hhsting
Senior Member
- Location
- Glen bunie, md, us
- Occupation
- Junior plan reviewer
I have full load amps of 40A for continuous load. However plans show going to have four current carrying conductors including the one fir the continuous load.
So 40x1.25 = 50A. Since I am going to have four current carrying conductors 50/.8 = 62.5A. So the minimum size conductor would be #6.
However I am provided with #3 awg due to voltage drop on 50A breaker. So would #6 gnd be big?
So in order to figure above I would proptional (#3 awg area/#6 awg area) or (#3 awg area/#8 awg)? #8 awg is for 50A mini conductor
So 40x1.25 = 50A. Since I am going to have four current carrying conductors 50/.8 = 62.5A. So the minimum size conductor would be #6.
However I am provided with #3 awg due to voltage drop on 50A breaker. So would #6 gnd be big?
So in order to figure above I would proptional (#3 awg area/#6 awg area) or (#3 awg area/#8 awg)? #8 awg is for 50A mini conductor