How to calculate circuit breaker rating?
820591
Add the loads in amperes of each of the appliances on that circuit.  Make sure that the total is less than the ampere rating of the circuit breaker.  You should probably only use about 3/4 of the amperes the circuit breaker is rated for.  For example, a 20 Amp breaker should only be loaded with about 15 Amps.\n\nIf only the wattage of an appliance is given, divide the wattage by the voltage to obtain the amperes.  For example, a toaster may draw 1200 watts.  1200 watts / 120 V = 10 Amps.  A light bulb may draw 60 watts.  60 watts / 120 V = .5 Amps.
455603
If it's for a simple hobbyist application they need to know only the resistance value and its power rating.\n\nA full specification would also include:\n\ntolerance,  thermal coefficient,  maximum voltage.
734859
Divide by the voltage and multiply by 1000.\n\nIf you are working with 3 phase power, divide again by sqrt(3)
821458
Use Ohm's Law.\n\nMake the network, put 10 volts across it, find how much current is being drawn across the 10 volts.\n\nE = iR so R = E/i, where E is 10 volts, and i is the current, amps, then R is the resistance in Ohms.
840199
The 80kA is the current the fuse can withstand i.e.80,000A. Very high currents can flow in the short time period from when a fault occurs and when the fuse blows to clear the fault. The fuse needs to be able to withstand this current, hence the rating so the design engineer can check that the potential fault current (maximum current that could flow) can not exceed the fuse rating otherwise the fuse may explode causing further damage. The potential fault current is dependent on the supply, transformers and wires etc increase the impedance and resistance and reduce the potential fault current.\n\nThe 100A is the fuse size. There are curves that will show you how long it will take for the fuse to blow at any current but basically a 100A fuse will carry 100A indefinitely. Small increases above the 100A would take many hours or days to blow. To blow within a second or so requires thousands of amps.\n\nHope this answers your question without too much text. Best answer votes always appreciated.
655128
Assuming that we have 2 series resistors R1 and R2. And the voltage applied on them is V.\n\nTo get voltage applied on R1 (V1)= V * R1/(R1+R2)\nand so V2= V * R2/(R1+R2)
118710
The meter must measure the power you use (and must pay for) using the formula P = IE (power equals current (I) times voltage (E)).  To do so it must accurately dedect the line voltage and electrical current entering your house.  The meter contains an aluminum disk that you can see spinning fast or slow depending upon how many (or few) electrical devices you have turned on.  Inside the meter there are two coils of wire (one on top and one on the bottom of the disk).  One coil senses instaneous voltage that can vary (and even droop in a brownout).  The other coil detects all the current you are using, if any.  With your main circuit breaker open there will be none.  When there is both voltage and current flowing, magnetic eddies are produced that measure and integrate the power being used by spinning the aluminum disk (like a small motor) which is connected to a set of gears that turn the meter numerals.  See the link for an illustration and detailed explanation.  Good luck.
978617
Try this web page:\n\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/electronic/zenereg.html\n\nIt shows the basic circuit and has calculator to figure out component values.\n\nThere are much better solutions out there for a voltage regulators than zeners... such as the three terminal regulators like 78xx or LM317 types...
177692
No matter what is the voltage rating, isolation or insulation is tested by using "MEGGER".\nThe value should be atleast above 1 Mega ohm in any case
261277
It's simple do this job. First of all you need to set your volt meter to a specific voltage level. Like if you want to check voltage of batteries, it ranges between 1.5 to 14 volts. So set your volt meter dail to DC (Direct Current) with this voltage specification.\nFor checking voltage in your house, its a 110-120 V or 220-240 V. This depends on your location. So now you have to set the dail to AC (Alternating current. THe symbol of AC is " ~ ".\n\nI', also a site here. It would solve the job definitly.\n\n\nhttp://www.ehow.com/how_16767_voltmeter.html
703486
Each welder would draw 113.6A (I=P/V). So just two of them would would add to 227A, which would probably trip the breaker...
823421
light duty is the problem - would be very easy to make one using any push on connector in the circuit ( you need a spring loaded connector with the pin being non conductive - a 'blade' shaped pin would be the easiest to rig )\n\nanother possibility is to use a switch of any type ( a - b ) that shunts the circuit to a open   circuit  ( this can be activated by a pull - and switched back manually )
458600
You say 3 phase, so the simple amps times volts doesn't work.\n\nThe complete formula is KW= 1.73 * amps * volts * power factor /1000\n\nThe 1.73 is the square root of 3.  The amps is your line current (same as phase current if it is wye, not the phase current of a delta) and the volts is the phase to phase voltage (not phase to neutral of a wye).  Power factor is 1 for simple resistive loads, and is less than 1 for inductive loads such as motors and fluorescent light ballasts.\n\nIf you take the power factor out of the formula, you get apparent power measured in KVA.
709498
The amp-hour rating is the current flowing out of the battery in Amps multiplied by the time in hours it is flowing until the battery is dead.\n\nYou could measure this by putting a resistor across the battery and an ammeter in series with it - measure the current - take readings periodically - until the battery dies. Average the readings then multiply the average by the number of hours it took to die.
504290
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.
115353
It doesn't affect it much at all, it is just a chunk of metal inside there. You would have probably just as much affect by adding 10 feet to the run of wire. \n\nUnless the breaker is old and gets hot from the increased resistance anyway. In that case you should replace it and then you will be fine.
898015
Nominal= 36ohms\nTolerance=5% \nMultiply nominal by .05 , then add that figure to nominal for Max,.subtract it from nominal for Min.,,and study your ohm's LAW!! ;-)=