And resistors in parallel exhibit an equivalent resistance of Rt = 1 / ( 1/R1 + 1/R2 + 1/R3 + 1/R4)
So in theory, if you use all four in parallel, the equivalent resistance should be (at 100ft)
Rt = 1 / ( 1/3 + 1/3 + 1/3 + 1/3) = 1 / ( 4/3) = 3/4 = 0.75
Not bad And assuming the same 2A @ 12V:
Vdrop = (2A)(0.75) = 1.5V
G.
Joel Jaeggli wrote:
On Wed, 12 Mar 2003, Gene wrote:
I guess it depends on the voltage too.
yeah at .0302 ohm per foot 24awg 3ohms per hundred feet... you're talking about a ~5volt drop so you put in 12 and get 7...
But if I remember my KCL/KVL and Ohm's Laws, what if you use the 4 unused lines in a standard PoE as the positive voltage rail and the shielding (assuming this is a shielded cat5) as the ground) to help reduce equivalent line resistance?
Aka "four heads...err...lines are better than one" idea.
-- general wireless list, a bawug thing <http://www.bawug.org/> [un]subscribe: http://lists.bawug.org/mailman/listinfo/wireless
