Do you need a capacitor in standard full wave rectification?
831976
Yes, you'll need a filter to get rid of the ripple. The voltage will be all in one polarity but will go from some peak to zero volts. You'll need a minimum of a capacitor in there if you want something close to a DC voltage.
942029
Use the sine wave as 2.8 peak to peak volts to a bridge rectifier. No clamping is possible without dc power source and capacitance.
961185
You cannot do this directly. Assuming you are talking about a motor, It may be possible to get the motor to turn using a capacitor but you will not get the rated torque (turning force). I'm sure there is plenty 3 phase theory on the internet if you want to see why they are so different\n\nOne way it can be done is by using an inverter that takes single phase input (providing the voltage you require is not higher than the voltage you have or you will then need a transformer as well). The inverter gives you the added value of being able to vary the speed but it may be cheaper to simply buy a single phase motor. As you didn't actually specify the device to be motor, be aware that VSD inverters may not work on other equipment.\n\nYou need to be wary of the size of the device you are trying to operate. Often 3 phase is used for higher powered devices that becomes in-practical to try and run on single phase anyway.
1126159
"Because it lacks a separate field circuit, an induction generator cannot produce reactive power.  In fact, it consumes reactive power, and an external source of reactive power must be connected to it at all times" [1].\n\nSo you are right that you will need a three-phase capacitor bank at the output terminals (delta-connected).\n\nHow to size the capacitors?\nThat depends on a few things.  Namely what kind of loads are going to be attached to it.  I presume this is going to be an emergency generator operating alone (most induction generators are used this way nowadays).  In that case, your capacitor bank will need to account for the generator consuming reactive power and the loads consuming reactive power.\n\nSo, if you know what you are going to connect to the generator, figure up an estimate of the total reactive power of your loads.  Then estimate how much reactive power your generator consumes (not sure how to do this; maybe estimate from your 7.5kW and 9.4kVA ratings?).\n\nFinally, to size your capacitors equate the total power consumed by gen and loads to this formula:\n\nestimated_3phase_power_consumed = 3*(line-to-line_volt)^2/X\nwhere X = 1/(2*pi*freq*C)\n\nSolve for C and there you go.  My source indicates that the output voltage for an induction generator can vary quite a bit with varying inductive load, so bear that in mind and maybe oversize the caps a bit.  Also, the types of capacitors should be able to withstand AC obviously and be rated within your expected operating temp, etc, etc.\n\nGood luck.
324170
the 20 ohm is shorted because the current can go through the switch without going through the reisistor. when it is closed, R =10+1 =11, V =3, Current = V/R = 3/11 amps\n\nbecause direct current cannot cross a capacitor, which is in cd. this is because a capacicotr is plates which are parallel and do not touch. obviously, if the plates do not touch, electricity cannot flow through them. alternating current can go through a capacitor, only because the capacitor keeps discharging
256398
You don't need much current at the input of an op amp, and 100 ohms with, say, 10 V (typical power supply V) across it dissipates a whole watt, so you don't want that low an R in a feedback loop or an input divider (power waste and heat being undesirable).  Also if you are using R-C circuits, a lower R means a larger C for given frequency response.  Too high an R though (in the megs) and you run the risk of picking up hum due to electrostatic coupling, and also loss of accurate gain due to leakage current.  So you'll see maybe 10K - 100K as a desirable range.
1166015
Semiconductors have different uses:\n1. an N-P junction diode can be used as a rectifier, ie, to convert AC to DC [partly or fully] using the tranmission capabilities of transmission in forward & reverse biased situations\n2. an N-P-N or P-N-P triode can be used for rectification with amplification\n3. Doped semiconductors can also be used to construct logic gates in microelectronic devices.
931090
AC in and DC out.\n\nA diode only allows current to flow in one direction and blocks it if it is wrong.\n\nthe diodes in the bridge are connected together so that the +V current flows in One direction. the bridge .....( I cant be bothered)\n\nwikipedia has the answer...\n\nhttp://en.wikipedia.org/wiki/Bridge_rectifier
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
940120
Not necessarily! \n\nOutput is usually measured in watts at the speaker connections, not volts.\n\nThe Line-out "standard" is 1v peak-to-peak on most systems except those designed to drive many speakers, (like restaurant background music), which are usually either 75v or 150v to avoid "line loss" in large systems with long runs, which then use special transformers at each speaker to convert back for the speakers.
1164919
you need to describe what you mean by "small" better.\nis it too small vertically, horizontally, or both?\nAlso, how good is your electronics background?\nDo you know how to discharge the CRT and large capacitors so you don't get zapped with 330 to 10,000 volts while attempting to repair it?\nYou also need to define "wavy" better.\nWith the tin amount of information you have given though, it sounds like the power supply is failing, but in which part is impossible to say.
470470
I'd like to know exactly what you're trying to do. Using the wrong type of capacitor or one with the wrong values could explode and burn. I'm talking about a pretty violent explosion at 110 VAC!  A Snubber will not work because that is now what they are for. A snubber is used on the gate of SCRs to prevent false retriggering especially on high power circuits. What you need is a Thermistor. This is a flat round device with two wires coming out of the edge. The thermistor will be put in series between the power switch and the capacitor. At normal room temperatures the thermistor might have about 50 Ohms of resistance. When it warms up, (almost instantly), the resistance drops to about zero. This action limits and controls the initial power surge into the capacitor when the switch is thrown thereby preventing damage to the switch. You need to know the voltage and current load to select the right thermistor otherwise it becomes smoke.\n\nPlese use caution: Capacitors can store huge amounts of power and will COMPLETELY DISCHARGE IMMEDIATELY.\n\nA large capacitor, when shorted out, can blow a 1/4 inch screwdriver in half! Sounds cool but the missing part is burning molten metal flying at you! Please be Careful!
95109
This might help\n\nhttp://www.ttec.co.tt/consumertips/conserveelectricity/default.htm
504081
You cannot run polarized electrolytics backwards, without destroying them. You can find dozens of different non-polarized types on-line from parts distributors such as Mouser, Allied, and Digi-Key. But if you want to get around the problem, there's still another way; Place two standard polarized electrolytics in series, "back-to-back" so that their negative terminals are connected to each other. Then, place a rectifier diode, such as a 1N4002, in parallel with each capacitor, so that the cathodes are in contact with the positive capacitor terminals, and the anodes are in contact with the negative capacitor terminals.  What you will have now made is a "bipolar" or "non-polarized" capacitor that is roughly equivalent in value to one of the two original polarized capacitors!  If, for example, both of the electrolytics were rated at, say 2200 MFD @ 25 VDC, the new "homebrew" NP capacitor will function as a 2000 MFD, 25VAC capacitor!  Reversing polarity on this "home brew" capacitor will not cause any problems, whatsoever!
365027
The capacity of a capacitor is measured in farads (F). Ususally they range from several picofarads (pF) to several microfarads (µF). For the measurement, look on the capacitor itself or if it is a disk capacitor then there is a system for figuring that out (look it up).
27954
You would need to know what voltage the capacitor was charge to. Volts * capacitance = Coulombs
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.