors (or balls) half is lost. Otherwise, W=U^2*C would be valid, which it is not.
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: capacitor math anomaly
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RE: capacitor math anomaly
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RE: capacitor math anomaly
Let's say we start with 1.0 joule in the first capacitor and 0 J in the second. Shorting them together, we end up with 0.25 J in each. This means that we have transfered 0.75 J out of the first capacitor, of which 0.50 J is lost. This means that 0.50/0.75 = 2/3 of the transfered energy is lost. Now, in this case, half of the originally stored energy (not the transfered energy) is lost. The same is true of the kinetic energy in my mechanical example.
Now in Brad's example, where the second capacitor is about 9% of the value of the first, he transfers out 2.02-1.70 = 0.32 J from the first capacitor, of which 0.15 J is stored in the seconc capacitor, so 0.17 J, or just over 1/2, of the transfered energy is lost. This is less than 10% of the originally stored energy.
If you work through the general formula as a function of the ratio of capacitances "n" (where C2 = n * C1), you get an expression for the fraction F of transfered energy lost of:
F = (n^2 + n) / (n^2 + 2n)
When n=1, F=2/3.
F can only get to 1/2 in the limiting case of n=0.
Curt Wilson
Delta Tau Data Systems
RE: capacitor math anomaly
To make it clear to me:
Amptramp Are you saying that the "lost" energy is "converted" into radiation?
Curt said:
"Actually, the measurements look correct to me."
Skogsgurra said:
"you should not be surprised if you see a 5 - 10 percent error"
Were you both referring to : "when discharging C1 form 19v to 17.48 volts though a load, it produces the same work as when C2 discharges through the same load from 17.48v to 0v."
Thanks
Brad
RE: capacitor math anomaly
Yes, that is what I am referring to.
Given your numbers and just doing the appropriate Q=C*V and E=0.5*C*V^2 calculations, I match your calculations and measurents for transfering charge from C1 to C2: about 0.15J transfered and 0.17 lost.
For transfering charge from C2 to ground ("infinite" capacitance), virtually the entire 0.15J is lost.
Curt Wilson
Delta Tau Data Systems
RE: capacitor math anomaly
Yes, in the absence of any circuit resistances (including ESR). As Gunnar mentioned:
The papers I noted indicate the typical lumped parameter circuit model does not account for radiation. It should.
If the circuit contains resistance, then the "lost" power will be shared by I^2*R losses and radiative losses.
RE: capacitor math anomaly
If your model does not include any loss or radiation mechamisms, than it will not reach steady state, it will continue to oscillate.
If your lumped model includes resistance (and no radiation), when the transient is finished, the integrated amount of I^2*R dissipated in the resistance is equal to the change in stored energy. If you take the limit as resistance approaches zero, the answer remains the same (current approaches infinite, duration becomes infinitessimal, integral of I^2*R remains finite and constant).
Radiation model already discussed.
A more familiar example would be rolling a ball off of a mountain into a valley between two mountains. Are we surprised that it comes to rest at a lower final potential energy? We shouldn't be...intuitively we know it will because in the real world there are losses. If there were not losses, the ball would roll back and forth forever converting between PE and KE. Just like our capacitor model if no losses it will never reach steady state.
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RE: capacitor math anomaly
So, a few percent, at least, in capacitor value and additional errors in voltage measurement can add up to a non-negligable total error. That was what I was referring to.
Your OP says: "According to the math, C1 discharged 17.03m coulombs @ .31 joules during this cycle directly into C2, which can only hold .16 joules @17.48v. What happened to the remaining .15 joules during this energy transfer?"
My answer is (still) that, including unavoidable input data and measurement errors, you have proved the "Two Capacitor Paradox*" with an error budget of only a few percent.
Well done! But no anomaly.
No paradox, either.
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: capacitor math anomaly
I am impressed!!!
But I'm still not clear if the problem below has been answered.
The problem still is if there are 50%+/-losses during the energy transfer, again, why are my experiments showing, with 1/2 the energy, that C2's total work output appears almost equal(approx 5% less, nothing like 50%) when compared to the C1's total work output.
Side note: I have also used a calorimeter with larger capacitors and different resistors/loads with proportionally the same results in the form of heat.
If anyone is interested, feel free to replicate my experiment and see if you have the same results. I am willing to help with add. info. concerning test setup if needed.
Again all theories welcomed.
thanks guys
Brad
RE: capacitor math anomaly
Then think of this: It is half of the transferred energy that is lost. Not half the energy available in the 'source'.
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: capacitor math anomaly
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: capacitor math anomaly
I'm having trouble understanding what your remaining question is. What do you consider "work output" to be?
Losses (what I think you are calling work output) are always greater than 50% of the energy transfered from the initial capacitor (see my formula above). In the first case (large to small capacitance), it's only slightly greater than 50%. In the second case (small to "infinite" capacitance), it's essentially 100%. But since this 100% is all of the just under 50% of the original, the amount of the loss is almost the same. That's all.
Curt Wilson
Delta Tau Data Systems
RE: capacitor math anomaly
Here's the problem from my perspective:
(note: watts used are for ease of illustration only)
1. C1 with starting voltage at 19v -discharge to 17.48v through load- 1 watt of work is produced by load.
This is my "standard" and is used to compare to C2's efficiency
2. C2 is charged to 17.48-discharge to 0v through same load-
.95 watts of work is produced by load.
My thought is: A capacitor is "50%" inefficient during discharge. This means that C1 discharged .31 joules to C2.
C2 can only store about .15joules at 17.48v. When C2 is discharged, (I hope I have this right) it should only be able to provide approx. .075joules to load to produce work.
Why are my tests showing results that do not match the math?
This is where all the questions started for me.
Thanks
Brad
RE: capacitor math anomaly
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: capacitor math anomaly
Sorry for the misuse of terms..
Lets try this-
C1 discharges through resistor
1btu of heat is produced
C2 discharges through resistor
.95btu of heat is produced
I hope that's clearer..
Thanks Brad
RE: capacitor math anomaly
How do you convert watts to BTU?
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: capacitor math anomaly
Brad: Have you really convinced yourself that if you discharge a capacitor through a resistive load to ground, only half of the stored energy is dissipated in the load, not all of it?
I think you may be tying yourself in knots because you are using the concepts of "efficiency" and "work" in the opposite manner from how they are usually used. The fraction of energy transfered that is dissipated in something like a load is not typically considered the efficiency -- it is one minus the efficiency. And what is dissipated is typically not considered work -- it is considered a loss that is not available to do real work.
When you discharge a capacitor to ground, you are converting all of the stored energy to heat. None is available to do useful work, so this process is considered 0% efficient. As I have said a couple of times, you can consider "ground" to have infinite capacitance if you want to treat this as a problem of transfering from one capacitor to another.
skogs: I repeat that the idea that these transfers are 50% efficient is simply wrong. Having gone through the analysis, not only is it not always true, it is never true. The losses (as a fraction of the energy transfered out of the first capacitor) must always be greater than 50%, and can go to 100%.
Curt Wilson
Delta Tau Data Systems
RE: capacitor math anomaly
It could be that I and some other of the participants in this thread have misunderstood the OP's question. If that is so, I apologize for being thick and persistant.
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: capacitor math anomaly
I hope this is a better way to explain my observations.
If not, please help me phrase it correctly.
I'm here to learn.
Let's try this one
C1 discharges .31 joules into resistor-
a certain amount of heat is produced.
C2 discharges (calculated to be)approx. .15 joules into same load
and produces 95% as much heat as C1.
Thanks Brad
RE: capacitor math anomaly
I cannot interpret that as anything else than a variation of the two capacitor situation. If you discharge a capacitor in a resistor, then most of the losses (I wouldn't call them losses, but resulting heat) are in the resistor and some in the capacitor's ESR and connecting wires. But that is not what we discussed from the start - has the subject changed somewhere in the thread?
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: capacitor math anomaly
You all are defiantly putting me on steep learning curve.
I have a feeling we may be getting to an answer.
Curt, am I understanding correctly that my observations may not be far from reality?
Also
Guys please do not get in any arguments concerning my questions- I like to look at this forum as conversation with a bunch of friends over a beer or even a diet coke.
Just sharing ideas in search for the truth...
Now, let me meditate on your last posts
I will be back this evening
Many thanks to all of you,
Brad
RE: capacitor math anomaly
The original question was:
What happened to the remaining .15 joules during this energy transfer?
This became an issue to me because the experiments did not "jive" with math.
You said:
"But that is not what we discussed from the start - has the subject changed somewhere in the thread?"
Maybe a little, I believe we went off on this tangent when I started to explain the reasons for my 1st question.
If I should start a new thread, let me know and I will do it.
Thanks Brad
RE: capacitor math anomaly
1. I think that we all agree that if we discharge one capacitor into another and let the two equilibrate, we end up with less stored energy in the two capacitors than was originally stored in the first capacitor. The difference in energy must leave the system.
2. If the discharge is essentially through a short circuit, the energy loss (of whatever amount) cannot be explained by dissipation through ohmic resistance, as in basic circuit models. (This is the "capacitor paradox".)
3. The papers than amptramp links explain that the "missing" energy is lost through radiative means, as you and several others already noted. This resolves the paradox.
So far, so good, but we have not talked about the amount of energy dissipated (whether through ohmic losses or radiation). You say it is always 50% of the energy transfered out of the first capacitor. I say it can range from 50%+ to 100%, depending on the ratio of the capacitances. (By the way, neither of the cited papers claim the 50% loss, and the equations they provide support my analysis.)
Let's start with the simple case of equal capacitances (C1 = C2 = C).
We start with Cap 1 at an initial voltage Vi (and Cap 2 at 0V). So the charge in the system is
Q = C * Vi
and the stored energy is
Ei = 0.5 * C * Vi^2 = 0.5 * Q^2 / C
Now we discharge Cap 1 into Cap 2 and let them equilibrate at a final voltage
Vf = Q / 2C = Vi/2 (by conservation of charge)
The stored energy is now
Ef = 0.5 * 2C * (Vi/2)^2 = 0.25 * C * Vi^2 = Ei/2
This is split evenly between the two capacitors, so each is now storing Ei/4.
The energy transfered out of Cap 1 is:
dE1 = E1i - E1f = Ei - Ei/4 = 3*Ei/4
The energy lost to the system is
dE = Ei - Ef = Ei - Ei/2 = Ei/2
The fraction of energy lost compared to energy transfered out of Cap 1 is
dE/dE1 = (Ei/2) / (3*Ei/4) = 2/3
So even in this very basic example, the idea that losses are always half of the energy transfered out of the first capacitor is disproven.
Yesterday I generalized the analysis by permitting C2 = n*C1. I repeated the above equations for this more generalized case and obtained an expression for the fraction of energy lost of
dE/dE1 = (n+1) / (n+2)
(Yesterday I had not canceled out the final common "n" term from numerator and denominator.)
With n=1, we get the result of 2/3 lost, agreeing with the earlier analysis.
This analysis is consistent with the equations in the papers amptramp linked, including the first equation in the first paper, which expresses the energy loss as a function of the two capacitances and the charge.
This analysis is also consistent with Brad's experimental results. In his first case, n is very small (0.089), so the fraction of energy dissipated is calculated as 52%, in very good agreement with his observations.
In his second case, n is very large (effectively infinite), so the fraction of energy dissipated is calculated as 100%, again in agreement with his observations.
Note that the starting amount of energy in Brad's second case is 48% of the amount transfered in his first case, so the amount dissipated in the second case (all of this 48% of the original transfer) is only slightly less than the amount dissipated in the original transfer (52% of the original transfer). Brad's observations and my math are in very good agreement.
I've been through my analysis several times now and can't find any problems with it. Can anyone else?
Curt Wilson
Delta Tau Data Systems
RE: capacitor math anomaly
RE: capacitor math anomaly
Your post 11 Feb 10 12:56 (the analogy with the colliding balls) is exactly what I meant. Same situation.
I may have been careless with definitions and wording - but I think that I have some company there in this thread.
I do not want to split hairs. All I wanted was to make the OP understand that energy 'disappears' - no matter how well you set up and build your circuit - and that one has to understand that before saying that there is an anomaly in euther the math or the measurements.
Sorry that my simplistic approach caused so much disussion. And, I do not think that your calculations are wrong. Not at all. I didn't even check them.
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: capacitor math anomaly
Curt Wilson
Delta Tau Data Systems
RE: capacitor math anomaly
The key is to realize that you are transferring the "charge" between the caps and not "energy". Only "stored" energy is then calculated based on the final charge and voltage of the caps.
Your total charge (Q=CV) is still conserved, like the momentum (m.v) in an inelastic collision, where the momentum is conserved but not the 'kinetic' energy. Final kinetic energy is calculated based on the velocity and mass after the collision. Total energy is still conserved in some other form but just not all in kinetic form.
The apparent error of 17.03 vs. 17.48 can be attributed to instruments and the deviation from ideal to real world.
Rafiq Bulsara
http://www.srengineersct.com
RE: capacitor math anomaly
I don't think he was aware of the '50% theorem'.
It has been a refreshing excersize and I still respect you a lot.
Good on ya.
Same to you Rafiq.
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: capacitor math anomaly
YOU GUYS ROCK!!!!!!!
Curt, you get the Gold Star...
If all of you are in agreement with Curt's equations and theory, I will consider this problem solved with the math to prove it.
In closing, I appreciate everyone's help and input.
Within 2 days, I now have the answer to a question that none of the local boys could answer. By the way, they all thought I was a "Nutcase".
Just getting started with the questions, till next time,
Thanks Brad
RE: capacitor math anomaly
If you think Curt deserves a star, please give him one by selecting "Thank cswilson for this valuable post!" at the bottom of any of his posts.
Thank you for the question, and welcome to the Forum!
RE: capacitor math anomaly
Thanks for the reminder
Just did it!
Also thanks for the info on the Capacitor Paradox
Very informative and inspired other questions...and answers.
Later on,
Brad
RE: capacitor math anomaly
1 - Charge will always be conserved. And it is easy to account for.
2 - Total energy will always be conserved... but it can be a lot more difficult to account for the energy transfer mechanisms that transfer energy out of your electric circuit into the environment.
Two energy transfer mechanisms to consider: 1 - resistance losses, 2 - radiation. Keep it simple and focus on the 1st one which is easier to analyse. Use a lumped circuit C / R / L model with ANY resistance at all.....I don't care how small the resistance is, just make it more than 0. When you analyse your lumped circuit model (which by definition does not have radiation losses), you will find that the different between initial and final energy stored in the capacitors is exactly equal to the integral of I^2*R losses in the resistor. You don't have superconducting leads do you? So you should not expect the total final energy stored in your caps to equal the total initial energy in your caps. You can certianly do the analysis for RC or RLC circuit and both will validate the statement. I can certainly provide that analysis to you if you want. But... does one really need to do an analysis to verify that energy is conserved?
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RE: capacitor math anomaly
C1:=11200E-6;
C2:=1000E-6;
We have total charge Qtotal, initially on C1
Qtotal:=C1*19 = 0.212800
Solve for q1 and q2 which are the final charges on the two capacitances.
We have two equations:
Equation 1:
eq1: q1+q2=Qtotal;
eq1: q1 + q2 = .212800
Equation 2:
V1 = V2 (final voltages are equal)
q1/C1=q2/C2;
The solution of Equation 1 and Equation 2 is:
q1 = .1953573770
q2 = .01744262295
Therefore we predict a final voltage:
V = q1/C1 = q2 / C2 =17.44.
That matches your result to 3 significant figures. Anything different would be unexpected (charge needs to be conserved).
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RE: capacitor math anomaly
I don't know if I am rehashing what was already said, but an important point to realize is that that your result is exactly what is predicted by conservation of charge within a tenth of a percent or so.
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RE: capacitor math anomaly
================Symbols=============
Qtotal = initial charge on C1
q1, q2 = final charge on C1, C2
V0 = Initial Voltage
Vfinal = Final Voltage
Cs = C1*C2/(C1+C2) is the equivalent series capacitance of C1 and C2
i(t) = current as function of t
I0 = initial current upon closing the circuit
Pr(t) = resistive I^2*R losses as function of t
Wr = total energy dissipated in resistance
Wc0 = initial stored capacitive energy
Wcf = final stored capacitive energy
DeltaWc = Wc0 – Wcf = change in stored energy
First solve the values final q1 and q2 as before:
eq1: q1+q2=Qtotal;
eq2: q1/C1=q2/C2;
This leads to:
q1 = Qtotal*C1/(C2+C1)
q2 = C2*Qtotal/(C2+C1)
Since V is proportional to q, we have:
Vfinal = V0*q1/Qtotal = Vfinal := V0*C1/(C2+C1)
Initial current upon closing the circuit is
I0 = V0/R
The current is given by
i(t) = I0 exp(-t/[R*Cs]) = V/R * exp(-t/[R*Cs])
Pr(t):= i(t))^2 * R = (V0^2/R)*exp(-2*t/R/Cs)
Wr = Int {Pr(t), t=0... Infinity}
Wr = [-1/2*Cs*exp(-2*t/R/Cs)*V0^2] evaluated at t=infinitiy minus same thing at 0
Since the argument at infniity is 0, we have:
Wr := 1/2*Cs*V0^2
Substitute in definition of Cs:
Wr := 1/2*C1*C2/(C2+C1)*V0^2
Initial stored capacitive energy is:
Wc0 := 1/2*C1*V0^2
Final stored capacitive energy is
Wcf:=(C1+C2)*Vfinal^2/2
Substitute in Vfinal = V0*C1/(C2+C1)
Wcf = (C2+C1)*[ V0*C1/(C2+C1)]^2 / 2
Wcf := V0^2*C1^2/[2*(C2+C1)]
Look for change in energy from initial to final
DeltaWc = Wc0 – Wcf
DeltaWc = 1/2*C1*V0^2 - V0^2*C1^2/[2*(C2+C1)]
DeltaWc =0.5 *V0^2 {(C1+C2)*C1 - C1^2} / (C2+C1)
DeltaWc =0.5 *V0^2 {(C1^2+C2*C1 - C1^2} / (C2+C1)
DeltaWc =0.5 *V0^2 {C2*C1 } / (C2+C1)
This is the same as Wr. Energy is conserved, the universe can rest. The interesting thing to note is that it doesn't matter WHAT value of R you choose, the end voltage will always be the same (by conservation of charge), and the total energy dissipated in resistor will always be the same (by conservation of energy), the only thing that changes is how fast the transient proceeds.
Again, you could add L into the circuit and the solution would be more complicated, but I can guarantee it will come out the same... because I know even without solving the circuit that total energy must be conserved.
Again this is one model, ignoring effects of radiation. But as expected whatever model you build, if it's a good model TOTAL energy must be conserved.
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RE: capacitor math anomaly
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RE: capacitor math anomaly
RE: capacitor math anomaly
At the risk of prolonging a ripe subject...
Attached is a spreadsheet to solve the R / L / C / C problem numerically.
The ResultsChart tab shows the results for C1=C2=1 F, R = 0.02 Ohms, L = 0.01H, Vc1(0) = 1.
You can see the plotted current, the two capacitor voltages, and the final total energy stored in circuit elements which decreases from 0.5J to 0.25J (as predicted from conservation of charge). The highest rate of decrease of stored energy occurs when current magnitude is maximum.
If you are interested to figure out how to use the spreadsheet (instructions included), you can try other values of the input parameters. The final stored energy is already known without any simulation (based on final voltages that satisfy conservation of energy), but at least you can see how the transient proceeds.
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