What happens to the energy stored in a charged capacitor when its plates are connected by a superconductor ?
814565
Assuming the plates are also superconductors and the dielectric between them is vacuum, the only place for the energy to go is by electromagnetic radiation. The connecting superconducting wire will have inductance so you will have a resonant "tank" circuit with a very high "Q" factor. What you get is damped oscillations at the resonant frequency. Depending on dimensions, it could take a while for all the energy to radiate away, but radiate it will.
1028768
Simply put, it's not a conductor. \n\nhttp://jchemed.chem.wisc.edu/JCEsoft/CCA/CCA1/MVHTM/R0540/S567.HTM
426575
Your question is confusing, but here I go.\n\nTo have zero current the circuit must be open.\nThat will give you the higher voltage.\nNow if you have a capacitor circuit, when you close the circuit will take some time to be zero current, while the capacitor charges.\nIn a inductor circuit the current is not possible to be zero because of the nature of the circuit, but will be near zero .\n\nIn a resistor circuit, the current never will be zero, the energy will become heat.\n\nso what are you trying to answer ???
152465
Umm,.. hmm.  You should probably google this, keyword: superconductors.  Its not really a new discovery.
606031
It's O.K. to stipulate there's no resistance - but not that the system can reach a static equilibrium state with no energy dissipation at all. The two assumptions are incompatible.\n\nIn the absence of resistance, the inductance, which can never be exactly zero even for a straight wire, would limit the current and also cause its direction to oscillate back and forth between the two capacitors until exactly 1/2 the original energy is lost through EM radiation. \n\nIf somehow there were no radiation, the system would keep oscillating forever with 1/2 the original energy being held in the magnetic field at the moment in each cycle when the capacitors were each exactly half charged. The rest of the time the capacitors would hold between 50% and 100% of the original energy.
299635
it does with the current what a capacitor does with the voltage. it stores current in open circuit, leting it then go with a continous growth. can be used as a source of magnetic field
340318
The energy is stored in the magnetic field of the coil. There are no other things to store in the coil. “The energy stored in an inductor” is the short form of saying “The energy stored in the magnetic field”.\n\nSimilarly, the energy stored in the condenser means, “The energy is stored in the electric field”.\n\nThe energy stored in a coil can be calculated using, inductance and current. It can also be calculated independently using magnetic flux and volume of the coil. Both are one and the same energy.
930867
Capacitors = stored energy, leaking dieletric (if hazardous), and/or reverse polarity may cause rupture/failure. Over-voltage can cause failure also. Always store de-energized and shorted.\n\nInductors = No hazard if de-energized. Inductors in service with an over-current condition can overheat and cause burns or failure of the component.
126646
Capacitors are charged by a potential difference being applied accross the dielectric causing electrons to flow into (not across) the capacitor.\nBasically, the voltage source is just cramming as many electrons as it can onto one side and sucking up as many as possible from the other.  Eventually, this process slows to a stop once the ballety has done all it can to develop a charge in the capacitor, at this point the capacitor is said to be charged.
742062
The short answer is no, a circuit cannot absorb heat. \n\nElectrons don't know which way they are flowing in a wire, and there will always be an ohmic loss. One exception is with superconducting materials, in which case electrical current flow does not generate heat, and its flow is essentially lossless.\n\nHaving said that, there are devices called thermoelectric coolers that work on the Peltier effect. These devices are constructed with chunks of material assembled between two plates, typically made of ceramic. \n\nThe device operates by creating a termperature difference between the two surfaces when you run a DC current through it.\n\nWhen electrical current is run through the material in one direction, it causes one plate to get cold, and the other to get hot. If you run the current in the other direction, the opposite effect occurs. The hot plate now gets cold, and the cold plate now gets  hot.\n\nUsually, current is run in one direction only. The hot surface is attached to a big heatsink to actually prevent it from getting too warm. This causes the cold plate to get even colder.  But even with that, the hot side gets hotter than the cold side gets cold. That is, there is a net consumption of electrical power which gets dissipated as heat throughout the overall circuit.\n\nBut this is also true of a kitchen refrigerator, which also gets cold when electricity is run through the device. (Don't forget though that the cooling coils on the back of the refrigerator get hot.). A TE cooler is simply a solid state refrigerator.\n\nThere are no devices that inherently get colder when electrical current flows through them.\n\nHope this helps,\n\n-Guru
631888
ahdunno is right in that mfd is microfarad.  (a 800 millifarad cap would be huge - still, 800 microfarad is a very large capacitance, so be extremely careful with that thing.  the energy stored in a capacitor is 1/2 C * V^2, which for 450V and .0008 Fd = 81 joules.  that amount of electrical energy can easily kill you.)\n\nanyway, if you connect a conductor across the capacitor terminals, the amount of current flow will be an exponential decay.  the time constant of the decay will be determined by the resistance of the conductor (as well as the internal resistance of the cap) so it's difficult to say exactly how much "current" will flow because it depends on several parameters that we don't know, and because one would need to specify *when* during the discharge you are measuring the current.  \n\nthere is a little applet here that you can use to calculate what the discharge will look like:\n\nhttp://hyperphysics.phy-astr.gsu.edu/HBASE/electric/capdis.html#c2\n\nyou'll need to find out the internal resistance (aka the "equivalent series resistance") of the capacitor and add it to the resistance of the conductor you use.  for the case of a "very rapid discharge" you'll have to use a very low resistance conductor, and there exists a limit to how fast you can discharge the cap, which is determined by its internal resistance.  the PEAK current which is at the very beginning of the discharge is simply I = 450V / R where R is just the total resistance of the discharge path.\n\nfinally, let me stress the safety point again.  if you are trying to get a "very rapid discharge" then you will need a very low resistance.  for example, 18 gauge copper wire has a resistance of 0.007 ohms / foot.  a typical ESR for an electrolytic is in the range of something like 0.05 - 0.01 ohms.  this means that the wire resistance itself is negligible.  if you go to that applet and plug your numbers in, for a resistance of 0.01 the peak current at discharge is 45,000 amps.\n\nread that again, and think about it.  45,000 amperes.  the wires in your house will melt and catch fire if you put 50 amperes through them.  an 18 gauge copper wire will easily vaporize at that current, and so will the capacitor.  so unless you want some serious burns and probably missing fingers, *don't* mess around with trying to get a "very fast discharge" with a big cap like this, unless you are very sure of what you're doing.  \n\nbelieve me - i've seen the insides of cabinets sprayed with molten copper and had their doors blown off when something shorted across a capacitor.  also, once we found a squirrel which got across the leads of a big cap and it had been cooked down to a hairy charcoal lump.  so please, please don't become a statistic.\n\ncheers !
34488
In electronics, it is a less-commonly used word for capacitor; more often found in non-electronic context, such as the automotive ignition system.\n\nA capacitor is a device that stores energy in the electric field created between a pair of conductors on which equal but opposite electric charges have been placed.\n\nThe ignition system of an internal-combustion engine is an important part of the overall engine system.
641823
http://en.wikipedia.org/wiki/Capacitance
345290
A voltage difference (from a battery?) is applied to the positive and negative terminals of the capacitor.  The battery pumps electrons onto one of the plates which are attracted by the opposite plate but can not jump the gap (arc) and are therefore stored in an amount (charge) depending on the plate area and voltage until discharged.
885785
I don't know what methods you are using in class, but here is the way I'd start:\nProblem 7.63:\nAt t<0, you have a current source feeding a resistor network.  You can calculate the initial voltage on the capacitor.\nAt t=0, the circuit is now a capacitor (with an initial voltage) discharging through 40 k-ohms in parallel with 10 k-ohms.  You can calculate the voltage after 400 micro-sec.\nAt t=400 micro-sec, the capacitor is now discharging through 40 k-ohms.  You know the voltage at the start of this stage, so you can calculate the voltage 1 ms later.\nAt t=1.400 ms, the capacitor is again discharging through a parallel resistor network.  Knowing the voltage on the capacitor at  the begining of this stage, you can calculate the voltage on the capacitor 0.2 ms later.\nKnowing the voltage on the capacitor at t = 1.600 ms, Vo can be found.  The 5 k-ohm and the 35 k-ohm resistors form a divider network.\n\nThere you have it, just look at the circuit at the various stages and calculate the voltage on the energy storage element.
238424
www.superconductors.org
179614
When a plate A with no charge is brought near a charged plate B, the plate A which was originally neutral gets charged on either side.\n \nIf  B was +vely charged, the near side  of A gets negative charges because of the electro staitic attraction and the farther side of it  gets positive charges because of electro static repulsion. \n\nThe charges  induced in plate A are called induced charges.\n \nNow there exists a potential difference between A and B. If the plate A is  earthed, all positive charges in it go to the earth  because of electro static repulsion, \n\nThe potential difference increases further.\n\nThere is air in between A and B which is insulating the -ve charge of B to go to the +ve charge of A.\n\nIf the potential difference is high, the air gets ionised and becomes a conductor of electricity and charges from A go to the plate B via the ionised air medium which we call a dicharge.