if my generator can output 30 amps 240, i can take the 240 and split it to
120 and output a total of 60 amps, 30 each leg, now how much current is
flowing on the neutral? i have a 10/3 stranded cable about 90 feet long, i
dont know if it will be 60 or 30. my service entrance cable all three look
the same size wire so i guessing it will be 30. thanks tony
thanks for the info, we just went from a 3500 generator, that ran constant
for 1 week then 1.5 weeks,because of 2 hurricanes, i knew that 10 gauge
stranded could handle the amps of the 3500 but the new 7550 puts out 62 amps
so i wanted verification that the neutral would not be overloaded. thanks
I'm not sure what sort of application this table deals with ampacity
wise, but it certainly aint normal wiring methods, or else I'm reading
it wrong. 12awg at 14.something amps? Better to consult your NEC
tables. The other data is interesting but in light of the 'max amps'
data might be questionable also.
It's a Linux world....well, it oughta be.
| > if my generator can output 30 amps 240, i can take the 240 and split it
| > 120 and output a total of 60 amps, 30 each leg, now how much current is
| > flowing on the neutral? i have a 10/3 stranded cable about 90 feet
| > i dont know if it will be 60 or 30. my service entrance cable all three
| > look the same size wire so i guessing it will be 30. thanks tony
| I know this isnt a TRICK question. But, with both 110 lines loaded the
| same, and resistive, there will be no current in the neutral, even though
| power is being delivered to the two loads.
Alright, I had to think about this one, and for the sake of the
unknowing, it should be explained. If a 220V load is being used, then
current flows from leg to leg, not leg to neutral, so the neutral carries no
current, as noted by older 220V circuits not using the neutral or ground,
just the two legs. If both legs are running 110V loads, however, then when
one leg is at maximum positive voltage, current will also be at a maximum
through that leg. However, the other leg will also be at maximum negative
voltage, so current will also be at a maximum on that leg too. If the loads
are equally balanced, then the neutral line that is shared between the two
(not the neutral for each leg) will have no current, as the current will be
going from leg to leg and not through the common neutral. However, since
current is rarely balanced between legs, only the difference between the
currents will pass through the neutral, and no more than the maximum current
of one leg (say, if one leg is maxxed out and the other one with no load)
then the common neutral will carry no more than the load from the loaded
It can get fancy with inductive loads, where currents and voltage do not
always line up evenly, but I'm too rusty on this one to make any more
If it's setup as a center-tapped neutral like normal, and you are
drawing an even 30 amps on each 120V leg at the end of the cable run,
the current on the neutral is zero. They balance each other out.
Worst case is 30 amps on the neutral if one hot leg is unloaded.
But you do not want that neutral to go open under any circumstances,
because very bad things happen when the loads get unbalanced - the
voltages start swinging.
Some old generators can be jumpered to provide true 60A 120V with
the windings in parallel. But then you can't use that 10/3 cable,
more like 4-3 - and voltage drop is a more serious problem...
--<< Bruce >>--
Bruce L. Bergman, Woodland Hills (Los Angeles) CA - Desktop
Electrician for Westend Electric - CA726700
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