calculation question

[stv292]

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

 

[telectrix]

the volt drop in the tables allows for the resistance of the line and neutral added together. what chris is saying is that the load is not all at 1 end of the circuit. say you've got 6 lights, each drawing 1A, evenly spaced. the first section of the cable carrys 6A, the 2nd carrys 5A , the 3rd carrys 4A etc., so the VD calculation is a bit more complicated than if it were a single load at the end of the cable. as a rough guide, you could calculate the VD for a single load, then divide by 2 to get an approximate value.

 

[i=p/u]

so when you installing a load of lights spread around cheshire, you actualy have more allowance for vd than allowed at the extremity of the circuit.... ive never done this sum, as you say ive only been taught the single load one

 

[i=p/u]

so i get 194v with higher resistance neutral conductor...

 

[i=p/u]

yes yes i see, so less voltage with more resistance, which i understand