HELP! Does anybody know the resisitivity of 5mm diameter 24SWG constantan wire? Thanks !!!!!!!!!!!!!!?
925581
Resistivity is independent of length or diameter.  It's 49 × 10^-8 ohm*m for constantan.
135604
Are you talking about the wire to supply the resistance heating wire or the resistance wire its self?\n\nThis is not something you should undertake in any way shape or form. It is something for a trained professional who has the experience and knowledge to install such a system safely.
264618
Hi, I think you're great, but i'm not going to type for 5 hours when all you needed to do was put "fibre optics" into the search engine. \nTry here:\nhttp://electronics.howstuffworks.com/fiber-optic1.htm
41902
usually 18-20 gage steel wire. though you can get them in plastic and or various other metals aluminum and or titainium for example though i cant see why youd want them unless you were to use them for a VERY non standard application
1066990
..Hi..check the link below:) ...I think you will not find any cheaper/quality ...\nkind regards!:) ...
1139339
The best way is to get it from the manufacturer.  If you don't know who made it, you can look up resistances from a bunch of places on the net if you know the wire gauge, but remember that copper resistance increases as it gets hot.  \n\nMost magnet wire can handle a few hundred volts with no troubles, but if you're doing some high voltage app, better be sure about the voltage rating.
929646
Depending on the output current you desire and what current the +24 VDC can supply a simple 3-terminal regulator like a 7812 should do the trick. It should be able to handle the 24 V input. You may need to add a 0.01µF Cap from Ground to the +5 VDC output to dampen any oscillations. There are usually a number of cookbook circuits in the data books. \n\nGood Luck!
278492
From an online Pipe company. Here is one that might be useful!. Good luck!
660156
http://www.physlink.com/education/askexperts/ae25.cfm
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.
398806
here is one of the datasheets available on the internet:\nhttp://www.ee.washington.edu/stores/DataSheets/74ls/74ls73.pdf
690607
It depends on wire diameter, but generally:\n\n20 mesh = .034" openings = 864 microns\n400 mesh = .0015" openings = 38 microns
926295
Convert all this to MKS units and find the cross-section of that wire.\n\nE (stainless-steel) is given to be 18 x 10^10 N/m^2.\nCross-section is from pi(r^2).  This must be in m^2, not mm^2.\n3.14159(0.125*10^-3 m)^2 = 4.909 x 10^-8 m^2 cross-section.\n\nYoung's modulus (E) is given as,\n\nE = (delta F / cross-section) / (delta length / original length)\n\nDelta F is the tension in the wire, thats the first part of solving this problem.  So rearrange the equation for delta F, renaming delta F as "tension" and substitute the value given in place of E:\n\nTension = (18 x 10^10 N/m^2) * (8 x 10^-5 m / 0.028 m) * (4.909 x 10^-8 m^2)\n\nTension in the wire is therefore 25 N.\n\nThe magnitude of force on the tooth is 25 sin 30 deg or 12.5 N\n\nWarning!:  While the reasoning is sound, double-check all of my math.  Also, note that I rounded the intermediate answer 25 N.
274304
Here's a good source of info:\nhttp://ece-www.colorado.edu/~kuester/Coax/connchart.htm
780548
Normally wire-size measurements are in millimeters and gauge. It will be easier for you to communicate with who you are ordering from if you do the computations to change it to millimeters and then transfer that to gauge of wire.\n\nI suggest you go to a refiner like David Fell & Co. (see link below.) The should be able to supply you with any gauge wire in gold that you need.
837447
PROBLEM ONE\nThe electric resistance R of a wire varies directly as its length L and inversely as the square of its diameter.  So you mean R=cte*L/d2 where the constant depends on the metal used.\nfor that alloy, C=Rd2/L=36*0.6*0.6/20=0.648\nIn the second wire, you use the same constant\nR=0.648*60/(1.2*1.2)=27 ohms\n\nPROBLEM TWO\nhere, period=T=cte*sqrt(L)\ncte=T/sqrt(L)=2/sqrt 6=0.816\nL=(T/cte)2=(1/0.816)2=1.22*1.22=1.5
837681
is it something like this\n\nhttp://www.tigerdirect.com/applications/SearchTools/item-details.asp?EdpNo=324225&CatId=82\n\nor checkout complete listing\nhttp://www.tigerdirect.com/applications/category/category_slc.asp?CatId=945\n\n.. besides easy job for any electrician > all the best\n_________________\n\nthe adapter has 2 ends\n1. primary\n100 - 250 volts ac\n2. secondary\n15 or 18 or 19.5 volts dc, depending on make / model\n\nthanx 4 clarification, hibiscus\nmy understanding was correct\nthe primary section is damaged\n\nis it 2 wire or 3 wire\nolder adapters were 2 wire & newer are 3 wire, which includes earth\n\nanother 2 links\n3 wire\nhttp://www.tigerdirect.com/applications/SearchTools/item-details.asp?EdpNo=1911192&CatId=945\n\n2 wire\nhttp://www.tigerdirect.com/applications/SearchTools/item-details.asp?EdpNo=1911196&CatId=0\nhttp://www.tigerdirect.com/applications/SearchTools/item-details.asp?EdpNo=1776690&CatId=0\n\nshould be childs play for any electrician to fix the cable & maximum 5 minutes job !