I am putting in a subpanel for my workshop. I need another 240v line, so I
am ripping out the two 120v lines and the 240v line I have already put in
and replacing them all with a subpanel. Wish I had done it right the first
time, but that boat has sailed.
I am running #6 copper. As I read it, it is good for 55a, and since there
is no 55a breaker, I can use a 60a. Is that correct?
On the other hand; the most I will ever ever use at one time is a bandsaw
and DC, which draw 27a together (which is why I need another 240v). I
happen to have a 30a breaker. Any reason not to just use that? It is over
the 80% rule, but I presume that applies to the wire rather than to the
breaker. If it trips, I can always replace it.
The #6 was only a few dollars more than the #8; so what the heck...
Nothing is wrong at all. The NEC specifically permits going up to the next
higher standard breaker size when the rated ampacity of the wire (55A in this
case) does not correspond to any standard breaker size.
Doug Miller (alphageek at milmac dot com)
That doesn't make sense to me at all.
Why allow the breaker to trip at a higher rating than the wire? Conceivably
the wire would fail without the breaker ever tripping. That presumes the
wire rating of 55 A means it is only capable of carrying 55A at 120V
Whether it makes sense to you or not, it *is* the Code. Article 240.4(B), to
What do you mean by "fail"? Melt? Start a fire? Neither will happen as a
result of applying a 60A breaker on a 6AWG copper conductor.
The vast majority of the time, a typical circuit with 60A overcurrent
protection isn't going to be carrying anywhere near 60A. In any event, the
difference between 60A and 55A on a #6 wire is not going to mean the
difference between the wire igniting vs. not igniting. I imagine that the NFPA
(authors of the NEC) decided that five amps on a wire that size wasn't enough
to worry about. You could always ask them.
Doug Miller (alphageek at milmac dot com)
replaced it with #10.
That is about the worst possible thing, since it comes on at full draw, and
stays on for a long period of time; yet nothing happened. Code is rather
conservative, which prevents many fires.
And don't forget, the wire is 90degree and the panel and breaker are
75degree; yet the 55a limit is as though they were 60degree. So the extra 5
amps is not a big deal.
You're also presuming that the heat generated by pulling 60 amps through a
#6 copper wire would cause it to reach it's flash point. I'm pretty
confident that the NEC has figured in considerable leeway
3) A halogen lamp. 2.5 amps (@ 120 volt)
Total (intermittently) 27.5 amps, plus maybe another light that was
on. Maximum 30 amps say all fed from a panel wired with #6AWG and a
50 amp (Square D) breaker from the main house panel about 40 cable
feet distant. Nothing blinked or went dim etc. Caught fire or popped a
Absolutely. The heat, if any, generated by the current flowing
through the wire is totally dependant on how much current is flowing,
and unrelated to the voltage applied to the whole circuit.
I think this will help a few people and won't help others at all.
E=IR (Remember that. :) ) In other words:
EMF=Current times resistance. In other words, in this case:
Voltage = 55 amps x R (resistance)
Power = Voltage x Amps. (Remember this.)In other words, in this case:
Watts = voltage x 55 amps.
But the voltage in the formula above is not the 110 or 220 that comes
from the wall. It's the voltage drop from one end of the wire to the
other, which is small, because the resistance of the wire is small.
The total voltage drop of an appliance plugged in to a wall is 110 or
220 or whatever, but almost all of that voltage drop is in the
appliance, and not in the cord going to it. It's the saw or washer or
water heater or mixmaster that is designed to do the work, and the
wire to it is designed to have low resistance and not impeded the
electrficity getting to the appliance. But, in this case we're
talking about running 6 gauge wire with a 60 amp breaker, even though
the wire is rated at only 55 amps. So how much heat (power or watts,
iirc) will be generated in that wire if the current goes up to just
shy of 60 amps, even though it is rated at 55 amps.
Look at the formulas at the top, and then continue here:
Since voltage = 55 amps x R, then:
Watts = 55 amps x R x 55 amps. In other words:
Watts = (55 amps) squared x R.
Where R is the resistance of the wire feedinn the appliance.
This is not the number of watts given off by a lightbulb, if there is
a lightbulb in the appliance, or used by the motor if there is a motor
in it. It's the number of watts given off by the wire feeding the
appliance. And it is totally unrelated to the voltage applied to the
appliance** **Except insofar as if higher voltage were applied, more current woudl
go through the same resistance wire and appliance.
Well, yes and no...depends on whether it is a constant current or
constant voltage source (or another way to consider it is in regards to
the chicken and egg :) )...
For a power circuit fed from the grid, one can consider it a constant
There is current flow through the circuit only because there is a
voltage drop from that source to the neutral.
How much voltage drop is in the wiring depends on the resistance of the
wiring, that is correct. But how much current flows through the wire
depends on the _total_ resistance from the source back to the neutral
(assuming a series circuit, the current through each component is the
same), not just the wire. So, given that there is a fixed voltage
supply and for a particular piece of equipment on the circuit in the
shop, the current will also be (nearly) constant and the voltage drop
across the wiring will be determined by that current. Consequently, of
the wiring loss, the situation looks more like a current source rather
than a voltage source.
But then again, one can analyze it as if there were a fixed voltage
impressed over the cable and arrive at the same numerical result...
V = IR
P = VI = (IR)I = I^2 R
Which is more fundamental; the VI or I-squared R form? All depends on
point of view... :)
Doug: no disrespect to you are any one ales in this forum,
I have no degree to speak of so I am not going to contradict you
But from my understanding insulation do not have anything
to do with wire Ampacity or in the other words capacity to
carry amount of current. Wire insulation rating is for soly purpose
where that wire can be used to be able to carry it's rated current.
example # 10 wire insulated with rubber is good up to maximum
140 Deg. F regardless if is carrying 30 amps. or 10 amps.
because at that temp. rubber will start to melt.
I use lot of glass cover wire and Teflon but # 10 wire is good
only for up 30 amps. higher amperage requires larger conductor
8 or 6 whatever I may need so I don't see insulation been factor here
I am sorry that is the way I see it
In wire sizes 10-12-14, the NEC limits the ampacity to 30-20-15 regardless
of insulation type, but in larger sizes the NEC allows different amperages
depending upon insulation type, as well as other factors
Well, I'm sorry, but your understanding is incorrect. Ampacity depends on
multiple factors, and insulation is one of them. I'll try to explain.
Exactly. And the rating depends on the type of insulation (among other
things) -- this is laid out clearly in Table 310.16 of the National Electrical
Which is exactly why wires with rubber insulation are limited to carrying
currents low enough to *not* heat the wire up that much. Wires insulated with
materials capable of withstanding higher temperatures are allowed to carry
more current because they can get hotter without damaging the insulation and
creating a hazard.
The insulation *isn't* a factor for AWG10 and smaller wires, because the NEC
specifically limits the allowable breaker rating for AWG10, 12, and 14
conductors to 30A, 20A, and 15A respectively, regardless of what insulating
material is used.
For wires AWG8 and larger, the ampacity *does* depend on the insulation. For
example, AWG8 wires with Type TW insulation are limited to 40A -- but with
Type THHN insulation, the allowable ampacity is 55A.
Doug Miller (alphageek at milmac dot com)
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