is there a formula to calculate a capacitors charge?
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.
97533
+12VDC, -12VDC, +5VDC, -5VDC
274986
Tf=(9/5)*Tc+32\n\nTf is temperature in Fahrenheit,\nTc is temperature in Celsius.
895782
Xc=1/2π f C\nAs you can plainly see, capacitive reactance is a function of both capacitance and frequency.\nIf you work the arithmetic, you will see that the reactance decreases as capacitance increases while freqency remains stable, thus is inversely proportional.\nYou can work through the various permutations and chart the various relationships.
1096749
Let us some salt having formula AxBy\n the subscripts will represent the charge as follows\n\nA charge is y\nB Charge is x\n\nEx: CaCl2\n\nCa charge is +2\nCl charge is -1\n\ngot it?
924475
= 9u-10u + 6v = 6v - u
292834
Sure, it's called a Non-Polarized Capacitor. Available at any electronics shop.
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.
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.
1095597
Reactance, like resistance, tells you how a capacitor responds (reacts) to a current or voltage. From the reactance, you can find the V/I relationship. It is defined as X=1/wC where w is the radian frequency and C is the capacitance of the circuit.\n\nIt will be better if you can let us know your education level and we can describe according to your need.\n\nhttp://www.kodkodkod.com/database.htm\nhttp://www.kodkodkod.com/index_files/submitquestion.htm\n\nRegards,\n\nKodKodKod Learning Consulting
483906
if you want a quick guess, use the rule of thumb, F=2*C+32\n\nif you want it exact, use the formula which is, F=1.8*C+32
227723
Regardless of what size capacitor you choose, there will be a small ripple on the output voltage, which of course gets bigger as the load increases.  You need to determine what is an acceptable ripple voltage for your load to calculate out a value.\n\nIf you want a totally smooth DC output, you'll need to first rectify and filter, then use a voltage regulator with a voltage set point that's a few volts below the voltage output from the capacitor.  Follow that by a smaller capacitor after the regulator to help catch any remaining transients and response after regulation.\n\nSomething to note:   Your capacitor's voltage rating should be much higher than the actual working voltage you will subject it to.  Eletrolytics like to blow up (literally) if you run them past their maximum rating, even for a short time.   Also remember this, the AC voltage that you measure at the output of the transformer (or the output of the rectifier, unfiltered) is the RMS voltage, not the peak voltage of the AC waveform.  Once you add the filter cap, your DC voltage will appear to be higher because the capacitor is charging to the peak voltage, which is basically the RMS voltage multiplied by 1.414   Keep this in mind as well when selecting the working voltage rating for your capacitor.
560737
En=-(13.6)/n^2eV\nE1=-(13.6)eV\nE3=-(13.6)/9eV\ndelta E=E3-E1=hv\nhv=12eV\ntherefore,v=(12*1.6*10^(-19))/(6.626*10^-34)\nv=2.8976*10^15hz
1114108
Formulas for what?\n\nYou might look for a copy of Engineering Formulas by Kurt Gieck. It was published by McGraw Hill. The 5th edition has the ISBN number 0-07-023231-8
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.
840963
Power Factor Correction:\nIf an inductive load with an active power demand P has an uncorrected power factor of cosf1 lagging, and is required to have a corrected power factor of cosf2 lagging, the uncorrected and corrected reactive power demands, Q1 and Q2, are:\nQ1 = P tanf1\nQ2 = P tanf2\nwhere tanfn = (1 / cos2fn - 1)½ \n\nThe leading (capacitive) reactive power demand QC which must be connected across the load is:\nQC = Q1 - Q2 = P (tanf1 - tanf2)\n\n\nThe uncorrected and corrected apparent power demands, S1 and S2, are related by:\nS1cosf1 = P = S2cosf2\nComparing corrected and uncorrected load currents and apparent power demands:\nI2 / I1 = S2 / S1 = cosf1 / cosf2 \n\nIf the load is required to have a corrected power factor of unity, Q2 is zero and:\nQC = Q1 = P tanf1\nI2 / I1 = S2 / S1 = cosf1 = P / S1 \n\n\n\nModern capacitor values:\nIn around 1970, a standardized set of capacitor base numbers was devised so that the value of any modern electrolytic capacitor could be derived from multiplying one of the modern conventional base numbers 1.0, 2.2, 3.3, or 4.7, by a power of ten. Therefore, it is common to find capacitors with values of 10, 22, 33, 47, 100, 220, and so on. Using this method, values ranging from 0.1 to 4700 are common in most applications. Values are generally in microfarads (µF).\n\nHow standardized values help:\n\nMost electrolytic capacitors have a tolerance range of 20 %, meaning that the manufacturer guarantees that the actual value of the capacitor lies within 20 % of its labeled value.\n\nA 10 µF capacitor, for example, may in actuality be as low as 8 µF whereas a capacitor marked 8 µF may in fact be as high as 9.6 µF. This causes an overlap of 1.6 µF. The standardized base values help this by alleviating overlaps due to tolerance, illustrated by the following table:\n\nThe left numbers are the base values. \nThe center numbers are the differences based on ± 20 % tolerance. \nThe numbers on the right illustrate the range of actual values by applying the tolerance. \nIn an effort to cut costs in the manufacturing process, the number of possible values was limited, because having overlapping values would be viewed as unneeded and redundant, since for most practical applications, having a 20 % (or even higher) tolerance did not adversely affect the functionality of the device. The reduction of variations in the product allowed companies to produce greater quantities of existing samples. This in turn, allowed for lower cost components.\n\n(Retrieved from "http://en.wikipedia.org/wiki/Electrolytic_capacitor_value"\nhttp://en.wikipedia.org/wiki/Power_factor_correction_unit)\n\nWHY CAN CAPACITORS "CORRECT" power factor on circuits involving motors? We know that in inductive apparatus, such as an a-c motor, current lags voltage. Capacitors draw current that leads the voltage-therefore, they offset the lag. \n\nBut why does current lead voltage in a capacitor? Back when we were learning basic electricity, a common explanation went like this: When the circuit is first closed, current must flow into the capacitor to build up a charge on the plates before voltage can build up on those plates. Hence, current must lead voltage. \n\nThat explanation leaves much to be desired. More often, "fundamentals" training begins with lead-lag relationships as "givens," and only the consequences of those relationships are explained in detail. \n\nWhat, then, is the real reason we say that, in a "perfect" capacitor, the current not only leads the voltage in time phase, but leads by exactly 90 degrees? Why does current lead, and why not some other angle? \n\nFORE MORE PL. VISIT:\nhttp://www.bowest.com.au/library/electric.html
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.