Load capacity of 200-amp panel

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On Tue, 27 Oct 2009 08:27:24 -0700 (PDT), in alt.home.repair, snipped-for-privacy@optonline.net wrote:

Two 200A 120V loads in series makes a 200A 240V load.

The only way you can get 400A out of a 200A service is with a step-down transformer.
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On Tue, 27 Oct 2009 08:27:24 -0700 (PDT), in alt.home.repair, snipped-for-privacy@optonline.net wrote:

Two 200A 120V loads in series makes a 200A 240V load.

The only way you can get 400A out of a 200A service is with a step-down transformer.
On Tue, 27 Oct 2009 08:27:24 -0700 (PDT), in alt.home.repair, snipped-for-privacy@optonline.net wrote:

Two 200A 120V loads in series makes a 200A 240V load.

The only way you can get 400A out of a 200A service is with a step-down transformer.
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On Oct 27, 8:36 pm, snipped-for-privacy@asgard.slcc.edu wrote:

I think we are in agreement, except for perhaps one point.
From your previous post, you clearly agree that you can in fact have two 120volt, 200 amp loads connected in series across the 240volt service. That is a perfectly balanced load. You now have 200 amps flowing in series through each load. In my world that is in fact "supporting" 400 amps of 120volt load. Lets say I had forty 10 amp, 120volt heaters. I could could clearly put twenty of them between one leg and neutral and twenty between the other leg and neutral and it would work. You now have a fully loaded balanced service. There is zero current flowing in the neutral and 200 amps flowing in the service. It works because the loads on one side are connected in series to the loads on the other side.
My whole point all along has been that the actual current in a 200 amps service is limited to 200 amps which clearly you agee with. And it has nothing to do with "parallel circuits", or power, voltage or anything else. It looks like the only difference we have is your definition of "supporting loads" may be stricter than mine.
And I think all of us are still waiting for a simple answer from Doug as to how many amps are actually flowing in a fully loaded 200 amp service cable circuit. I've asked that several times now and still have no answer, despite having fully answered all his questions.
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On Wed, 28 Oct 2009 02:24:20 -0700 (PDT), in alt.home.repair, snipped-for-privacy@optonline.net wrote:

Your semantics create confusion. In NO CASE can you flow more than 200 amps through any single load. To say that you have many 120V loads drawing a total of 400A is WRONG and obscures the fact that two balanced 120V loads is absolutely identical to a 240V load.
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Thanks for the kind words Smitty. :)
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snipped-for-privacy@optonline.net wrote:

Very good. Now divide 48kVA by 120V and tell me what you get.
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On Oct 26, 7:18 pm, snipped-for-privacy@milmac.com (Doug Miller) wrote:

Again, this has been answered here repeatedl, so I don't see why you keep asking.. One more time, it's obviously 400 amps.
Also, you only get those 400 amps if the load is balanced so that it appears as a series load. The 200 amp current flows in one hot and out the other. If you had a single 120V 400 amp load, it would sit between one hot leg and neutral, where the capacity is limited to 200 amps and the cables would melt. Gee, I wonder why? Could it be because the actual current in a 200 amp service circuit is only 200 amps?
You can divide and get any answer you want. I could divide 48KVA by 10volts and get 4800 amps. So a 200 amp service could support a total 4800 amp, 10 volt load too. But how much max current is actually flowing in the service cable entering the house? Exactly the same as always, 200 amps. If you believe otherwise, please tell us what currents are flowing in each of the three conductors.
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snipped-for-privacy@optonline.net wrote:

Which is exactly what I've been telling you for the last four days: a 200A 240V service will support 400A of 120V loads. I'm glad you finally figured it out.
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On Oct 27, 2:04 pm, snipped-for-privacy@milmac.com (Doug Miller) wrote:

That has never been in dispute. What has been is how many physical amps are flowing in the service cable circuit of a 200 amp service? Here's a hint: Try answering this simple physics question without refering to voltage or power.
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You're right. If the the two 120V loads are perfectly balanced, then it's the equivalent to two 120V 0.6 ohm loads in series across 240V each pulling 200 amps, and you can disregard the neutral or even disconnect it. The neutral is there to hold the voltage on each leg or side to 120V when the loads aren't perfectly balanced.
>The 200 amp current flows in one hot and

True
No, the main breaker would open long before any melting

A whole bunch for a few microseconds. A breaker isn't instantaneous.
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Yes, I agree, assuming there is one. In our hypothetical case I was ignoring any breakers.

We have been discussing continous loads at the service max, not transients. That 200 amp service can support a total load of 4800 amps at 10 volts, or 2400 amps at 20 volts. As I said before, you can slice it and dice it anyway you want, but you still have 200 amps max of current flowing in the service.
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Yes, I agree, assuming there is one. In our hypothetical case I was ignoring any breakers.

200A thru each main-breakered leg - yes. But I'm not sure where the 4800A @ 10V is coming from. Do you think the transformer secondary windings will sustain at 4800A? Are you now taliing about continuous or transient??
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Take twenty four .05 ohm resistors. Wire them is series. In series they equal 1.2 ohms, Connect them across the 240V, 200 amp service hot lines. You now have 10 volts across each resistor and 200 amps flowing through the circuit. So, you're supporting twenty four 200 amp, 10volt loads. Each resistor sees 10volts and 200 amps. Taken together that's 4800A at 10V. How much current is flowing in the service? Still 200 amps. Showing once again that this feature works because to get loads of more than 200 amps out of a 200 amp service, the loads have to appear in series.
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On Sat, 24 Oct 2009 23:55:15 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

That makes no sense. 1A counted twice is 1A. No amount of counting changes what exists.
I'm alone in my room. Therefore, counting the number of people in the room shows 1. Now, I look in a mirror and count again. Now there's TWO people in the room.
You have ONE 50-foot rope. Every time you see rope count it. Now you have 10 ropes.

You still haven't shown where that TWO amp current is.
Is it in the first load? There's just 1A there. Is it in the second load? There's just 1A there. Is it in the first supply wire? There's just 1A there. Is it in the second supply wire? There's just 1A there. Is it in the neutral wire? There's ZERO current there. Is it in the air? There's heat there, but no current.
WHERE is it?
[snip]
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1A in _each of two parallel circuits_ is a total of 2A. [...]

The two loads are in parallel, not in series. 1 + 1 = 2.

Two parallel circuits, 1A in each. 1 + 1 = 2.

1 amp in each of the two hot legs -- which with respect to 120V loads, are effectively two separate parallel circuits.
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[snip]

Of course. The problem is WHERE is the 2A.

No. That would require identical sources. These sources are not identical, but opposite. The difference can be up to 339V (the peak value for 240V RMS).
In the circuit you describe there is ZERO current in the neutral. This is the same as the neutral not being there (current in that wire is 0A in either case). What you have is a SERIES circuit. It is the same current going through both resistors, therefore addition is not appropriate here.
This reminds me of a story I heard a few years ago. Three 12-year-old boys wished they could vote. Since the voting age was 18, the boys decided that if they went together they'd get 2 votes.
12 + 12 + 12 = 18 * 2
The fact that you can do arithmetic doesn't mean it's appropriate to do so.
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Doug Miller wrote:

snipped-for-privacy@optonline.net wrote:

....with a common neutral wire which must provide a return path for both circuits.

As long as the neutral wire is rated for an amperage capacity of 400A...if it's not, and you try feeding 400 amps through a wire only rated for 200A max, what do you think will happen? ;-)
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You badly misunderstand how this works. In a 240/120 residential service, the neutral carries only the unbalanced current (the difference between the currents in the hot legs, not their sum): if 50A is being drawn on one hot leg, and 90A on the other, the neutral carries only 40A. If one hot leg is carrying 200A, and the other 199A, the current in the neutral is *not* 399A -- it's 1A. And if both hot legs are loaded exactly equally, whether that's 1mA each or 200A each, the current in the neutral is zero.
For 200A service, the neutral does *not* need to be rated for 400A. The most it can ever carry -- if one hot leg is fully loaded, and the other is unused -- is 200A, and if the loads are even halfway close to being balanced across the two legs, most of the time the current in the neutral is far less than that.
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On Sat, 24 Oct 2009 18:17:14 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

You have a SERIES circuit (considering that the neutral is effectively disconnected).
[snip]
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Wrong. The neutral is "effectively disconnected" *only* if the loads on the two legs are exactly the same. The two legs function as two parallel circuits with respect to 120V loads. Obviously they are indeed in series WRT 240V loads.
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