ARaman@massey.ac.nz (A.V. Raman) (09/26/89)
This problem has puzzled me for 2 years
Can someone please help me out:
________________
| |
V --- ----- C
--------- -----
| |
-----RRRRR------
Consider the RC circuit above.
The aim is to determine an equation for the energy expended in the resistor
R at t = infinity.
If we denote the energy at t = infinity by E, then
E = integral from t = 0 to t = infinity (power (t) dt)
where power = R * square (i)
since i = f(t) = V/R exp (-t/RC)
E = integral from t = 0 to t = infinity (square (V) / R exp (-t/RC))
which implies E = C * square(V) / 2
which is perfectly logical considering the fact that E = total energy -
energy stored in the capacitor.
But note that the energy expended in R does not depend on R itself.
That won't be quite as astonishing if not for the following reduction
of the above circuit. What if R = 0, V = 1 and C = 1 for example?
On creation of such a circuit, the charge that will flow out of V into C
is 1 coulomb.
By definition then, the energy transferred from V to C is
charge * pressure = 1V * 1C = 1 joule
But what is the energy stored in the capacitor?
It is half C*square(V) = half * 1 * square(1) = half joule.
Where has the remaining half joule gone?
Beats me!
--
/*----------------------------------------------------------------------*/
Anand Venkataraman - Systems group, Computer Center, Massey University,
Palmerston North, New Zealand
INTERNET: A.Raman@massey.ac.nz Ph: +64-63-69099 x7943 NZ = GMT + 12larry@kitty.UUCP (Larry Lippman) (09/27/89)
In article <318@massey.ac.nz>, ARaman@massey.ac.nz (A.V. Raman) writes:
-> This problem has puzzled me for 2 years
-> Can someone please help me out:
-> ________________
-> | |
-> V --- ----- C
-> --------- -----
-> | |
-> -----RRRRR------
->
The summary says it all. I ain't gonna say another word. :-)
<> Larry Lippman @ Recognition Research Corp. - Uniquex Corp. - Viatran Corp.
<> UUCP {allegra|boulder|decvax|rutgers|watmath}!sunybcs!kitty!larry
<> TEL 716/688-1231 | 716/773-1700 {hplabs|utzoo|uunet}!/ \uniquex!larry
<> FAX 716/741-9635 | 716/773-2488 "Have you hugged your cat today?" jeffw@midas.WR.TEK.COM (Jeff Winslow) (09/27/89)
This was hashed over extensively within the last year on these groups. The answer involves radiation, the impossibility of creating a resistanceless and inductanceless connection, and the breakdown of simple circuit equations in geometric configurations that try to approximate such a connection. That is - how are you at applying Maxwell's equations? The law is not violated. I have faith, oh Brother Ohm! :-) Jeff Winslow
jhallen@wpi.wpi.edu (Joseph H Allen) (09/27/89)
In article <318@massey.ac.nz> ARaman@massey.ac.nz (A.V. Raman) writes: Circuit: +--C--V--R--+ I assume that initially no charge on C when the | | circuit is connected. +-----------+ >which implies E = C * square(V) / 2 >But note that the energy expended in R does not depend on R itself. Yes. >That won't be quite as astonishing if not for the following reduction >of the above circuit. What if R = 0, V = 1 and C = 1 for example? An impossible situation. What this means in the ideal case is that some amount of energy is being transfered in 0 time. There is no V which can do this (even if the wires are superconducting). Also, if V is really strong, a superconducting wire would stop being a superconductor since the magnetic field is large. >By definition then, the energy transferred from V to C is >charge * pressure = 1V * 1C = 1 joule >But what is the energy stored in the capacitor? >It is half C*square(V) = half * 1 * square(1) = half joule. >Where has the remaining half joule gone? Half the energy transfered ALWAYS becomes heat. In any "clutch" I.E., a device which transfers energy from one side to another until both sides are equal, 1/2 the energy goes into heat. This is true for both electrical systems and mechanical ones. If you have one flywheel which is spinning and one which isn't and you connect them, 1/2 the energy of the system is lost to heat. No matter how you connect them (I.E., no matter what R is).
liud@guille.ece.orst.edu (Dongtai Liu) (09/28/89)
I once had a similar example that puzzled me for some time. My circuit consistedof two capacitors and a switch. One capacitor has initial voltage and the other has no. the swith is then closed and the voltage on both capacitors becomes 1/2 of the initial voltage on the first capacitor (remember, v is proportioanl to q/c, since q=cont, c=twice as much, v->1/2 as much). conservation law seems to be violated again, since E=0.5*C*V*V, there is only 50% energy left in this simple system. I know the answer; but I would like to give you a clue first. When you close theswitch (or let R=0 in your circuit and connect the battery), wouldn't you hear a click from a radio nearby?r
aephraim@garnet.berkeley.edu (Aephraim M. Steinberg) (09/28/89)
A clear way of seeing that this calculation is faulty without even bothering to complain about the impossibility of zero-resistance and zero-inductance (it seems to me that the same error could be made as easily with a finite R) is to ask why you multiply 1C by 1V. When you start moving that coulomb, there is not yet 1 volt of potential on the other side. It ends up looking like a triangle with base 1C and height 1V, whose area is of course 1/2. By setting R=0, you are simply taking lim t-->0 of an integral, but that doesn't give you the right to drop the integral and just multiply your final values.
jeffw@midas.WR.TEK.COM (Jeff Winslow) (09/28/89)
In article <1989Sep28.013457.28172@agate.berkeley.edu> aephraim@garnet.berkeley.edu (Aephraim M. Steinberg) writes: >A clear way of seeing that this calculation is faulty without even bothering >to complain about the impossibility of zero-resistance and zero-inductance >(it seems to me that the same error could be made as easily with a finite >R)... If you have a finite R, the "missing energy" is dissipated as heat in the resistor. I worked it out on the net when the subject came up before - wish I'd saved it. But it's pretty easy. Jeff Winslow
michael@fe2o3.UUCP (Michael Katzmann) (09/29/89)
In article <318@massey.ac.nz> ARaman@massey.ac.nz (A.V. Raman) writes: >This problem has puzzled me for 2 years >Can someone please help me out: > > ________________ > | | > V --- ----- C > --------- ----- > | | > -----RRRRR------ > >Consider the RC circuit above. > >The aim is to determine an equation for the energy expended in the resistor >R at t = infinity. > >If we denote the energy at t = infinity by E, then >E = integral from t = 0 to t = infinity (power (t) dt) >where power = R * square (i) >since i = f(t) = V/R exp (-t/RC) >E = integral from t = 0 to t = infinity (square (V) / R exp (-t/RC)) > >which implies E = C * square(V) / 2 >which is perfectly logical considering the fact that E = total energy - >energy stored in the capacitor. > >But note that the energy expended in R does not depend on R itself. > >That won't be quite as astonishing if not for the following reduction >of the above circuit. What if R = 0, V = 1 and C = 1 for example? > >On creation of such a circuit, the charge that will flow out of V into C >is 1 coulomb. >By definition then, the energy transferred from V to C is >charge * pressure = 1V * 1C = 1 joule >But what is the energy stored in the capacitor? >It is half C*square(V) = half * 1 * square(1) = half joule. >Where has the remaining half joule gone? >Beats me! > Oh but we forgot something didn't we! To charge the capacitor (charge to flow), we have a current! A flow of current creates a magnetic field, which stores energy (1/2 L * I * I). Those lines between the plates are allowed to have zero resistance but as all RF people know even wire has inductance! Without the resistor we have an LC oscillator. The energy stored in the in the inductor (wire) will be added to the capacitor and then returned to the inductor. The voltage source will just look like DC offset and a short to the oscillation! --------------------------------------------------------------------- email to UUCP: uunet!mimsy!{arinc,fe203}!vk2bea!michael _ _ _ _ Amateur | VK2BEA (Australia) ' ) ) ) / // Radio | G4NYV (United Kingdom) / / / o _. /_ __. _ // Stations| NV3Z (United States) / ' (_<_(__/ /_(_/|_</_</_ Michael Katzmann Broadcast Sports Technology. 2135 Espey Ct. #4 Crofton Md. 21114 USA Ph: +1 301 721 5151
ankleand@mit-caf.MIT.EDU (Andrew Karanicolas) (09/29/89)
In article <318@massey.ac.nz> ARaman@massey.ac.nz (A.V. Raman) writes: >This problem has puzzled me for 2 years >Can someone please help me out: > > ________________ > | | > V --- ----- C > --------- ----- > | | > -----RRRRR------ > >Consider the RC circuit above. > >....... >That won't be quite as astonishing if not for the following reduction >of the above circuit. What if R = 0, V = 1 and C = 1 for example? >.... >Where has the remaining half joule gone? >Beats me! > The capacitor problem once again! The difficulty with what you are asking is that you are not considering _how_ the voltage is applied. If it is applied in a step function manner, then you cannot consistently talk about the energy as you will be implicitly evaluating an integral E = int(-inf, inf){ v(t)*i(t) }dt v(t)=V*u(t) i(t)=C*V*delta(t) where delta(t) is the Dirac delta function. As a result, the integrand involves a product u(t)*delta(t) which is not defined. If you want, I can send you references for more information. It is not a paradox, it is not E&M radiation, it is a mathematical issue. -- =================================================================== Andrew Karanicolas MIT Microsystems Laboratory ankleand@caf.mit.edu MIT EECS Cambridge, MA 02139 ===================================================================
phil@diablo.amd.com (Phil Ngai) (09/29/89)
In article <318@massey.ac.nz> ARaman@massey.ac.nz (A.V. Raman) writes: |By definition then, the energy transferred from V to C is |charge * pressure = 1V * 1C = 1 joule |But what is the energy stored in the capacitor? |It is half C*square(V) = half * 1 * square(1) = half joule. |Where has the remaining half joule gone? |Beats me! Congratulations, you've come up with an example which shows that the circuit model of reality has idealizations which can lead to incorrect answers. Any model of reality will have corners where they are not useful. Part of your job as an engineer is to know when you are getting close to one of those corners and what to do about it. In this case, you have to go to a lower level and use an E&M model. -- Phil Ngai, phil@diablo.amd.com {uunet,decwrl,ucbvax}!amdcad!phil "Should the US send assault rifles to Colombia? How about small arms?"
yarvin-norman@CS.YALE.EDU (Norman Yarvin) (09/30/89)
In article <4298@wpi.wpi.edu> jhallen@wpi.wpi.edu (Joseph H Allen) writes: >Half the energy transfered ALWAYS becomes heat. In any "clutch" I.E., a ^^^^^^^^^^^^^^^^^^^^^^^ >device which transfers energy from one side to another until both sides are ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ >equal, 1/2 the energy goes into heat. This is true for both electrical ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ >systems and mechanical ones. If you have one flywheel which is spinning and >one which isn't and you connect them, 1/2 the energy of the system is lost to >heat. No matter how you connect them (I.E., no matter what R is). Not true for all "device"s. Take the two-capacitor problem, for instance. Connect an inductor across the charged-up cap. This will form an LC oscillator. At a certain point in the oscillations, all the energy will be stored in the inductor. At that point, add the second cap in parallel to the first. The inductor will then charge up both caps, with no loss of energy. For mechanical systems, one would use springs, perhaps. We can transfer energy in arbitrarily small increments using inductors. Then again, jhallen is probably right for any system where the only energy storage devices are the two between which energy is being transferred.
liud@guille.ece.orst.edu (Dongtai Liu) (10/01/89)
I have seen some interesting explanation about the capacitor problem. Here I would like to atack the problem in another way. Whenever you talk about the combination of a capacitor and a zero resistance, you expect a infinite current (since i=c(dv/dt)), which means there is a high frequency. OK. that's enough. Do you still remember the assumption of KVL? so, the good old "network theory" is nolonger valid here, since the "components" can nolonger be specified as they apea "should" be. THat is why we need Mr Maxwell even if Mr. Ohm exists.