How would you work out the average charging current of a capacitor?
438763
Potential difference (V) across the capacitor is directly in proportion to the charge (Q) in coulombs, calculated by Q =CV   Remember that C must be in Farads, so 1 microfarad is 1x10^-6 not 1, etc.\nSo values up your pd or Volts axis on the graph can be multiplied by capacitance to yield charge (Q).\nThe change in charge Q through the time period in question divided by the length of the time period (in seconds, not milliseconds etc) gives the average rate of movement of charge in Coulombs per Second (= Amps) over that period.\n(Also, the gradient at any point of the trace gives the current at any instant, but that is not what you asked.)
575172
V=IR, voltage = current times resistance. If you know the current flowing in the circuit, just plug that and the known resistance into the formula to get the voltage. If you don't know the current, you need to measure it. One way to measure it is to measure the voltage across the resistor and plug that and the resistance into the equation and calculate it. But if you are measuring the voltage across the resistor, then you don't need to calculate it, do you?
502062
I think following link will help you.Good luck for your future projects.\n\nhttp://en.wikipedia.org/wiki/Capacitors
496124
Calculate the electricity.
1081330
I = 60 x e/s\ne = elementary charge\n60 x e/s = 60 x 1.6 x 10^-19 C/s = 9.6 aA\naA = attoampere.\natto =10^-18
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 !
293488
use a clamp ammeter to measure the line current. then check the resistance of the cable your using. you can get the voltage drop by multiplying the square of current to resistance reading...
786148
Electronic engineering - circuit analysis:\n\nSolving a circuit for the current and voltage according to Kirchhoff's Laws and these device equations:\nResistor   : v = R * i\nCapacitor: i = C * dv/dt\nInductor   : v = L * di/dt\n\nAnything complicated you generally solve by using the Laplace transform, in which case the above equations simplify to\nfor V,I = Laplace transforms of v,i:\nResistor   : V = RI\nCapacitor: I = sCV\nInductor   : V = sL\n\nThen you solve the system of equations for V,I and apply a reverse Laplace transform.
1050520
Think about the simpliest capacitor. It is clear that two metal plates that don't touch won't let a DC current pass, right? But the spacing is very thin so that the plate's charges can effect one another. When you first put current (even DC) to a discharged capacitor, it charges up and for a brief time it does conduct current. How long? Well it will charge to 2/3 of the total voltage in RC seconds (where R is the resistance of the circuit). So when AC current is "fast" enough, the cap is constantly in a charge/discharge state and current flows. The effective resistance (reactance) of the cap is 1/(2*pi*F*C). So the resistance goes down as the capacitance and the frequency rises.\n\nThe inductor, on the other hand is just a coil of wire. Again, it is easy to imagine that it will conduct DC. But when you first apply current (even DC) to a coil, the magnetic field will interact with the coil and produce an opposing voltage. This only occurs when the magnetic field is forming or collapsing (remember, magnetic fields that do not move do not produce electricity). So right when you change the current, the inductor will "resist" the change, but if the current doesn't change, it is just a wire. The reactance of a coil is 2*pi*F*L, so the resistance increases with frequency and inductance.\n\nNow, combining inductors and capacitors is another interesting subject, but you didn't ask about that....
15184
Using Ohm's Law, V=IR, we have 10=I(5).  Solving for I, the current is 2 Amps.
357536
First you will need to work out what the two half equations are (Hint, have a look at a table of standard reduction potentials - these not only give Eo for each half - equation, but also the half - equation of course). URL below if you need it.\n\nSecond, record the values of each half equation, then use the formula:\n\n Eo cell = Eo cathode - Eo anode \n\nto get the cell voltage. (If you are not sure how to do this, try the link below or re post)\n\nThirdly, use the Nernst equation (should be either in your lecture notes or your text, if not try Googling "Nernst Equation") to calculate Eo under the conditions stated in part b of your question.\n\nI hope this helps.
1054541
It is a capacitor tester.  It charges the capacitor and then measures the current that leaks through it.
770048
you have  power = voltage*current   \n\nhere current  = 1500/120 = 12.5 amperes \n\nAn ampere = 1C/s = 6.24 10^18 electric charge of one electron\n\nso in 600 s you have  = 12.5 *600*6.24 10^18  = 4.68 10^22
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.
644653
I thought a capacitor stored up charge in a dc circuit.  If it was ac then it would be charging and discharging 50 times a second - so it could not store a charge, but the current just switches through albeit in alternating directions..
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.
833466
Volts times amps = watts.