how to calculate load rating for an electric cable size?
595330
If you are in the US, there is a building codebook that has tables for all wire sizes and their ratings.  Its called the NEC handbook
169135
Simply put, the generator voltage (volts), frequency (Hz) and power (watts) and phase (single/three) are called ratings. These are selected to suit the load that will be connected. and the environment in which it will be used
263411
I'm assuming you're using copper for your bus bar? In this case more is better. I would run multiple solid copper bus bars to carry 4000 amps and if done properly, a simple poly enclosure would work fine with very little space needed between the two.
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...
130012
There are a few things one should consider when engineering a busbar solution.  A general rule of thumb for sizing copper busbar used for 60 Hz applications is 1000 Amperes per square inch.  For example, 1/4" x 2" = 500 Amperes continuous.  If you'll be bolting busbars together, the general "overlap ampacity" rule of thumb is 200 Amperes per square inch of overlap.  These are the two basic rules for sizing copper busbar in most applications.  \n\nMore information is necessary to answer the question, such as frequency, maximum rated transformer current, desired heat rise, material type (copper / aluminum).  There are other things to consider besides conductor size such as, bracing the busbars to achieve the desired interrupting rating, insulating the busbar to achieve the proper BIL level, etc.\n\nBusbar engineering isn’t quite as simple as it looks and if done wrong, it could lead to severe injury or death.  This information is very general in nature, and consequently, is not applicable to every situation.  If you’ve never done this before, you should hire an experienced and qualified bus fabricator.  This information is very general in nature, and consequently, is not applicable to every situation.
844099
It depends on the applied voltage.  Divide the voltage by the resistance which gives you the amperage.  Now multiply the amperage  with the voltage, and that is your wattage.\nW = V²/R\nThis is straight forward for DC; with AC things become a little more difficult, as because of the fluctuation of the voltage, the average is only .707 of the nominal voltage.
946535
The standard for voltage drop in a circuit is 2%.  Remember that the current must go back to the source so if you are running 3 Phase you must multiply 750ft by 1.732 to get the proper length.\n  So, using ohms law -voltage drop=current (in amps) X resistance (in ohms).  #000 gage copper wire has .00073 ohms/ ft.  Using 200 amps in this formula, If you work out the math you will find that a 30 amp service is about all you can safely install at that distance.  Even at 30 amps you are running the risk of over heating and shortening the life of your equipment. \nexample:  750ft X 1.732 X .00073 ohms per ft = .9 ohms total\n\nSo:  .9 ohms X 30 amps = 27 volts dropped\n\nThus:  if your service is 240 volts your voltage drop will be a little over 10 % or 213 volts at the point of use..  Some equipment may endure this.  I would not chance it.
871196
Don't you just need to balance forces in orthogonal directions and then use these components to get the forces along the cable?\n\nSince the sign is stationary, vector sum of forces must be zero.  Sketch a free body diagram and label known forces.
535050
Well, first a cable operator needs to have authority to operate from the locality, FCC, and probably other government offices.  However, if let's just say you have a feed from your cable provider and happen to want to have 300 different outlets in your house (big house) here is what you need to do.  \n\n1.  Make sure the drop coming into your house is perfect, and new.\n2.  At the point it enters, it will need to be split with an 8-way splitter.\n3.  Buy eight, 8-port amps like this. http://americanesuperstore.stores.yahoo.net/c-0329.html\n\nThis gives you 64 outlets at the right level, repeat if necessary.
331292
See the other postings of this question.  The equation is\n\nhttp://members.aol.com/logan20/elecspon.html
140428
x\n\nthe answer below is just for voltage - see my graph above - it is quite accurate. here it is again, larger: X
420078
Here's a link to the standard American Wire Guage table.  It has it all listed:\nhttp://www.eng.auburn.edu/~tdenney/ENGR1110/wireawg.html
1160447
Keith P clearly messed up something if he thinks 100 ft of 14-2 weighs under a pound.\n\nMy reference below has weight directly on the chart.  For example, 14AWG solid is 13 lbs/1000ft.  14-2 w/g has 3 of these conductors, so it is 39 lbs/1000 ft.  You can do the other calculations yourself.  In case you don't already know, the 8 and 6AWG romex cables have 10AWG ground wires, not 8 or 6.  Also, the 8AWG and 6AWG conductors are stranded.
333276
This is a little too complex a question for free answers.  But:\n\nLower NUMBERS in gauge correspond to larger AREA of the wire cross section.  The larger the area, the more amerage can be safely carried by the wire.  Voltage is not really an issue at common household voltages, as most modern wire is rated to 600V and household doesn't exceed 220V.\n\nIn-wall wiring is solid copper.  You will generally match the wire gauge to the circuit breaker size as follows:\n\n14 gauge = 15 amp breaker (15 amp max)\n12 gauge = 20 amp breaker\n10 gauge = 30 amp breaker\n\nThe use of watts is misleading; it's AMPS that matter.  But watts * volts = amps, so you can deduce amerage from reading the wattage value, and knowing the current in your home.  The in-wall wiring specs are not generally limited by wire LENGTH, at least not in the way extension cords are.\n\nExtension cords, unlike in-wall wire, are not solid copper.   However, they're also less likely to burn down your house.  They are rated by wire lngth and gauge.\n\nYou can use the links I've provided here to see required gauge sizes.  Most indoor space heaters are 1500 watts (about 12 amps) though  some are 1750 amps.
686255
your problem is statically determinate. so, use static equations to calculate the tensile force in the cable. maximum sag will occur at the center of the cable where the load should be placed.\n\nyoure asking for deflection and the force. it cannot be. the maximum allowable sag should be used to calculate the tensile force, either, calculate the sag using the allowable tensile force. this can be solve based on the area of the section and the allowable stress.\n\nI hope I was able to answer your question. better if you could send me the figure. my email is exaatto2000@yahoo.com or rbra_engrs@yahoo.com\n\nruel
736596
It depends on the current.  You want to drop 6 V across the resistor to get the voltage to the load to 6 V.  The voltage drop across the resistor is V = R*I.  The resistor value you need is R = V/I.  Once you know the resistance, you need to determine its power rating.  The power dissipated in the resistor is P = V*I, where V is the voltage across it and I is the current through it.  Or, you can use P = (I^2)*R.  You should choose a power rating at least 1.5 times the power you've calculated so it doesn't run too hot.
708021
The symbol for electrical load is Z called impedence.\n\nThis consists of the inducticve reactance, capacitative reactance and ohmic resistance.