Household Wiring Question

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tk909

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Edmonton, AB
When you have a main disconnect in a house, say a 200A 2P breaker, is that 200A per pole, or 200A overall (100 per pole)?

Or effectively is that saying 200A at 240V / 400A at 120V?
 
tk909 said:
When you have a main disconnect in a house, say a 200A 2P breaker, is that 200A per pole, or 200A overall (100 per pole)?
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It?s 200 amps on Pole #1, and it?s 200 amps on Pole#2, and it?s 200 amps overall. 200 + 200 = 200.
tk909 said:
Or effectively is that saying 200A at 240V / 400A at 120V?
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That?s another way to say it.
 
Its a 200 amp: But

According to the Power law if its good for 200 amps. at 240 then its good for 400amps. at 120 since you can take off 200 from line #1 to neutral and 200 from line #2 to neutral , you can draw a total of 4 hundred amps from the this panel.:)

Editied to say this 200amps.+200amps.=200amps only because the breaker and conductors are only worth 200 amps.

With 200 amp. main and 200 amp. conductors. My last and final answer to your trick question 200A at 240V or 100amp. per pole at 120 volts.Because with a balanced load on each line of 100 amps. the neutral would be zero amps. and each line would have a load of 200 amps.
 
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200a ?

200a ?

200A X 240V=48000. 48000/2 ( 2 being L1 & L2)=24000.
24000 per line / 120V = 200amps. Neutral carries imbalance between L1 & L-2 and under normal conditions would not exceed 200amps even if on leg had no load.
 
Thanks for the replies, I really appreciate the help.

so if I'm understanding this correctly:
I have a 2P 200A Main Panel,
I can allow for a load of, to 80% Canadian Electrical Code continous load, 38.4kW (200A * 80% * 240) if 38.4 kW was evenly distributed between the two poles, i.e., 80A of continous current through each L1 and L2?
 
ronaldrc said:
My last and final answer to your trick question 200A at 240V or 100amp. per pole at 120 volts.Because with a balanced load on each line of 100 amps. the neutral would be zero amps. and each line would have a load of 200 amps.
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No, with a balanced load of 100a on each line, the neutral current would be zero, and each line would have a load of 100a.

100a @ 240v is twice the power as 100a @ 120v.
 
tk909 said:
so if I'm understanding this correctly:
I have a 2P 200A Main Panel,
I can allow for a load of, to 80% Canadian Electrical Code continous load, 38.4kW (200A * 80% * 240) if 38.4 kW was evenly distributed between the two poles, i.e., 80A of continous current through each L1 and L2?
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No, 80% of 200a @ 240v is 160a.

160a @ 240v is the same as 2 x 160a @ 120v.

38.4Kw evenly distributed between two lines is 160a @ 240v total, or 160a on each of two 120v lines.

It would take 320a on a single 120v line to supply the same 38.4Kw.

By the way, the same 38.4Kw supplied by a 208/120Y 3ph would be 106.67a.
 
Larry

How can I draw 100 amps. from line #1 to neutral and 100 amps. from line #2 to neutral and not be drawing a total of 200 amps?
 
ronaldrc said:
Larry

How can I draw 100 amps. from line #1 to neutral and 100 amps. from line #2 to neutral and not be drawing a total of 200 amps?
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100a=L1 @120v
100a=L2 @120v
200a=L1+L2 @120v
100a=L1+L2 @240v
 
Let me put this another way. Circuit loading will all be resistance only

If I load Line #1 to the neutral with 100 amps. and I load line #2 with 100 amps. to neutral this would be a total of 200 amps. And this would also be a perfectly balanced circuit. We would have zero current flowing on the neutral.

Now lets cut the neutral, now how many amps are flowing from line #1 to Line #2?

Okay since no one is going to argue it probably would be 100 amps.
 
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ronaldrc said:
Larry

How can I draw 100 amps. from line #1 to neutral and 100 amps. from line #2 to neutral and not be drawing a total of 200 amps?
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That depends on what you mean by "a total."

When each line is carrying 100a at 120v, the total can be looked at as 200a at 120v. No problem there. But, we're splitting that 200a in half because we're doubling the voltage. Each line only has to carry half of the total load.

That's why each line's load is only 100a.

Let's say we're supplying only 240v (L-L) loads. If they total to 100a, that means each line carries 100a, not 200a, right? In the hypothetically-balanced system with line-to-neutral loads, there's no neutral current, so it behaves the same way.

In other words . . .

100a @ 120v + 100a @ 120v = 100a @ 240v

OR

100a @ 120v + 100a @ 120v = 200a @ 120v

. . . but not both.


Now, if we had a balanced line-to-neutral load of 100a on each line, and at the same time, a line-to-line load of 100a, the total current on each line would be 200a, not 300a.

If line #1 has 60a, line #2 80a, and there's a line-to-line of 50a, then L1 = 110a, L2 = 130a, and neutral = 20a.

Is that a better explanation?
 
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ronaldrc said:
If I load Line #1 to the neutral with 100 amps. and I load line #2 with 100 amps. to neutral this would be a total of 200 amps. And this would also be a perfectly balanced circuit. We would have zero current flowing on the neutral.
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Well, to the neutral bus, it would still be 100a*, but the bus itself would have a complicated web of little currents in it, interconnecting the L1 neutral-conductors' holes to the L2 neutral-conductors' holes, depending on how you happended to have landed them.

*100a in and 100a out

It is true that the supply's neutral conductor would seen no current.

Now lets cut the neutral, now how many amps are flowing from line #1 to Line #2?

Okay since no one is going to argue it probably would be 100 amps.
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No fair! You didn't give us enough time to answer! :mad: :)grin:)
 
Larry

I respect your opinion very much or I wouldn't have given what you said a second thought.

When you said it would be 100 amps. I had to go back to ohms law and voltage ratios.We would be doubling the resistance since it would be in series if you cut the neutral.There by reducing the current too half

You where right and I was wrong about the 100 amps. But I can't see that just by numbers alone, everyone is not geared that way.

But you have to admit without giving it any serious thought that my logic made a little sense.

Any way I was wrong about a couple of thing yesterday I am only human.

I do appreciate you correcting me. I was miss leading some and I appologize for it.

Ronald :)
 
ronaldrc said:
How can I draw 100 amps. from line #1 to neutral and 100 amps. from line #2 to neutral and not be drawing a total of 200 amps?
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Because the word ?total? is meaningless.

What happens when you have a 100 amp load on one side (call it Phase A) is that, for the ?first half cycle,? current goes from the source to the ?Phase A load? via the ?Phase A hot conductor,? and travels back to the source via the ?neutral conductor.? If you also have a load of 100 amps on the other side, then during THAT SAME half cycle the current goes from the source to ?Phase B load? the via the ?neutral conductor,? and returns to the source via the ?Phase B hot conductor.? As a result, the common neutral conductor is getting 100 amp going one way at the same time as another 100 amps going the other way, and the net neutral current is zero. So as a practical matter the 100 amps leaving the source on the ?Phase A hot conductor? will, in fact, return to the source via the ?Phase B hot conductor.? On the other half cycle, it goes the opposite way, and the neutral still has zero current.
 
Charlie

I agree with you on the flow. I didn't really give this any serious thought when I posted.

I did not take into consideration that we are doubling our resistance from 1.2 Ohms to 2.4 Ohms I personnelly think this is the main factor
I don't think after Larry corrected me on it, it is not that complex.

I don't have a problem of making a mistake I do it daily.

And if it hadn't 40 plus years since I was in Electrical school I would have
thought at 100 amps. at 120 volts would be 50 amps. at 240.

Or in this case the total amps. per line would be 100 amps.
 
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Yes I understand that. Thanks :)

Editied to insert this drawing


100_amp_example.jpg
 
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