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Yes sorry not thinking what you were using. I did V=IR current is 6A, resistance is (19.5mohm/m*56m) =1.092 ohms so volt drop = 6*1.092 =6.552V
hi
the osg say's i can use 56m of cable of 1.0/1.0 installation method 103,101 e.tc. for lighting, on a radial at 6A. the mV/A/m is 44.
so 44x6x56/1000=14.7v that is 6% of lighting not 3%.
what am i missing
isit not just in line conductor you work out vd???
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