other things to consider -- copper has a positive temp coefficient, which means the power dissipation will increase until equilibrium or failure also, the magnetic field set up in the wire at these current levels can actually pull the wires together... -----Original Message----- From: Bob Ammerman [mailto:rammerman@ADELPHIA.NET] Sent: Wednesday, May 29, 2002 10:10 AM To: PICLIST@MITVMA.MIT.EDU Subject: Re: [EE]:formula/link for current capacity of wire (AC) ----- Original Message ----- From: "Thomas C. Sefranek" To: Sent: Wednesday, May 29, 2002 2:21 PM Subject: Re: [EE]:formula/link for current capacity of wire (AC) > On 29 May 2002 at 11:45, Micro Eng wrote: > > > ok..power guys... > > > > Whats the formula, given that I have the current and voltage, to calculate > > the wire size that is required. > > > > I'm talking large amounts of current...like 833Amps @120VAC > .05093 / 1,000 * 833 * 40 = 1.7 volts loss for 4/0, just over 1%. > > Wire size is not determined by voltage. > How much POWER are you willing to loose in the wire? > How much VOLTAGE loss can you stand? > > > > I know for 200A, I need a 4/0 wire. > Perhaps: > > 4/0 wire is .05093 ohms per 1,000 feet. > For 40 feet, (20' there and back) you get .002 ohms > @ 200 amps, the voltage drop is .4, 200 A * .4 V = 80W Note that 80W of heat are going to be generated. This is about 2W / foot which shouldn't make the wire too hot... Bob Ammerman RAm Systems -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details. -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.