What Formula Used To calculate the size of Electrical cables if we know  the Amperage/?
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Look at this site:\nhttp://www.kevinboone.com/cableselection_web.html
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inner surface area of a conduit of inner diameter d and length l = pi*d*h\nherd d = 10mm= 0.01 m\nand h = 5m\nso inner surface area = pi*0.01*5\n= 0.157  m squared
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By ohms law\nV = IR\n  = 3 amp * 3 ohms\n  = 9 amps/ohm\n  = 9 volt(since amp/ohm is given a special name volt)\nThe volt, symbol V, is named after the Italian scientist Alessandro Volta, who invented the voltaic cell and the electric capacitor.
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V (Voltage) = I (Current) x R (Resistance)\n\nOr (although you probably don't need it):\n\nJ (current density) = sigma (conductivity) x E (electric field)\n\nI always use wikipedia to look up stuff like this, it's very helpful for all kinds of info.
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Voltage is the amount of "potential difference" present - an analogy with water would be the pressure of the water\n\nCurrent - measured in Amps - (Coulombs per second) the flow of electrical charge - an analogy with water would be the thickness of the pipe.\n\nVolts x Current = Watts\n\nJust as\n\nWater pressure x pipe thickness = amount of water pumped\n\n\nSo a 12volt bulb taking 2 amps will be 24 Watts\nA mains bulb at 240Volts with 0.5 amp will be 120 Watts
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voltage = current x resistance\nE = I x R  It IS Ohm's law. An easy way to remember the formula and it's configurations is to draw a circle and divide it in half horizontally. Above the line write "E". In the lower half divide the semicircle vertically and write "I" on the left, "R" on the right. It should look something like this:\n                                                     E\n                                                   I    R\nNow just cover up the one you need. and either divide, or multiply.
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T = mg.\n\nThe length of the cable is irrelevant in this example. The beam is hanging off the cable.\n\nThe tension in the cable therefore is m*g, which is 2450 N.
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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
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A length of a single wire:\nL^2 = 72*72 + 54*54 (from a right triangle)\nso L = 90 ft\nFour sets of wires -> he needs 360 ft
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175.76 Amperes...\nusing m=z*i*t\nz=63.546/96500\nt=24*60*60\nm=10*1000
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Amiya\n\nThe busbar sizing is the easiest of all in electrical engineering\n\nYou will need to understand the basis\nThe bus bar sizing is limited by the allowable current desity in the copper bar\n\nIf you want to size for a 1000A current, then you will take the 99.99% pure copper as the bar material. The current density allowable depends on the ambient temperature and the ventilation system adapted inside the panel. However upon assuming a reasonable local values for it, just devide the current by the current density of copper, you will get the allowable copper area of cross section\n\nfollowed by this, you can find out the cross sectional availability of the copper bars. If one is not meeting your area, you can stack two similar bars. You can not stack two different sizes of bars due to their difference in resistances. \n\nI think this background is just enough for the basics.\n\nLet us go to specifics.\n\nDo you want to design manually or using a software?\n\nManual designing is outbid by software.\n\nTry the free software below. I just tried to download and checked the operation of the software. It works excellent. I recommend you to try this.\n\nhttp://www.cda.org.uk/download/d11.exe\n\nStill if you insist for a manual calculation, Pl download the link below\n\nhttp://www.socomec.fr/documentation/documents/Catec-gb.pdf\n\nCheck the last page for busbar designing\n\n\nIf you find my answer fitting to your question,\nJust rate my answer. I will get 10 points for the best answer. This will take me to level 2 in yahoo answers. I can help answering others more effectively at higher levels.\n\nGood Luck\n\nGood Luck.\n\nHope i have solved your problem
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T= L + 2B\n\nT is total surface area\nL is Perimeter x Height\nB is base area
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Draw all this out so you can see it - it won't be to scale, but don't let that fool you.....\n\n\ndiameter = 50cm (radius = 25cm)\nchord = 20cm from diameter (thus it's 20cm from the center)\n\nOK, you'll need to assume that the line segment forming the diameter touches the chord at point B on the circle - i.e. we'll say the diameter is line segment AB and the chord is line segment DB. \n\nC is the center of the circle, so we have radius CB = 25cm. \n\nAlso, we'll have line segment CE drawn perpendicular from the center (C) to point E on the chord, forming a right angle and thus *bisecting* chord DB.\n\nYou have the hypotenuse of triangle ECB (radius CB) as 25cm, and side EC as 20cm (distance from the center to the chord - always perpendicular).\n\nLine segment EB will be the second side of a right triangle, so your formula:\n\nEC^2 + EB^2 = CB^2\n20^2 + EB^2 = 25^2\nEB^2 = 25^2 - 20^2\nEB^2 = 625 - 400 = 225\nEB = 15cm\n\nSince EB = half the length of the chord DB, DB = 2*15cm = 30cm.
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Power is directly related to amperage by the voltage, P=I*V.
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Length of wire = 9in+9in+11in+11in\n                  =40in\n\nLength of Square = 40in divide by 4\n                          = 10in
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See the other postings of this question.  The equation is\n\nhttp://members.aol.com/logan20/elecspon.html
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I'd say that you already know.  If my memory serves (at 6 AM) it is an equation where you can determine the unknown if you have two values.  The equation can be expresssed multiple ways.  Ohms are given in impedance or interference/resistance, and I think the other values are volts(or watts) and amperage (amps).  Whether it's home, commercial or electronics, the law applies to the flow  of electron in any circuit.