how capacitor support voltage?
362318
Capacitors store the applied voltage when the source voltage is removed.  Think of it like a bucket of water filled to the top with water running in it.  If you stop the running water the bucket will still contain the full amount, (the applied voltage).  Put a small hole in the bucket a the water will drain out slowly.  The hole represents a chosen resistance, the escaping water represents the current, and the dropping level the decreasing voltage or capacitors discharge.
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.
349042
It depends on the circuit you ar designing and again the part of the circuit. \n\nFor example if it is a power supply smoothing capacitor it would be a large value in micro farrads. \n\nBut if you want a capacitor for removing high frequency spikes, then it would need a small capacitor in pico or nano farrads.\n\nYou have to work with the formulae for the part of the circuit you are dealing with, and there are loads of formulas.
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.
914458
i=C(dV/dt)\n\nwhere\n i= the instanious current through the cap\nC= the cap siz in farads\ndV= the change in voltage\ndt= the change in time\n\nI use this equation in my op-amp class where were are doing PWM power supplies and motor drivers right now.
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 !
844099
It depends on the applied voltage.  Divide the voltage by the resistance which gives you the amperage.  Now multiply the amperage  with the voltage, and that is your wattage.\nW = V²/R\nThis is straight forward for DC; with AC things become a little more difficult, as because of the fluctuation of the voltage, the average is only .707 of the nominal voltage.
969454
You can make such a graph yourself if you know the capacitance at one displacement.\n\nFor parallel plate capacitance transducers, capacitance is inversely proportional to the distance between the transducer plate and a nearby conductive "ground" plane. This is an hyperbolic function of separation that goes to infinity as the separation approaches zero. So, if you reduce the separation to   one-half what it was initially, the capacitance will double. Reduce the separation to half of that new distance and the capacitance doubles again. And so on, until the plates are too close together to hold off whatever voltage is applied across the plates to measure the capacitance.\n\nStrictly speaking, this formula only applies to plates with infinite surface area. Real plates will have a "fringe effect" at their edges that will prevent them from obeying the inverse distance separation rule perfectly, but this becomes less of a factor for small displacements about a nominal initial "zero" position. Even though the actual capacitance versus separation relationship is hyperbolic, a small enough segment of any curve is approximately linear.\n\nSee the site for a fairly clear practical discussion of capacitors.
143587
Here is something that may or may not help\n\nIf you take a largish non polarized capacitor\n≥45µF\nand you apply AC 120V 60Hz to one side of it\nwith no ground reference! \nyou can then place \na small wattage low voltage lamp across the \ncapacitor terminals and it will light.\n\nThis is a way to send power over a single conductor.\nI have done this.\nI think it would also work with a coil instead of a cap.\n\nAs far as I can tell\nTesla wanted to use this principle to \npower the Earth\nusing the earth as the single conductor.\n\ndo you mean generate electric power at the other site?\ndo you mean energize an electric generator?\nmaybe for a very focussed wave and relatively small distance or a tiny amount of power.\nBut that is sort of why we use wires so much!\n\ndon't RFID cards transduce a wave and send it back\nI mean technically that is an example right?
827050
Yupp... but unfortunately is a hardcopy on my desk. Perhaps you may straight to tell us what\n\n1. You are designing from HT to Transformer 11kV / 415V\n2. You are designinig from Transformer to LV ??\n3. What is the kVA of the transfomer?
547848
To get voltage doubling, you basically charge up a capacitor on one cycle and on the reverse cycle, run in series with that capacitor so you have 2X the voltage. For a full wave voltage doubler, you need to have an innovative arrangement of rectifiers so that you can get full cycle charge at all times.\n\nA example of a genuine full-wave voltage doubler may be fonud here:\n \nhttp://www.kwarc.org/bulletin/99-04/tech_corner.htm
469560
you need to find one that gives u enough of the voltage and ampage, as most of the universal ones, do set them selfs to what your laptop needs, and are generally cheaper than the manufactures ones (you just need to know if your laptop falls in the range of what it can supply.\n\n\nif you want manufacture adaptors you could look here for toshiba adaptors http://www.portables.co.uk/acatalog/Toshiba_ac_adapters.html
1177002
yeah you can use a switch (voltage switch) or if you can use binary logic, need to know how much voltage/current you are dealing with
232108
Ideally a capacitor has an infinite resistance (which is a DC measurement). However the AC equivalent of resistance, reactance (usually denoted as X), tells you how much AC current a capacitance passes with a given AC voltage applied across its terminals. Reactance depends on the AC frequency. It is X = 1/(C*W), where C is capacitance and W (omega) is 2*pi*frequency. Similarly to Ohm's law, you can find average current from E = I*X. Unlike Ohm's law, this is true for average current but not for instantaneous current, since E and I are 90 degrees out of phase.
1174183
my advice is not to fiddle with the inside of that box... if you touch the capacitor, it will likely kill you... just buy a new one... when i got a replacement, it ran me $50..
475454
he discovered the capacitor and defined capacitance.\nhe gave the formula.\nCHARGE(Q)=C(capactiance)* voltage(v)\nthe unit for measuring capacitance of a capacitor is FARAD.\ncapacitor is a linear component that stores charge.\nit consist of two metal plates separated by a dielectric material eg->glass,mica,fibre..etc,\n\ncapacitance of a capacitor is 1 farad if it can store a charge of  1 coloumb when i volt of current flows through it.
1100598
voltage = current x resistance\nE = I x R  It IS Ohm's law. An easy way to remember the formula and it's configurations is to draw a circle and divide it in half horizontally. Above the line write "E". In the lower half divide the semicircle vertically and write "I" on the left, "R" on the right. It should look something like this:\n                                                     E\n                                                   I    R\nNow just cover up the one you need. and either divide, or multiply.