Hello Sir,please explain how a  Two-pole QO-GFI  Circuit Breaker with Three-wire 120/240 load wowk (trip)?
97466
there is a metal strip inside that the current flows through. If too much current flows, the strip is actually made of two metals one that expands dirfferently than the other, then it will bend enough to cause the circuit to trip mechanically.\n\nThe GFI part measures the current out and the current back through the neutral with electronics. If they do not match by more than something like 4 miliamps then the current is flowing through ground. This is bad if you happen to be between ground and the circuit so to protect you it trips the breaker.
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You can find the current flowing through a Wheatstone bridge by using Kirchoff's Loop Rules and Junction Rules.      \n\nThe sum of the Voltage Sources - Voltage Drops = 0 in a closed circuit.  \n\nThis setup can be solved using a matrix or a CAS. \n\nGood Luck
431144
The only reason you get shocked when you touch one phase and ground in the commercial power distribution system is because the commercial power distribution system uses a grounded system where the earth itself is used instead of a wire.  Commercial power distribution I believe (and I may be wrong on this) uses Wye (pronounced "Y") connected three phase power with the center grounded allowing the earth itself to be used in place of the fourth wire.  In a balanced 3-phase system (all loads draw the same amount of power from all three phases) the center is not required, but few people have three phase blenders, so only a single phase of the power is sent to an outlet along with the neutral which is grounded.  If you look at Naval electrical distribution systems, they are ungrounded.  Navy schools teach that the use of an ungrounded system raises reliability because two faults to ground on two different phases are required to disable the electrical distribution system, and that in ideal conditions one can touch a single phase and ground and not be shocked.  Berkeley challenges this and I have included a link to their web sight.  I believe the truth is somewhere in the middle because I have touched a single phase and ground and been shocked.  It was a 440V wire, and I definitely felt it, but it was not nearly as sever as being shocked by an outlet (115V) so ungrounded reduced the severity, but did not eliminate the shock.  The electricity that I felt was probably from capacitance in the insulation.  To any one who cares, I managed to shock myself while removing a dead front fuse.  Dead front fuses are "idiot proof", but you know what happens every time they make something that is "idiot proof"?  They make a better idiot to go with it.  (:
742046
For three phase AC powered loads, three separate power lines provide power to the load. The voltage waveforms on these three lines differ only in their timing, or phase relationship. A fourth wire may provide a neutral connection to the load. In the case of balanced loads, where the current in each of the three phases is equal, the current in the neutral wire will be zero. The neutral wire may therefore be eliminated, as it is not necessary for the purpose of carrying current. The vector sum of the currents drawn from the three phase lines will be zero with a balanced load. A non-zero sum of the current indicates an unbalanced load and that a current is returning to the generator, or current source, via an unintended path. This is the ground fault current.\n\nThe ground fault current for a three-phase system, without neutral, may be measured by performing a vector summation of the phase currents. This summation may be performed by a current sensing transformer with the phase currents passing through the common aperture in parallel. The magnetic field induced in the transformer core is then the vector sum of the three individual phase currents and the voltage induced in the secondary winding placed on the transformer provides a signal representing the vector sum of the currents, the ground fault current.\n\nThe electrical characteristics of ground faults in solidly grounded WYE systems allow for simple detection of ground currents by monitoring the phase and neutral currents for a simple unbalance in those currents. The principle is that the current going to the load on the normal phase and neutral conductors should return via those same conductors. Any missing current is deemed a fault to ground on the load side of the detector and, when that unbalance exceeds a set threshold, a ground fault is determined. The basic ways of measuring this unbalance for solidly grounded four-wire systems use one ground sensor surrounding the phase and neutral conductors, four individual current transformers connected in a WYE configuration, or four voltage sources connected in a delta configuration.
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You'll need to provide a little more information in order to have your question answered properly.  AC or DC? Input voltage?  Fixed or variable output voltage? If variable, over what range? If fixed, what is the desired output voltage?  Output current?  How critical is the load regulation?\n\nAs you can see, the answer to your simple question can be a bit complicated :)
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?
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less than 0.3  for power sations\n0.5ohm for substation\n1 to 2 for homes
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Most AC electrical sources can be converted to other phases, voltages and currents but not to increased power (P = IE) either using semiconductors, transformers or motor generator sets (and transformers).  Usually 240 Volt AC is obtained using two phases of a three-phase distribution system where a single phase is 110 Volts nominal (perhaps 120 Volts actual or stepped up) to double voltage to 240 Volts.  However, in the lab, many different phases, voltages and currents may be required for special applications and suitable converters must be found (or built?).
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The AC Switch Block allows a conventional 5V signal to control a much larger AC line. The switch can handle up to 16 Amps of current provided it is kept cool enough. At 110V, this is more than 1500W. We suggest that this limit be approached very carefully and recommend staying far beneath this.\n\nOnce an AC line is connected to the AC terminals, several places on the board will become live. This makes it absolutely essential that the board be mounted in a way that will prevent short circuits and accidental human contact. This is true even for rough prototypes or trial hook ups. Treat AC line voltages with much respect.\n\nHow Do I Use This?\nWith your AC device unplugged, first determine which cable from your AC device's power cord should be switched. It's best to switch the "live" or "hot" wire. On any AC outlet, there are usually 3 slots (sometimes 2). The right slot, which is slightly smaller than the left slot, is the "live" slot - check which prong plugs into this slot when you plug the AC device in to confirm which wire is live.\n\nOnce you've determined which wire is live, snip it in two somewhere along its length, strip about 1/8" insulation away from both the open ends, and tin the exposed wires. Make sure to leave enough slack on the cord for the two tinned ends to be inserted to the two AC connectors on the AC Switch Block.\n\nA small 5V signal applied to the D0 input will control the switch that turns the AC device on and off. Run a wire from the 0V connector on the AC Switch to the ground of the device providing the 5V signal. Double check that all your connections are correct, and plug in the AC device to the wall - now, a 5V signal on the D0 input turns the device on and a 0V signal turns it off.\n\nApplications\n-Automate Your Household. Many household appliances such as lights, fans, blenders, drills, vacuums, etc. run on AC power. Now you can use the behavior from a Poly Block an Intro Module, or any other device that put out 5V, to control these kinds of devices.
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Try this. I found it on line and it might help solve the problem. http://www.tselectronic.com/gc/video/surge_block.html?tse_Session=a1810461ed6dc5852c42a495bed5dfd8\nI don't think dharden1205 has any idea what he's talking about. there could be a carrier voltage to help with signal transmission, but it should be low enough as not to harm your modem. If you've got it running off a power supply (UPS), then it's already isolated from the direct supply in your home. This shouldn't be happening. Call back your provider and insist they come out and test your cable for transient voltage. You have options to cancel your service and go elsewhere right???
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The RMS (root-mean-square) value of AC voltage is NOT dependant on the frequency.  For a sine wave, it is .707 (half the square root of 2) multiplied by the peak voltage.\n\n120 and 240 volt circuits common in the US are derived from the center-tapped transformer, the center called the neutral.  This gives you 120V between each phase and neutral, and 240V between the phases.  An analogy that might make more sense is with batteries.  You can put two 9V batteries in series to get 18V, but you still have 9V across each battery.\n\nI think what you want to know from your question, is why do we have both 120V and 240V.  You can argue 120V is safer than 240V, but it is more historical that 110-120V is used in the US (many other countries, but not all, use 220-240V).  The reason for having 240V for some equipment is with a higher voltage it will draw less current, so smaller wires can be used.  Similar to power transmission using high voltages.\n\nThe safety ground, the 3rd round pin on a normal plug, is connected to the neutral at the electrical service.  The neutral is grounded (to the earth), and so is the safety ground, but that connection to the earth has nothing to do with its purpose.  This safety ground connects metal parts of equipment to the neutral, so if a fault in the equipment causes voltage on those metal parts, it will trip the circuit breaker rather than give you a shock (if you were touching it WHEN the fault occurred, you could get a shock before the circuit breaker tripped).  Another thing about the safety ground, it never carries current under normal conditions, and for that to be true it can only be connected to the neutral at the service.  NEVER connect a safety ground to a neutral at another point.\n\nAlso, never take the cover off your circuit breaker or fuse panel if you don't know what you are doing.  Suggesting someone take off the cover and poke around with a fluke is a bad idea.
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It means there was a power surge in the system and the main fuse box either in your house or at the main power station flipped to the off postion...also known as being tripped. Normally a safety device or circuit breaker will 'trip' when it is detecting too much load...eg a faulty appliance, too much appliances plugged in (use of too many adaptors), or hot electrical joint (loose), undersized protection etc.
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Divide by the voltage and multiply by 1000.\n\nIf you are working with 3 phase power, divide again by sqrt(3)
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Kilowatts  Equations\nDirect Current  =        I x E/1000\nAC / 1phase 115v or 120v =I x E x PF/1000\nAC / 1phase 208,230, or 240v= I x E x PF/1000\nAC 3 phase All Voltages     =    I x E x 1.73/1000\n\nE = Voltage \nI = Amps \nPF = Power Factor
186229
Here are some sites that may help.  Both have circuit examples utilizing LVDT's.  Hope this helps.
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The website book section on AC circuits linked below is probably your best bet to get a full understanding of this topic.  Click on the power factor chapter; it is pretty good.\n\nNevertheless, my quick answers to your questions:\n\npower factor arise only in 3phase?\n** Nope.  It must be considered in both single and multiple phase AC circuits\n\ndifference between kW and kVA? \n** One is the unit for real power (kilo Watts) and the other is the unit for apparent power (kilo Volt-Amperes).  kVAR (kilo Volt-Amperes Reactive) is the unit for reactive power (basically power that is stored and released in inductive/capacitive loads).  Real power and reactive power sum up to equal apparent power (see the triangle diagram in the linked page).\n\nHow is the power factor caused?  What is the effects?\n** The power factor is the cosine of the impedance angle of your load (or the apparent power angle of the load, same thing).  Power factor varies between zero and one.  Where one indicates all real power (resistive impedance) and zero indicates all reactive power (inductive or capacitive impedance).  In between zero/one is a combination between the two.  Generally, we like to have a power factor near one; otherwise supplying power to large reactance loads can be wasteful and can lead to unsatisfactory voltage drops at the load.\n\npower_factor = cos(voltage_angle - current angle) = cos(impedance_angle)\n\nWhat is lead/lag/unity?\n** unity => pf = 1 (all resistive load, voltage and current in phase)\n** lagging => pf between 0 and 1 (corresponds to inductance in the load, current angle LESS than voltage angle)\n** leading => pf between 0 and 1 (corresponds to capacitance in the load, current angle GREATER than voltage angle)\n\nWe need the lead/lag distinction in power factor, otherwise we don't have any idea if we are talking about an inductive/capacitive load.\n\nHope that helps.
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well, these systems are different for historical reasons - something was decided on as a standard in one part of the world and something else, at the same time, in a different part of the world.  it's just like driving on the right vs. left.. they're different, and incompatible, and it's a massive pain in the rectum.\n\nbut like driving, changing either system over to the other requires overcoming a massive amount of social inertia and a huge infrastructure cost - everyone would have to replace their appliances/cars, power-distribution systems/traffic signals would all have to be replaced.  and because of the nature of these systems, you cannot "phase in" a change - it would all have to switch at the same time in one ugly go.\n\nfunctionally there is no real difference in using 50 or 60 hz.  \n\nhowever, a different voltage has its advantages and disadvantages.  240V distribution is considerably less safe than 120V - get across 120V and you can be electrocuted for sure, but it's far far more dangerous to be electrocuted by 240V.  so safety is one factor that favors a lower voltage.\n\non the other hand, for a fixed requirement of energy usage, a higher voltage will require a lower current.  this is easy to see if you look at the equation for power in an electric circuit, P = VI.  double the voltage and you halve the current.  \n\nthey tell you not to plug a lot of energy-hungry appliances into a single outlet because doing so increases the amount of current that flows through your household wiring - too much can cause the wire to get hot and melt, possibly starting a fire and certainly causing costly damage.  so a higher voltage power distribution system will be generally cheaper (less current = smaller wires = less material costs - this is part of the reason that long-distance transmission lines are stepped up to 120 kV) and less prone to failures from overloading.  so that is an argument that lends itself to choosing a higher voltage.\n\nso, both have their ups and downs - ultimately i think the choice is pretty arbitrary, but the different standards are going to cripple any movement towards a global society.