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Posted by Sam E on October 25, 2009, 9:50 pm
[snip]
>>You have a SERIES circuit (considering that the neutral is effectively
>>disconnected).
>>[snip]
>Wrong. The neutral is "effectively disconnected" *only* if the loads on the
>two legs are exactly the same.
Which they are (either in the 200A+200A example or the 1A+1A one).
> The two legs function as two parallel circuits
>with respect to 120V loads.
In a parallel circuit BOTH ends of the loads are connected together
(or at least to identical voltages). Neither is true here.
>Obviously they are indeed in series WRT 240V
>loads.
Strangely, I get the idea that you actually know this stuff.
In this 200A service there are THREE current-carrying conductors.
Each of these conductors is of the proper size to carry 200A. OK so
far?
You say (when this service is fully loaded) that two of these
conductors is carrying 200A (for a total of 400A, as you say).
Then where is that 400A going? The only remaining conductor is the
neutral, a big enough conductor for 200A (yes, this 400A was at 120V
but current is still current and voltage doesn't change the
conductor's current capacity).
Somehow I'm imagining a bridge that can handle 200 cars per minute,
but that can be 400 if half the cars are blue :-)
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Posted by Doug Miller on October 25, 2009, 11:08 pm
>[snip]
>>>You have a SERIES circuit (considering that the neutral is effectively
>>>disconnected).
>>>[snip]
>>Wrong. The neutral is "effectively disconnected" *only* if the loads on the
>>two legs are exactly the same.
>Which they are (either in the 200A+200A example or the 1A+1A one).
And that almost never happens in real life, either....
>> The two legs function as two parallel circuits
>>with respect to 120V loads.
>In a parallel circuit BOTH ends of the loads are connected together
>(or at least to identical voltages). Neither is true here.
Wrong -- both are true.
>>Obviously they are indeed in series WRT 240V
>>loads.
>Strangely, I get the idea that you actually know this stuff.
>In this 200A service there are THREE current-carrying conductors.
>Each of these conductors is of the proper size to carry 200A. OK so
>far?
OK
>You say (when this service is fully loaded) that two of these
>conductors is carrying 200A (for a total of 400A, as you say).
400A @ 120V, or 200A @ 240V, yes.
>Then where is that 400A going? The only remaining conductor is the
>neutral, a big enough conductor for 200A (yes, this 400A was at 120V
>but current is still current and voltage doesn't change the
>conductor's current capacity).
>Somehow I'm imagining a bridge that can handle 200 cars per minute,
>but that can be 400 if half the cars are blue :-)
Cute. Just answer these questions; assume a 240V 200A service.
What is the maximum power that service can provide?
If all the loads supplied by that service are 120V loads (e.g. blender,
toaster, light bulbs, range hood, stereo, TV, computer, etc.) what do you get
when you divide that maximum power by 120V?
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Posted by on October 26, 2009, 10:49 am
On Oct 25, 11:08=A0pm, spamb...@milmac.com (Doug Miller) wrote:
all.invalid> wrote:
> >[snip]
> >>>You have a SERIES circuit (considering that the neutral is effectively
> >>>disconnected).
> >>>[snip]
> >>Wrong. The neutral is "effectively disconnected" *only* if the loads on=
the
> >>two legs are exactly the same.
But a balanced load is exactly what was shown in the simple circuit
example above that he understood and is discussing.
> >Which they are (either in the 200A+200A example or the 1A+1A one).
> And that almost never happens in real life, either....
Which matters not a wit. Unless of course you are trying to get
close to the maximum capacity of the service. If it's totally
unbalanced, guess what? You get 200 amps at 120V, or exactly half
the power capacity of the service. Gee, I wonder why? Could it be
that it's because the service can only handle 200AMPS? And that with
a 200 amp unbalanced load at 120V, 200 amps is coming in on one hot
and it's all going back on the neutral?
> >> The two legs function as two parallel circuits
> >>with respect to 120V loads.
> >In a parallel circuit BOTH ends of the loads are connected together
> >(or at least to identical voltages). Neither is true here.
> Wrong -- both are true.
Wow, it's getting really strange here. Of course, by definition, a
parallel circuit is one where the ends of the individual elements are
connected together. A series circuit is one where elements are
connected one after the other, in series.
> >>Obviously they are indeed in series WRT 240V
> >>loads.
> >Strangely, I get the idea that you actually know this stuff.
> >In this 200A =A0service there are THREE current-carrying conductors.
> >Each of these conductors is of the proper size to carry 200A. OK so
> >far?
> OK
> >You say (when this service is fully loaded) that two of these
> >conductors is carrying 200A (for a total of 400A, as you say).
> 400A @ 120V, or 200A @ 240V, yes.
And there you go again, inserting voltage into a question of
amperage. Amperage is a measure of the charge, ie electrons passing
through the conductor and IS NOT LINKED TO VOLTAGE.
> >Then where is that 400A going? The only remaining conductor is the
> >neutral, a big enough conductor for 200A (yes, this 400A was at 120V
> >but current is still current and voltage doesn't change the
> >conductor's current capacity).
> >Somehow I'm imagining a bridge that can handle 200 cars per minute,
> >but that can be 400 if half the cars are blue :-)
> Cute. Just answer these questions; assume a 240V 200A service.
> What is the maximum power that service can provide?
You;ve asked that question multiple times and it's always been
answered the same: 48KVA
Now answer his question that you avoided. Apply Kirchoff's law and
tell us where current is flowing in a 200 amp service cable that
totals up to 400 amps. All of us here agree and can account for
200amps. So explaing the missing 200.
> If all the loads supplied by that service are 120V loads (e.g. blender,
> toaster, light bulbs, range hood, stereo, TV, computer, etc.) what do you=
get
> when you divide that maximum power by 120V?
If it's a balanced load, you get 400 amps because half the load is in
SERIES with the other half. As I've outlined about 6 times now, you
have 200 amps coming in on one hot, going through the loads in series
and then out the other hot. 200 amps is flowing in the service. If
you say it's 400, then why isn't it 2 amps that flows in a 120watt
light bulb plugged into an outlet? 1 amp comes in one wire, 1 amp
goes out the other wire. Yet the world agrees that only 1 amp is
flowing, not 2.
If it's a totally unbalanced 120V load, then you can't just divide the
power by 120 as youu imply, because you have 200 amps flowing in on
one hot, and 200 amps flowing out on the neutral. So you have a
120V, 200 amp load and only a power of 24KVA.
No matter how you slice and dice it, there is a max of 200 amps
flowing in the service. Since you believe otherwise, outline the
current flows as I have here and how it adds up to greater than 200
amps flowing in the service conductors.
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Posted by Doug Miller on October 26, 2009, 7:21 pm
trader4@optonline.net wrote:
>You;ve asked that question multiple times and it's always been
>answered the same: 48KVA
Now divide 48kVA by 120V and tell me what you get.
>> If all the loads supplied by that service are 120V loads (e.g. blender,
>> toaster, light bulbs, range hood, stereo, TV, computer, etc.) what do you get
>> when you divide that maximum power by 120V?
>If it's a balanced load, you get 400 amps
Which is exactly what I've been telling you for the last three days. Glad you
finally figured it out.
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Posted by Sam E on October 26, 2009, 8:12 pm
On Mon, 26 Oct 2009 03:08:49 GMT, spambait@milmac.com (Doug Miller)
wrote:
>>[snip]
>>>>You have a SERIES circuit (considering that the neutral is effectively
>>>>disconnected).
>>>>[snip]
>>>Wrong. The neutral is "effectively disconnected" *only* if the loads on the
>>>two legs are exactly the same.
>>Which they are (either in the 200A+200A example or the 1A+1A one).
>And that almost never happens in real life, either....
True, although it happens in examples, such as the ones used here.
>>> The two legs function as two parallel circuits
>>>with respect to 120V loads.
>>In a parallel circuit BOTH ends of the loads are connected together
>>(or at least to identical voltages). Neither is true here.
>Wrong -- both are true.
IF those 2 legs have identical voltages, the difference between them
is 0 (that's what "identical" means). In that case, 120V loads should
be OK but 240V loads would get nothing.
>>>Obviously they are indeed in series WRT 240V
>>>loads.
>>Strangely, I get the idea that you actually know this stuff.
>>In this 200A service there are THREE current-carrying conductors.
>>Each of these conductors is of the proper size to carry 200A. OK so
>>far?
>OK
>>You say (when this service is fully loaded) that two of these
>>conductors is carrying 200A (for a total of 400A, as you say).
>400A @ 120V, or 200A @ 240V, yes.
That's some strange reality. In this one voltage and current are
different things, and you can't change one into another with
arithmetic.
There is still no 400A at any voltage.
>>Then where is that 400A going? The only remaining conductor is the
>>neutral, a big enough conductor for 200A (yes, this 400A was at 120V
>>but current is still current and voltage doesn't change the
>>conductor's current capacity).
>>Somehow I'm imagining a bridge that can handle 200 cars per minute,
>>but that can be 400 if half the cars are blue :-)
>Cute. Just answer these questions; assume a 240V 200A service.
I notice you ignored mine. You have yet to show any non-imaginary
location of that 400A.
>What is the maximum power that service can provide?
48KW. Of course we were talking about CURRENT.
>If all the loads supplied by that service are 120V loads (e.g. blender,
>toaster, light bulbs, range hood, stereo, TV, computer, etc.) what do you get
>when you divide that maximum power by 120V?
That would be 400A. Of course that's only in your imagination since
the math is invalid (120V is obtained by splitting the service into 2
separate halves, each of which is only 24KW).
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>>disconnected).
>>[snip]