Too much for 200A

[Matt Fellman]

Without doing a full-blown load calculation what's your first reaction to the following:

Service is 200A

3800 sq ft, 1966 built house
2 electric water heaters, 1 @ 50, 1 @ 65 gal
2 ovens, 1 cooktop with griddle - all electric
Electric dryer
A larger than typical amount of lights and plugs, including extensive outdoor lighting, lighting and power for a pool (not heater - it's solar and gas), water features, etc.

This is a custom built place and the original owner had quite a thirst for gadgets.

I'm pretty sure it's too much. I'm recommending a load calculation (in addition to a host of other things). I'm just wondering what you guys think.

[Nick Ostrowski]

Matt, I'd say as long as they aren't cooking a Thanksgiving meal, doing the laundry, have their guests all showering, all at night, they should be OK ;).

I have no idea. Like you, I'd leave it up to an electrician to figure out.

[Patrick Norton]

Matt,

200 amps appears to be adequate for the house. They are definitely using all of their 200 amp service.

[Matthew Barnicle]

I probably would not have written it up. Judgement call.

[Scott Patterson]

Without doing a full-blown load calculation what's your first reaction to the following:

Service is 200A

3800 sq ft, 1966 built house
2 electric water heaters, 1 @ 50, 1 @ 65 gal
2 ovens, 1 cooktop with griddle - all electric
Electric dryer
A larger than typical amount of lights and plugs, including extensive outdoor lighting, lighting and power for a pool (not heater - it's solar and gas), water features, etc.

This is a custom built place and the original owner had quite a thirst for gadgets.

I'm pretty sure it's too much. I'm recommending a load calculation (in addition to a host of other things). I'm just wondering what you guys think.

200 amps can supply everything you have listed and then more. Don't forget that several of those items will be on 220 and might even have their own panel. I bet the pool is on a separate panel.

[Jim Robinson]

New house or existing? If it's older and not blowing the 200 amp main, it's not too much.

[John Arnold]

...If it's older and not blowing the 200 amp main, it's not too much.

Jim - I try to keep in mind that my client(s) may have different needs than the sellers. I'm not saying the 200 amp service is too small, just that no history of the main blowing (and how would you find that out?), in itself, might not tell you that much about performance going forward.

[Matt Fellman]

There are several other panels but they're all being supplied through the service so I didn't mention it. Does that change the overall setup or impact the load calculation?

There are 2 @ 50amps and 1 @ 70amps. I figured since it's all going through the service there is no difference.

[Patrick Norton]

Matt, as long as the service wires and breaker sizes to the subpanels are correct, that is fine.

[brian schmitt]

200 amps can supply everything you have listed and then more. Don't forget that several of those items will be on 220 and might even have their own panel. I bet the pool is on a separate panel.
scott,
what does 220 and seperate panels have to do with anything? the 200 amp service is it.you can have 100 other panels on that 1 service and all you are gonna get is 200 amps? it may be adequate but op didn't specify heating loads and motor loads etc. iwould ask for load calcs on a new residence of that size

[Scott Patterson]

scott,
what does 220 and seperate panels have to do with anything? the 200 amp service is it.you can have 100 other panels on that 1 service and all you are gonna get is 200 amps? it may be adequate but op didn't specify heating loads and motor loads etc. iwould ask for load calcs on a new residence of that size

Yea, that wasn't real clear. I should have said that many times you will find that a pool will have its own service equipment or main.

[Jon Randolph]

While that does seem like a lot, remember that though we list the service as 200 amps, you can concievably pull up to 199 amps through each leg prior to tripping the main. Each 220 volt breaker is getting the power from both legs, not just one. 199 + 199 = 398 amps, hit 399 or exceed 200 from either leg and the main trips.

Think of the main overload disconnect as a 220v circuit breaker. With 10 gauge conductors (30 amps) the appliance can concievably pull 29 amps through each leg prior to the breaker tripping, but if either leg exceeds the 30 amp rating, the breaker goes.

[John Steinke]

To answer the question, without 'doing a load calculation,' is a lot like making a weather report without looking through the window. That's the only definitive way to answer the question.

In more immediate terms, you can try to pick a 'peak' time, and actually measure the amps flowing through the service drop.

Absent either approach, an 'answer' is little more than trying to count cards at the blackjack table.

[Michael P. O'Handley]

Hi,

I've seen 125 amp panels in big old high end houses around here built in the 60's that have handled the same sort of loads just fine.

ONE TEAM - ONE FIGHT!!!

Mike

[Jerry Peck]

Each 220 volt breaker is getting the power from both legs, not just one. 199 + 199 = 398 amps, hit 399 or exceed 200 from either leg and the main trips.

Not quite a good way to describe it.

There are two 'legs' to the 240 volt circuit, however, the same current is flowing through each one (with the exception of the off balance current which if flowing through the neutral, this is the current difference between the two hot legs).

That means (assuming a balanced load) that if there is 200 amps flowing through one leg, there will be 200 amps flowing through the other leg - the current is NOT additive.

Now (assuming an unbalanced load) there is 200 amps flowing through one leg, but only 100 amps flowing through the other leg, the other 100 amps is now flowing through the neutral - STILL not additive.

A 200 amp breaker will carry 200 amps ... period, not 200 + 200.

[Jon Randolph]

A 200 amp breaker will carry 200 amps ... period, not 200 + 200.


Doesn't each leg of the drop from the PoCo have the capability of carrying 200 amps (assuming that everything is adequately sized, of course)?

So... Pulling 99 amps on one leg and 101 amps on the other will throw the 200 amp main.

[Matt Fellman]

To answer the question, without 'doing a load calculation,' is a lot like making a weather report without looking through the window. That's the only definitive way to answer the question.

In more immediate terms, you can try to pick a 'peak' time, and actually measure the amps flowing through the service drop.

Absent either approach, an 'answer' is little more than trying to count cards at the blackjack table.

I realize the shortcomings of my question... I was really just trying to run it by the board to see if I would be negligent in not calling it out. It sounds like I wouldn't and ended up leaving it off the report.

I think I just get used to seeing new houses 'over-wired', particularly in comparrison to the 1960's. A new house of this size with this equipment would likely have 2 panels, whether needed or not.

[Richard Abrams]

I read your description and if nothing has been added to home since it was built No problem but that is highly unlikely. Circuitry is your problem almost any new appliance will require dedicated grounded service so you will need more breakers, More breakers more load. This house today as you described would probably need at least a 250 ampere panel or with circuitry 2- 200 side side rthe second panel fused at 125 amp.:)

[Jerry Peck]

Doesn't each leg of the drop from the PoCo have the capability of carrying 200 amps (assuming that everything is adequately sized, of course)?


Jon,

First ... *it is the same current* ... going through each of the two hot conductors.

I'll try to show it below:

___________________ Phase A ->
& |
& 120 volts
& |
& -------------------- Neutral ->
& |
& 120 volts
& |
___________________ Phase B ->

Say Phase A leg carries 80 amps -> to the panel

and Phase B leg carries 100 amps -> to the panel

The 80 amps circles through the transformer from A to B, the 20 amps difference circles through the transformer from Neutral to B

You do not have 180 amps, you have 100 amps - it is just split up as described above.

[Jon Randolph]

Just like in pre-school :confused:, it is much easier to grasp with pictures.


Thanks Jerry:o

[Tim Voss]

Actually you do have the equivalent of 400A available to drive 120v loads. As far as power goes, 200A of 240v is equal to 400A of 120v.

[Jerry Peck]

Actually you do have the equivalent of 400A available to drive 120v loads. As far as power goes, 200A of 240v is equal to 400A of 120v.

"Actually you do have the equivalent of 400A available "

Actually, you do NOT.

Just try to push 400 amps through a 200 amp breaker (well, okay, maybe *IF* that is an FPE main :D ).

Nope. No way Jose.

Tim,

You are mixing "amps" and "watts".

240 volts at 200 amps = 48,000 watts
120 volts at 400 amps = 48,000 watts (but try to force that 400 amps through that same 200 amp breaker - ain't gonna work, not even considering that the conductors are going to overheat and probably melt down too)

[Tim Voss]

That's why I said "equivalent". You could put 200A of 120v loads on one phase and 200A of 120v loads on the other phase and the breaker is fine. Thus, you are driving 400A of 120v loads.

[Jerry Peck]

That's why I said "equivalent". You could put 200A of 120v loads on one phase and 200A of 120v loads on the other phase and the breaker is fine. Thus, you are driving 400A of 120v loads.

"you are driving 400A of 120v loads"

That's what I keep trying to explain ... you will not get 400 amps through a 200 amp breaker as long as the breaker is working correctly.

Which is why I also said " You are mixing "amps" and "watts". "

You should be using the term "watts", not "amps".

[Tim Voss]

I'm not confused about watts and amps. I'm using the fact that 200A at 240V is equal to 400A at 120V. Like you said in your post:

240 volts at 200 amps = 48,000 watts
120 volts at 400 amps = 48,000 watts

I can put four 100A single pole (120V) breakers in that 200A panel and run four 100A motors at 120V. That is 400A (at 120V). The main breaker will only have 200A going through it (at 240V) and it will be fine. If you put a meter on each motor, you would measure 100A, times 4 = 400A.

[Jerry Peck]

I'm not confused about watts and amps. I'm using the fact that 200A at 240V is equal to 400A at 120V. Like you said in your post:

240 volts at 200 amps = 48,000 watts
120 volts at 400 amps = 48,000 watts

I can put four 100A single pole (120V) breakers in that 200A panel and run four 100A motors at 120V. That is 400A (at 120V). The main breaker will only have 200A going through it (at 240V) and it will be fine. If you put a meter on each motor, you would measure 100A, times 4 = 400A.

"I can put four 100A single pole (120V) breakers in that 200A panel and run four 100A motors at 120V. That is 400A (at 120V). "

Nope, ain't gonna work.

You ARE mixing the terms "amps" and "watts" ... either that or you have a complete lack of understanding of electricity and wiring - I don't know which for sure, but I'm beginning to think it is the latter.

If you put "four 100A single pole (120V) breakers in that 200A panel " you will only be able to run *TWO* (2) 100 amp 240 volt motors (using those 4 breakers as double pole breakers), *OR* you will only be able to run *TWO* (2) 100 amp 120 volt motors (using those breakers as single pole breakers) - either case - that is only *TWO* (2) 100 amp motors.

Try to run 4 100 amp motors and they will either smoke and burn up, trip those "four 100A single pole (120V) breakers", or trip the 200 amp main.

You ARE NOT going to be able to run 400 amps through a 200 amp main if the main is working properly.

[Peter Plein]

Sorry Jerry, Gotta call you on this one. The 200 amp 240 volt service will indeed run 4 100 amp 120 volt loads. 400 amps at 120 volts equals 200 amps at 240 volts (assuming the loads are evenly distributed). Motor loads are by nature inductive and not a good example to cite. If we were to use a resistive load such as incandescent lighting or single phase electric heaters. You can indeed load this panel with 20 single pole breakers evenly distributed feeding 20 cicuits at 120 volts. Of course some of the items mentioned in the original post are not 120 volt loads. We are not given enough information to do a proper load calculation. If the heating system is NOT electric my seat of the pants opinion is that 200 amps at 240 volts is sufficient. The fact that the main has never tripped is not really important. If the loads ever exceed 80% of that service it is not big enough and yet never trip the main even if that 40 year old breaker is still functioning correctly.

[Jerry Peck]

Sorry Jerry, Gotta call you on this one. The 200 amp 240 volt service will indeed run 4 100 amp 120 volt loads. 400 amps at 120 volts equals 200 amps at 240 volts


"400 amps at 120 volts equals 200 amps at 240 volts "

Yeah (wattage wise), *WHEN RUN AT 240 VOLTS*, but when those 4 100 amp motors ARE RUN AT 120 volts (which is what Tim keeps saying) - *that's 400 amps*.

Like I keep saying ... ain't NO WAY you are going to pull 400 amps through a 200 amp main, or 200 amp rated equipment for that matter, not without damaging something in it.

Tim keeps saying he can - not going to do it.

Maybe Tim is not the only one who knows less about electrical wiring and electricity than they think?

Read what Tim keeps writing - he keeps saying 100 + 100 + 100 + 100 = 400 and he can run that through a 200 amp main ------- NOT GOING TO HAPPEN.

[Tim Voss]

Peter, I agree that motors was not a good example. I was trying to make it simple and not get into 80% load, power factor and inductive loads and such.

[Jerry Peck]

Peter, I agree that motors was not a good example. I was trying to make it simple and not get into 80% load, power factor and inductive loads and such.


Makes no difference ... motors ... big honking resistors ... you are NOT going to pull 400 amps through a 200 amp service - not without burning something up, and not without tripping the 200 amp (provided, I keep saying, that it is a good functioning main, one which will actually trip at or close to 200 amps).

[Tim Voss]

Jerry,

I think the difference is where you measure the amps. I said in my post it is only 200 amps (at 240v) going through the main breaker, but it is 100amps (at 120v) measured at each of the 4 single pole loads. If you add that up for 4 loads, that is indeed 400amps (at 120v).

I'll refer back to your diagram:

___________________ Phase A ->
& |
& 120 volts
& |
& -------------------- Neutral ->
& |
& 120 volts
& |
___________________ Phase B ->

Let's say it's just one giant motor pulling 200amps at 120v on each phase.
120v motor on Phase A pulling 200 amps, and back to Neutral.
120v motor on Phase B pulling 200 amps, and back to Neutral.

Neutral current cancels out, 400 amps of 120v motors being run. Nothing smoking, nothing tripping, only 200 amps through main breaker.

The key is, you can not run ONE 400amp (120v) motor. The current has to balance out, but you can run TWO 200amp (120v) motors for essentially 400amps of available 120v current.

[Bruce King]

Good discussion, you are both right, Jerry is just saying that 400 amps can not flow through the main breaker and Tim is saying you can get four 120V 100A loads to work if they are wired correctly while staying within the total wattage.


The link shows the advantages of having a neutral in order to drive 120V loads seperately.

The figure that applies to this discussion is a little over half way down the web page.


Single-phase power systems : POLYPHASE AC CIRCUITS (http://www.allaboutcircuits.com/vol_2/chpt_10/1.html#02168.png)

[Tim Voss]

Yes, that's exactly what I was saying. That's a great link, Thanks Bruce.

That's why in my first post I said "equivalent" of 400A of 120v power. I never said I was putting more than 200A through the main breaker at 240v.

[Richard Moore]

Good discussion...

Y'know, I don't think it is a good discussion. Not with comments like "You ARE mixing the terms "amps" and "watts" ... either that or you have a complete lack of understanding of electricity and wiring - I don't know which for sure, but I'm beginning to think it is the latter." and "Maybe Tim is not the only one who knows less about electrical wiring and electricity than they think?".

I can't see anything wrong with what Tim posted.

The main in a 200-amp 240-volt panel is actually 2 separate 200-amp breakers tied together, just like any two-pole breaker (ignoring quad mains). I don't have a diagram but, for simplicity, we'll assume that all breakers are single pole serving only 120-volt circuits. If you only turn on the breakers attached to one leg you can pull 200 amps (at 120v) through one half of that main breaker. Turning on breakers on the other leg and drawing current up to an additional 200 amps (at 120v) will not trip the main breaker. It's no different from being able to use a 20-amp double pole breaker to power two 20-amp 120v circuits or a single 20-amp 240v circuit.

All Tim was saying, correctly, was that you have 200-amps at 240v available OR a total of 400-amps at 120v available. Both would produce the same wattage (48kW) and spin the meter at the same rate as that senses the cumalative current on both legs.

[Jerry Peck]

Boy ... do I feel foolish, silly, and stoopid.


Jerry is just saying that 400 amps can not flow through the main breaker

Bruce,

That is *EXACTLY* what I was saying ...
(I read it as Tim was adding up the 100 amp motors, as would be the case when they were all on the same phase.)

However, that was *NOT EXACTLY* what Tim was saying.
(I went back and re-read it and Tim was saying, trying to tell me, that he was talking about using both phases equally.)


and Tim is saying you can get four 120V 100A loads to work if they are wired correctly while staying within the total wattage.

Talk about having to take 40 lashes with a wet noodle and go back to my corner ... :o

Bruce, thanks for splashing that bucket of ice cold water in my face and waking me up ... :eek:

[Jerry Peck]

All Tim was saying, correctly, was that you have 200-amps at 240v available OR a total of 400-amps at 120v available. Both would produce the same wattage (48kW) and spin the meter at the same rate as that senses the cumalative current on both legs.


Richard,

"All Tim was saying, correctly, was that you have 200-amps at 240v available OR a total of 400-amps at 120v available."

Don't start there again, that's what started all of this and that is the "wrong part".

You DO NOT have "a total of 400-amps at 120v available".

All you have available is 200 AMPS regardless of the voltage.

Going beyond that in the discussion is where I went wrong, on that part, stated the way you just stated it, I was correct.

Let's not go where I was correct again as it may lead to a turn of the discussion which I will miss again and be wrong again. :)

No matter how you look it it, regardless how high or low the voltage is, - you only have 200 amps available - we all agree on that, right?

[Bruce King]

When you have 120V loads only, the 240V 200 amp main breaker is now performing as two 120V 200 amp breakers.


Now each side of the main breaker has 200 amps going through it, so it is 400 amps total. There are no free amps or watts available, if you have four 100 amp 120V loads you can bet the power company is sending you 400 amps using two 120V legs. The difference is that it is only 120V loads. The rating of the breaker as 200A is just a naming standard, it really is a 48kw breaker.

The breaker only sees wattage and converts it to heat in order to activate the trip mechanism. Since the wattage is the same as 200a/240V the breaker does not trip with 400 amps/120V.

A 200 amp breaker can trip with very low current passing if a bad connection is present nearby to transmit enough heat into the breaker.

Added with edit: These large breakers use a magnetic current sensor and not heat to sense current.
Should have put "equivalent load amps" to the reference to 400 amps but I should have used wattage and described it differently.

[Jerry Peck]

When you have 120V loads only, the 240V 200 amp main breaker is now performing as two 120V 200 amp breakers.

Correct.


Now each side of the main breaker has 200 amps going through it, so it is 400 amps total.

You have 200 amps flowing in one breaker, through the circuit, which includes the loads, then out the other breaker - you still only have 200 amps total in the circuit.



There are no free amps or watts available, if you have four 100 amp 120V loads you can bet the power company is sending you 400 amps using two 120V legs.

You could bet that, but it is a bet you would lose.

The power company is only going to be able to squeeze 200 amps through that 200 amp breaker.



The difference is that it is only 120V loads.

Nope. As stated above in previous posts, if you connect those 120 volt loads between one phase and the neutral, you will trip the breaker trying to pull 400 amps through a 200 amp breaker, however, if you connect those 120 volt loads using both phases equally, now that 200 amps (still only 200 amps) is doing twice as much work as you effectively have 240 volt loads connected at the center point.


The rating of the breaker as 200A is just a naming standard, it really is a 48kw breaker.

The breaker only sees wattage and converts it to heat in order to activate the trip mechanism. Since the wattage is the same as 200a/240V the breaker does not trip with 400 amps/120V.

Huh?

The breaker does not care whether it is being used on a 120 volt circuit or a 240 volt circuit, it is only 'monitoring' the current flowing through it.

A properly functioning 200 amp breaker is going to trip at 200 amps (or close to it). You will not get 400 amps through a functioning 200 amp breaker - that is what I was saying above in my other posts and that is still correct.

Watts? It does not have any idea what a watt is nor does it care.

[Richard Moore]

"No matter how you look it it, regardless how high or low the voltage is, - you only have 200 amps available - we all agree on that, right?"

Jerry,
If I was simply describing the 240 volt service, then yes, but it's not "regardless" and in the context of Tim's posts it seemed very obvious, at least to me, that he was describing the total 120 volt amperage available if you added the available 200 amps on each leg. I really don't get how you can agree, finally, that it is possible to run a total of 400 amps of 120-volt load by using both legs and then say that that would not be available.

So no, I don't think "we all" agree, but at this point it really is just semantics and, as I think most understand the basic principles, it's not worth a continuing argument. The only reason I got involved in this in the first place was because I found the "personal" attacks on Mr. Voss's knowledge and/or intelligence unwarranted and distasteful. Sadly, I see you have your wet noodle out again, but still haven't apologized to him for those remarks.

[Jerry Peck]

in the context of Tim's posts it seemed very obvious, at least to me, that he was describing the total 120 volt amperage available if you added the available 200 amps on each leg. I really don't get how you can agree, finally, that it is possible to run a total of 400 amps of 120-volt load by using both legs and then say that that would not be available.

No ... no ... no ... you are only getting *200 amps*.

[Bruce King]

Jerry, How many maximum watts do you think you can get from the service with only 120V balanced loads on both legs?




How many RV'ers out there have 50 amp 240V/120V on their rig?

Same thing applies, you can get 100 amps of 120V balanced distribution.

The only RV equipment that uses 240V is old dryers or ones that have residential equipment installed. The manuf's use 120V RV equipment such as 120V dryers.

Added with edit: should read "equivalant of 100 load amps"

[Jerry Peck]

Jerry, How many maximum watts do you think you can get from the service with only 120V balanced loads on both legs?

Bruce,

That is exactly what I pointed out above - the misuse of the terminology.

We are talking amps and now you say "watts".


How many RV'ers out there have 50 amp 240V/120V on their rig?

None.


Same thing applies, you can get 100 amps of 120V balanced distribution.

Now you are back to amps again.

Make up your mind and use those terms in the correct way.


The only RV equipment that uses 240V is old dryers or ones that have residential equipment installed. The manuf's use 120V RV equipment such as 120V dryers.

Huh?

RV power is connected such that there is no 240 volts available - all loads, and wiring, are 120 volts only.

Think of it as a multi-wire circuit.

See my drawing to help explain where there is not 400 amps.

[Bruce King]

Jerry, I'm starting to worry about you : )


Watts can be used in place of amps with simple math and is actually preferred in the industrial world so that no one gets confused on which voltage is being talked about, as is the case here.
Added with edit: since we KNOW what voltages we are discussing here!


Your picture is nice, just add the loads present on each leg and you have your "400 amps", (410 amps as shown on the picture)



50 amp RV's are fed with the exact same hookup as a 4-wire 240V range.

The RV does not distribute the 240V but it has it available in the control box if you wanted to add a 240V outlet. The RV uses the 240V 50a circuit as two 50 amp 120V busses that distribute to various loads in the RV. 240V is present inside the RV in the control box, stick a meter across the two 50 amp busses and wham it will read 240V, but you know that, you just want to argue.


Heres a few links on the easy 50 amp 240V RV connection for anyone who might not know. The one that electricians screw up is the big 30A 120V RV outlet, they wire 240V to those all the time and smoke RV's.

http://www.myrv.us/Imgs/PDF/50-amp%20Service.pdf

Well, what is 50 amp service? (http://www.bobhatch.com/electricStuff/whats_it_mean.htm)



I know you had a 50 amp RV and fully understand that it connects to 240V shore power so what is your problem? I see you changed the subject to only the inside distribution where I posted about the feed AND the interior distribution.

Post us some code and quit picking fights on here about theory, hahahaha.

[Jerry Peck]

Watts can be used in place of amps

No, power ("watts") equals amps (current) times resistance (load), "watts" and "amps" are two different animals. "Watts" is the 'power used', "amps" is the current flow through the load using that 'power'.


with simple math and is actually preferred in the industrial world so that no one gets confused on which voltage is being talked about, as is the case here.

Huh?

Starting talking "watts" and it has no relationship to voltage other than based on the fact that voltage equals power ("watts") divided by amps (current).


Your picture is nice, just add the loads present on each leg and you have your "400 amps", (410 amps as shown on the picture)

Okay, now that is what I was waiting for ... :)

There is only "200 amps" there. See drawing attached.


The RV does not distribute the 240V but it has it available in the control box if you wanted to add a 240V outlet. The RV uses the 240V 50a circuit as two 50 amp 120V busses that distribute to various loads in the RV. 240V is present inside the RV in the control box, stick a meter across the two 50 amp busses and wham it will read 240V, but you know that, you just want to argue.

Er ... Bruce ... didn't you see where I stated "Think of it as a multi-wire circuit."?

What is a multi-wire circuit?

Right, you just described it.

What I stated was "RV power is connected such that there is no 240 volts available - all loads, and wiring, are 120 volts only."

RVs ARE connected "such that there is no 240 volts available" and "all loads, and wiring, are 120 volts only".

RVs ARE connected that way. The exception (every rule has an exception - that's what 'they' say, it is almost always true) would be 'some' ultra high end RVs ... 'some' ultra high end RVs are wired to used the 240 volts of the supply.

RVs are supplied by three basic wiring configurations:
- 20 amp / 120 volt
- 30 amp / 120 volt
- 50 amp / 120 volt multi-wire (giving two 50 amp 120 volt circuits)
- - (with those rare exceptions using the 50 amp / 120 volt multi-wire as the supply for a 50 amp / 240 volt service connection)

Those exceptions are possible because EVERY multi-wire circuit has 120 volt circuits with 240 volts between them - the 240 volts is just not used, in multi-wire circuits, using the 240 is not allowed.

Heres a few links on the easy 50 amp 240V RV connection for anyone who might not know. The one that electricians screw up is the big 30A 120V RV outlet, they wire 240V to those all the time and smoke RV's.


Well, what is 50 amp service? (http://www.bobhatch.com/electricStuff/whats_it_mean.htm)

Bob (or maybe it was Mary) is 'mostly right' in their comments.

This is where they are wrong:
"I have heard that there is an RV park some place in Arizona that the park owner has tied 1 leg of power using a 4/0 wire for the hot to both busses in a RV receptacle panel. He has then put in a double pole 50 amp breaker. By doing this he is really delivering 50 amps at 120 volts to the customer."

If the circuit were wired that way, it would be like using parallel conductors, in which case you could still pull 50 amps through EACH leg because EACH leg as a 50 amp breaker.


Post us some code and quit picking fights on here about theory, hahahaha.

- 210.4 Multiwire Branch Circuits.
- - (A) General. Branch circuits recognized by this article shall be permitted as multiwire circuits. A multiwire circuit shall be permitted to be considered as multiple circuits. All conductors of a multiwire branch circuit shall originate from the same panelboard or similar distribution equipment.
- - (C) Line-to-Neutral Loads. Multiwire branch circuits shall supply only line-to-neutral loads.
- - - Exception No. 1: A multiwire branch circuit that supplies only one utilization equipment.
- - - Exception No. 2: Where all ungrounded conductors of the multiwire branch circuit are opened simultaneously by the branch-circuit overcurrent device.

[Bruce King]

Jerry, you should be kicked off this board for taking part of my sentence about amps and watts and posting it as a quote to make it look wrong. This is pure BS and you know it.


You provide absolutely no help on this board that can not be easily supplied by many other inspectors. You actually hinder the message board with this type of crap because many inspectors visit this site and do not participate because of you.

You really screwed up this time ole boy. hahahaha : )

No hard feelings, I worked with several guy's just like you for 20 years, I have to admit I liked to jerk their chains too. : )

[Jerry Peck]

Jerry, you should be kicked off this board for taking part of my sentence about amps and watts and posting it as a quote to make it look wrong. This is pure BS and you know it.


*IT* *IS* *WRONG*, no BS about it.

No math will make one interchangeable with the other.

They are two different and distinct terms with two different and distinct meanings.

"and you know it", at least you should know it.

[Bruce King]

Jerry, when you know other variables, as we both do here (voltage), current and wattage *CAN* be easily interchanged wih math, watts = volts x amps.


Four loads connected as balanced 120V with 100 amps each can be seen at the main, 200A 48,000 watt breaker as any of these:

one 48000 watt feed
one 48kw feed
one 48000va feed
one 48kva feed
one 240V feed at 200 amps
two 120V feeds at 200 amps each, two x 200= 400 amps total derived from the main. 400 x 120 = 48,000va or 48,000 watts

Stop the BS Jerry, now you have the math.

But, here we go again, YOU already knew all of this!
Your goal is to get the last post, so go ahead and post some more BS! hahahahahaha!

added with edit: should have used "equivalent" to 400 load amps to prevent misunderstanding.

[Jerry Peck]

Jerry, when you know other variables, as we both do here (voltage), current and wattage *CAN* be easily interchanged wih math, watts = volts x amps.

No, you still cannot interchange watts and amps or volts.

Not unless you are going to say something like 48,000 watts (240 volts times 200 amps) and interchange it with 200 amps (48,000 watts / 240 volts) - simply doing what you did does nothing and tells nothing. Now, if you want to start writing all that out when you start trying to interchange the two DIFFERENT terms, by all means, do so, but you didn't, so don't try to crap (BS) your way out of what you said.


Four loads connected as balanced 120V with 100 amps each can be seen at the main, 200A 48,000 watt breaker as any of these:

one 48000 watt feed
one 48kw feed
one 48000va feed
one 48kva feed
one 240V feed at 200 amps
Those would be correct, EXCEPT - it is not a 48,000 watt breaker. It is a 200 amp breaker, it knows nothing about watts.

two 120V feeds at 200 amps each, two x 200= 400 amps total derived from the main. 400 x 120 = 48,000va or 48,000 watts
That one is not correct. No where in there do you have 400 amps to work with.

You could say it this way:
two 120 volt 100 amp loads connected together with the center point grounded across 240 volt making one 240 volt 100 amp load, which equals 24,000 VA (watts so-to-speak), there are two of these circuits for 48,000 VA total.

No where in there is there 400 amps.

You keep coming up with some fictitious 400 amps which just are not there.


Stop the BS Jerry, now you have the math.

Bruce, YOU stop the crap (BS), YOU know there is no 400 amps in there anyplace. YOU do the math - I made it REAL EASY for you in the above.


But, here we go again, YOU already knew all of this!
Your goal is to get the last post, so go ahead and post some more BS! hahahahahaha!

No MY GOAL is to get YOU TO UNDERSTAND IT, that no where in there is there 400 amps.

Whether I post last trying to explain it to you, or whether you post last saying 'I finally got it.' makes no difference to me. Or even if someone else posts last trying to explain it to you. I'm trying to help you understand what you apparently do not understand - not that I know everything about this, as was evidenced further up the thread- but maybe you cannot bring yourself to say that?

[paul hardy]

Jerry lets look at it this way 2002 NEC table 310.16 allows 4/0 aluminum XHHW (typical dwelling unit service wire) 180 amps ( 310.15 b allows it to be used for 200 amp service or feeder) so each phase has a allowable ampacity of 200 amps.This being said in theory phase A could have 200 amps of 120 volt loads only and a 4/0 service would not be overloaded also phase B could have 200 amps of 120 volt loads only and a 4/0 service would not be overloaded. That being said you would have 400 amps of 120 volt loads only allowable, phase A at 200 amps and phase B at 200 amps and if they were exactly balanced 0 load on the grounded conductor.

[Jerry Peck]

Jerry lets look at it this way 2002 NEC table 310.16 allows 4/0 aluminum XHHW (typical dwelling unit service wire) 180 amps ( 310.15 b allows it to be used for 200 amp service or feeder) so each phase has a allowable ampacity of 200 amps.This being said in theory phase A could have 200 amps of 120 volt loads only and a 4/0 service would not be overloaded also phase B could have 200 amps of 120 volt loads only and a 4/0 service would not be overloaded. That being said you would have 400 amps of 120 volt loads only allowable, phase A at 200 amps and phase B at 200 amps and if they were exactly balanced 0 load on the grounded conductor.


Paul,

"That being said you would have 400 amps of 120 volt loads only allowable, phase A at 200 amps and phase B at 200 amps and if they were exactly balanced 0 load on the grounded conductor."

That's the part that you and Bruce are not getting ...

The current going in through the Phase A conductor is the same as that in the Phase B conductor, not additive, not two 200 amp current, *the same 200 amp current* ... not "400 amps".

Okay, let's try this.

Take a swimming pool filled with water, install a submersible pump which can pump 200 gallons per minute. This is the 'power source'.

Take a hose suitable in size to allow that 200 gallons per minute to flow through it and connect one end to the pump and the other end to a tee to a load which uses 100 gallons per minute to rotate a motor, then, from that tee, run another hose to another load which also uses 100 gallon per minute to rotate a motor. Got it so far?

Okay, now, connect the first 100 gpm motor to another 100 gpm motor, then connect the second 100 gallon per minute motor to another 100 gpm motor. Still following me?

Finally, connect another hose from the farthest 100 gpm motor, add a tee, connect in the nearest 100 gpm motor to the tee, and then connect the tee to a hose draining back into the swimming pool.

Got it?

This is what we have:

Hose out carrying 200 gpm, teeing off 100 gpm with 100 gpm continuing on.

The first teeing off of 100 gpm goes through two 100 gpm motors back to the return line.

The other 100 gpm goes through the other two 100 gpm motors back to the return line.

How many gpm are going through the supply and return line?

400 gpm?

Or is it just 200 gpm?

It is just 200 gpm.

Same set up with electricity and current flow. You have 100 amps, the same 100 amps, going through two 100 amps loads, with two such paths of 100 amps each, for a total of 200 amps.

NOT "400 amps".

Got it?

[Bruce King]

Jerry,

I forgot one thing, at the beginning of this discussion someone mentioned "equivalent to 400 amps".

I should have included that word "equivalent" in my posts but thought we were all discussing practical things and not the actual waveform behavior of the service.

Technically there are 200 amps flowing within the 240V source but we were trying to use the system as an example to show that you can use it as only a 120V system and get FOUR 100 amp loads to operate without tripping the 200 amp breaker.


I hope you do agree that the power cost to run four hundred amps of 120V loads is the same as 200 amps 240V because it really is.


Anyone that did not learn something from this thread is not being honest with themselves. I know I have learned how to also view the weird setup we were discussing as a 240V system even though we were using only 120V loads in the example.

[Bruce King]

Here's a quote from Paul Dickerson (a NACHI member) that explains it very well.

Lets follow the electrons. For 1/120th of a second, 200 amps will flow into the panel through Leg A, the same 200 amps will then flow through the 120V loads connected to leg A, the same 200 amps will then flow to the neutral bus, the same 200 amps will then flow through the 120V loads connected to leg B, then the same 200 amps will flow out of the panel through leg B. Then the current reverses and takes the same path backwards. It is all the same electrons.

[Brandon Chew]

I'm not going to pick sides; I'll just try to explain what I believe to be true, in very simple terms.

power = watts = volt amperes = volts X amps

Absent a fault in the system, and neglecting very small losses due to heat, power leaving the transformer is equal to the power returning back to the transformer.

In a 120/240V three-wire "single phase" residential wiring system, the 120 V coming in on the A leg from the transformer, and the 120 V coming in on the B leg from the transformer are 180 degrees out of phase with each other.

With your 200 amp main breaker installed, you can pull 120 V at 200 amps (24 kw) from the transformer on the A leg. At the same time, you can pull 120 V at 200 amps (24 kw) from the transformer on the B leg, for a total of 48 kw. 240 V branch circuits are formed by tapping into both the A & B bus bars in the panel, after the main OCPD.

If you pull 48 kw off of the transformer you must send 48 kw back to the transformer. Some goes back out of phase to what's coming in on the A, some goes back out of phase to what's coming in on the B, and if there is any unbalanced current, it goes back on the shared neutral. If the loads on A & B are perfectly balanced, there should be no current flowing on the shared neutral back to the transformer.

It should be possible to hook up four 120V loads each drawing 100 amps of current from the transformer at the same time without blowing the 200 amp main if they are balanced on the A & B. The power draw from each of those loads is 12 kw, or 48 kw total.

You can't equate amps to power without also specifying the voltage at which the amps are being measured. In this example you are in fact pulling a total of 400 amps off the transformer at 120 volts (48 kw). If you throw clamp meters on the service conductors you should measure 200 amps at 120 volts on the A, 200 amps at 120 volts on the B, and none on the neutral.

Unbalance that 48 kw load and you'll blow the main on whichever leg is trying to pull more than 200 amps at 120 volts (24 kw). The main OCPD will not let you pull more than 200 amps at 120 volts through either the A or the B legs.

If the system is wired and working properly to stabilize the voltage, you shouldn't measure more than 200 amps at 120 V (referenced to ground) on any individual conductor that runs from the transformer to the main OCPD as long as the duration is less than the trip point of the OCPD. This should be true whether the load is balanced or unbalanced.

[paul hardy]

Jerry what you show with your pump is from phase A to the grounded conductor the power supply (4/0) in my example is allowed 200 amps there are two power supply's phase A and phase B the main is only passing 200 amps per phase but there are two phases of 200 amps each. If I were to connect phase A with two 100 amp loads 120 volt phase A would carry 200 amps and the grounded conductor would carry 200 amps no overload now add phase B with two 100 amp loads 120 volt phase B would carry 200 amps and in theory the grounded conductor would balance at 0 amps.

[Jerry Peck]

Jerry,

I forgot one thing, at the beginning of this discussion someone mentioned "equivalent to 400 amps".

I should have included that word "equivalent" in my posts but thought we were all discussing practical things and not the actual waveform behavior of the service.

It is still NOT "equivalent to 400 amps". It would be, as Brandon explained, hopefully better than I have been, "equivalent" to 48,000 watts ... but that gets into interchanging the terms "watts" and "amps" again, which I also tried to explain, and Brandon also explained.


Technically there are 200 amps flowing within the 240V source but we were trying to use the system as an example to show that you can use it as only a 120V system and get FOUR 100 amp loads to operate without tripping the 200 amp breaker.

Not only "technically" speaking but "practically" speaking also. Which is what I was doing.


I hope you do agree that the power cost to run four hundred amps of 120V loads is the same as 200 amps 240V because it really is.

Agreed.


Anyone that did not learn something from this thread is not being honest with themselves. I know I have learned how to also view the weird setup we were discussing as a 240V system even though we were using only 120V loads in the example.

Agreed.

[Jerry Peck]

Bruce,

Paul almost has it right.

This:


Lets follow the electrons. For 1/120th of a second, 200 amps will flow into the panel through Leg A, the same 200 amps will then flow through the 120V loads connected to leg A, the same 200 amps will then flow to the neutral bus, the same 200 amps will then flow through the 120V loads connected to leg B, then the same 200 amps will flow out of the panel through leg B. Then the current reverses and takes the same path backwards. It is all the same electrons.

Should read as follow:

Lets follow the electrons. For 1/120th of a second,
1) 200 amps will flow into the panel through Leg A,
2) the same 200 amps will then flow through the 120V loads connected to leg A,
3) the same 200 amps will then flow to the second set of 120 volt loads connected to the first set of 120 volt loads,
4) the same 200 amps will then flow to leg B,
5) then the same 200 amps will flow out of the panel through leg B.
6) Then the current reverses and takes the same path backward. It is all the same electrons.

As set up for our discussion - *no* current will flow on the neutral bus or neutral conductor.

Now, in the photo I posted with the extra 10 amp load, another 10 amps will be flowing through the neutral conductor and the top phase leg, which we can call Leg A. Which makes 210 amps, which will likely trip the breaker (which is why I put it there, to help show that "400 amps" would trip the breaker as even "210" will trip the breaker - but that "400 amps" discussion now seems to be history, as well it should be).

[Jerry Peck]

Jerry what you show with your pump is from phase A to the grounded conductor the power supply (4/0) in my example is allowed 200 amps there are two power supply's phase A and phase B the main is only passing 200 amps per phase but there are two phases of 200 amps each.

First, no neutral current will flow in the neutral when there are those four 100 amps loads being discussed.

Second, you do not have "two phases of 200 amps each", you have 'two phases with the same 200 amps in each' ... therein lies the big difference.

Thinking there are *two* sources of 200 amps leads one to think there is 400 amps. Understanding that there is only one source, with two legs, which have *the same* 200 amps in them, hopefully allows one to understand that there is only 200 amps.

Sometimes electricity is difficult to understand, that is why I switched to using the water example - virtually everyone understands water flow, not everyone gets electrical current flow.

My water example would be representative of DC, for AC, simply switch the supply and return hoses 120 times per second (I figured it would be easier to understand without complicating matters with 'switching the supply and return hoses 120 times per second' :) ).

[Tim Voss]

This is getting a little silly, but I'm going to stick by 100+100+100+100 does indeed = 400

Physics doesn't lie, so if Power = Volts x Amps and I have 240v x 200a then I also have 120v x 400a

I can wire up a 99% efficient AC to AC voltage converter (transformer) that will transform my 240v x 200a service into 120v x 396a to run those 2 giant 200amp 120v loads I have. I really do have the equivalent of 400amps. But I don't need to, because the utility company has already given me a 120v neutral tap.

It's true you will never see more than 200 amps on any given wire (unless you use the transformer mentioned above), but the amps ARE additive, even though they are the same amps. That is because the same amps are going through two 200a 120v loads, so they are doing twice the work as one 120v load. Thus, they are "equivalent" to 400amps at 120v.

[Carl Eisen]

Another way to look at it is as a wave on a oscillascope. You have two waves of the same amplitude but peaking at opposite times of each other thus cancelling each other out. At no point can Phase A and Phase B become an additive value.

[James Duffin]

The way I see it is that the power company charges by the KW. So if you have a 200 amp 240 load your cost will be for 48 KW.If you have two 200 amp 120 volt loads you are still paying for 48 KW. Your cost for each load will be the same and both loads can be fed from the same 120/240 volt single phase pane with a 200 amp main breaker.

[Jerry Peck]

This is getting a little silly, but I'm going to stick by 100+100+100+100 does indeed = 400

Yes, "100+100+100+100 does indeed = 400", HOWEVER ... you do not have "100+100+100+100", you only have "100+100" and that does indeed = 200


Physics doesn't lie, so if Power = Volts x Amps and I have 240v x 200a then I also have 120v x 400a

Nope.

You do have 240 volts X 200 amps. but you do not have 120 volts X 400 amps.

You have 120 volts X 100 amps four times, but you do not have 400 X anything.

(sigh)

[Jerry Peck]

Another way to look at it is as a wave on a oscillascope. You have two waves of the same amplitude but peaking at opposite times of each other thus cancelling each other out. At no point can Phase A and Phase B become an additive value.

Actually, what you have is *one* waveform, simultaneously in Phase A and Phase B.

The current flow in Phase A and out Phase B for 1 120/sec, then flows in Phase B and out Phase A for the next 1 120/sec, then it repeats that process for the next 2 120/sec. At this point in time, 1/60 of a sec later, you have completed one cycle, one waveform which rises from 0 to +170 then drops to 0 to -170 then rises back to 0.

Think of the water hoses being switched 60 times a second, what you are describing would be both water hoses being used on at the same time, creating no potential difference and thus no flow (voltage is potential, which is what creates current flow).

[Jerry Peck]

The way I see it is that the power company charges by the KW. So if you have a 200 amp 240 load your cost will be for 48 KW.If you have two 200 amp 120 volt loads you are still paying for 48 KW. Your cost for each load will be the same and both loads can be fed from the same 120/240 volt single phase pane with a 200 amp main breaker.

Correct.

[Tim Voss]

Now I'm not sure what you are saying. You say I have 120v x 100a x 4. Yes, I agree. But you also say I have "100+100=200", don't agree.

What about this? Can I light 400 1amp 120volt light bulbs with my 200amp 240volt service?

[Jerry Peck]

Now I'm not sure what you are saying. You say I have 120v x 100a x 4. Yes, I agree. But you also say I have "100+100=200", don't agree.

Tim,

This is the difference in how you are saying it.

You are saying 100+100+100+100=400 (if they were added up that way, you would be correct, but they do not add up that way).

What you really have is 100+(that same 100 goes through the next load)+100+(that same 100 goes through the next load)=200


What about this? Can I light 400 1amp 120volt light bulbs with my 200amp 240volt service?

ONLY IF there are 200 between Phase A and Neutral and 200 Between Phase B and Neutral.

That would make 1+(that same 1 flows through the next lamp)+the same thing 199 more times=200

______________ Phase A <- 1 amp current flow
|
1 amp Lamp
|
______________ Neutral -0- amp current flow
|
1 amp lamp
|
______________ Phase B -> 1 amp current flow

Add as many pairs of lamps as you wish, there will be an additional 1 amp current flow per pair of lamps wired that way. 200 pairs of those lamps would = 200 amps.

Add one single lamp between Phase A and Neutral and there will be an additional 1 amp current flow, this will be the same for each single lamp added between a given Phase A and Neutral.

Now add a single lamp between Phase B and Neutral and that will 'pair up' with one of the single lamps on Phase A and Neutral not adding any current flow - all it will do is take 1 amp from the Neutral and run it through Phase B - same 1 amp though.

[Tim Voss]

All I'm saying is that if I can light 400 1amp lights, I have the equivalent of 400amps of 120v power.

If you read this link provided by Bruce, Single-phase power systems : POLYPHASE AC CIRCUITS (http://www.allaboutcircuits.com/vol_2/chpt_10/1.html#02168.png) it explains quite nicely how putting 2 120v loads in series on a 240v circuit works.

I know, I know, I know I only have 200amps on the 240v main breaker wire, but if I meter the current going out to each lamp, it is 1amp. 1amp times 400 = 400amps of equivalent power at 120v. You agree I'm lighting 400amps of 120v lights? I don't see how you can say I don't have the equivalent of 400amp of 120v power? 240x200 = 120x400

The main point I was trying to make when I said you have the equivalent of 400amps of 120v power, is that when you do a load calculation for all 120v loads on a residential 240v 200amp service, you add the amps up to 400, not 200 to make sure you are within load limits (of course there is 80% load factor to consider as well). Like the RV service mentioned above, for all 120v loads on a 240v service, you can calculate double the amps of loads you would calculate for 240v loads.

If this is still not clear, then what about the fact that I can put an AC-AC voltage converter on my 240v 200a panel and pump out 120v at 400a (minus negligable conversion losses) without blowing the main breaker? Obviously the equivalent of 400amps at 120v is there.

[Bruce King]

Tim,

Give it up, Jerry does not deal in "equivalent" or any discussion that is not black and white. He looks for the smallest error or misphrased post so he can correct you, its his hobby, and he is pretty darn good at it.

I do appreciate his input on here though as I do everyones, its just sad that 98 percent of inspectors do not participate on message boards.

[Jerry Peck]

All I'm saying is that if I can light 400 1amp lights, I have the equivalent of 400amps of 120v power.

That's not all you've been saying.

You've been saying you could do that through a 200 amp main.

You can do what you just said through a 400 amp main, yes, but not through a 200 amp main. All you do is connect each lamp phase leg to neutral - no problem.

The problem comes in when you try to do that through a 200 amp main, and, then, *again* ... no, you are incorrect - you do not have 400 amps ... "equivalent" or not, notwithstanding.

If you connect them as has been described, you will only have 200 amps "equivalent" or "real".

Now, if you were to do as James did and talk in power only (watts, kw, etc.), then you are ignoring 'voltage and amps' and talking only about 'power', in which case you would have the same 'power' usage ... just not the same "amps", which is what you keep bringing up.

Regardless how many times you try to come up with 400 amps on that circuit with a 200 amp main, the answer will always be the same - "No, that is incorrect."

[Jerry Peck]

Give it up, Jerry does not deal in "equivalent" or any discussion that is not black and white. He looks for the smallest error or misphrased post so he can correct you, its his hobby, and he is pretty darn good at it.

Bruce,

Incorrect assumptions again (you do seem to make a lot of incorrect assumptions about why other people do things, is that because that is why you do them?

I don't look "the smallest error or misphrased post so he can correct you", if you make an incorrect post, it is likely because you are misunderstanding what is being discussed, doing so here present little threat of you being 'taken to the cleaners' by your clients or 'the trades' ... doing so 'out there' can lead to you being 'wrong' and not being able to back up what you said, or you having to say 'what I meant to say was ... ', which does not make you look good either.

Better to learn here (or wherever) what is correct and what is not, than to learn in front of your client.

It is however, ultimately your choice.

Bruce,

You were kind enough to, as I stated, "Bruce, thanks for splashing that bucket of ice cold water in my face and waking me up ... ", now, though, refill the bucket, add more ice, then splash over your face, then splash the rest on Tim's face, maybe it will help the two of you like it did me? :cool:

Let me know if that wakes you two up like it did me, thanks. :)

[Brandon Chew]

Actually, what you have is *one* waveform, simultaneously in Phase A and Phase B.

The current flow in Phase A and out Phase B for 1 120/sec, then flows in Phase B and out Phase A for the next 1 120/sec, then it repeats that process for the next 2 120/sec. At this point in time, 1/60 of a sec later, you have completed one cycle, one waveform which rises from 0 to +170 then drops to 0 to -170 then rises back to 0.

Think of the water hoses being switched 60 times a second, what you are describing would be both water hoses being used on at the same time, creating no potential difference and thus no flow (voltage is potential, which is what creates current flow).

I don't think that is correct and here is why.

[Note to reader: comments about transformer in the next paragraph were revised and additional discussion of "phase" as applied to voltage, current, and power in this system was added by me in post #86.]

To keep it simple and not go into three phase power, a typical transformer supplying a residential 3-wire single-phase 120/240V system has three windings inside it. On the primary side is one winding on a conductor from the power company's distribution system. On the secondary side are two windings, each one wrapped in opposite directions, that create two 120V wave forms (A & B) that are 180 degrees out of phase with respect to each other.

A major reason for generating the A & B wave forms 180 degrees out of phase of each other is so that you are able to create a 240V potential between the two legs and have a dual voltage system. Having done that, a big side-benefit is that you are able to have current flowing out on the A and coming back on the B at the same time that current is flowing out on the B and coming back on the A. This is what results in the neutral having zero current flow when the loads on the A & B are perfectly balanced. The maximum unbalanced condition would result in a maximum current of 200A @ 120V flowing on the neutral back to the transformer, which allows us to size the neutral conductor to carry 200A @ 120V. If the A & B waveforms were generated in phase with each other, the neutral would have to be sized to carry 400A @ 120V back to the transformer.

Refer to the diagram attached below. Maximum potential of each wave form with reference to zero (neutral) oscillates between +/- 170V (120V is an average value) as Jerry noted. The waves are 180 degrees out of phase to each other. This creates a maximum potential difference of +/- 340V (240V average) when you measure between the peak of one wave to the other. Both waves begin at the same time and they cross the zero line every 1/120 second.

[note to reader: there is incorrect info in the next two paragraphs about current flow. I have retracted these two paragraphs and have posted a correct description of current flow in post #86]

Jerry, the analogy between water flowing and electricity flowing is a good one for most things. It breaks down when you transition from discussing DC current to AC current. Thinking in terms of water makes it difficult for people to grasp the concept that electrons can be flowing along a conductor in two directions at the same time. It's the two windings in the transformer generating two 120V currents out-of-phase by 180 degrees that is the key to making this physically possible.

The water analogy (two hoses drawing water from the transformer at the same time) does work to illustrate what happens once the loads on the A & B become unbalanced (which is the common real-world situation in residential panels), or if the A & B are being generated in phase with each other. With A & B 180 degrees out-of-phase with each other, at any given point in time that current is being drawn through the A leg to the house, the capacity of the A leg to return the current that is being drawn from the B leg back to the transformer is limited to an amount that is relative to the current being drawn by the A leg. The excess draw from the B leg has to go back on the neutral. If the A & B are being generated in phase with each other, neither the A or B could carry any flow back to the transformer, and all of the flow from both of them would go back on the neutral at the same time. When A & B "hoses" were both flowing at their maximum rate of flow in phase with each other, the neutral "hose" would need to be twice the size of either one of them.

If there are some 120V loads attached to the A leg and other 120V loads attached to the B leg, or if you have 240V loads attached to both legs, then current is flowing from the transformer through both the A & B at the same time. Current flow does switch directions every 1/120 second, but it's switching directions in the A & B at the same time.

On three-wire single-phase 120/240V residential service that is sized for 200A:

the conductor on the A leg is sized to carry up to 200A @ 120V
the conductor on the B leg is sized to carry up to 200A @ 120V
the neutral conductor is sized to carry up to 200A
the main OCPD is sized for up to 200A @ 240VThe OCPD monitors the current flowing (amps) on each leg and will trip, killing the power to both legs, if either the A or the B pulls more than 200A @ 120V.

Through a "200A" residential service, you have the ability to pull up to 400 amps of power at 120 volts off the transformer if the 120V and 240V loads on A & B are perfectly balanced, but you cannot pull more than 200 amps (at 120V) through either the A or the B.

[Final note to reader - See post #90. It is best to describe the power being pulled from the transformer at maximum load as "48,000 watts by a 120/240V 3-wire 1-phase system", and the household system is sized and rated for "200 amps". Although you can pull 200A @ 120V through the A & B legs at the same time, you should not measure more than 200A anywhere in the system. From post #86, the current waveforms on the A & B legs are mirror images of each other. When currents of opposite signs (+/-) are added together they cancel each other out, which is why, on a balanced system the grounded conductor carries zero amps. The current you measure on the grounded conductor is showing you the practical result of adding the two 200A current flows on the A & B together.]

[Jerry Peck]

Brandon,

This is the way most single phase power transformers are wound ( Single Phase Transformers - Power Transformer Co (http://www.powertransformer.us/singlephasetransformers.htm) - see drawings and text).

The current is flowing through the secondary winding in one direction for half of the cycle, then reverses for the other half of the cycle. This creates one current path, in one direction, and one waveform for that current.

Taping off the center give 240 volts to both phase legs or 120 volts between either phase leg and the grounded center point.

When installing the loads as has been discussed, there is no grounded center point current flow (no neutral current), thus, all of the current is going in the same direction at the same time, thus one waveform for the current.

If you were to introduce two opposing waveforms, there would be no current flow. You need the electro-motive force provided by the difference in potential, and two opposing waveforms would cancel each other out, creating no potential difference which is needed to drive the current flow.

[Michael P. O'Handley]

Aargh!

Reading and trying to follow this is making my head begin to throb the way it did in high school algebra (For anyone that doesn't know it, I'm a math moron.).

Is there anyone in the business that you both (Peck and Chew) would recognize as being THE authority on this subject and who's opinion on this point you both would accept without challenging or endless debate?

This is important stuff and the danger is that anyone reading this thread can decide who they want to believe is correct and will then follow that persons guidance even if that person is 100&#37; wrong. Well, this isn't stuff that you can have a different opinion about; this is real honest-to-goodness quantifiable stuff that's subject to scientific absolutes, so one of you has to be wrong. That means anyone accepting the wrong person's interpretation could potentially get themselves, or their clients, killed. That especially goes for me, 'cuz despite 12 years in this business I'm still not "comfortable" around this subject. I don't like 50% odds of being wrong and screwing this up. I like odds that are well in favor of me being right, so I won't get myself 'lectricuted.

I think we need someone else to step in here who's THE guru. Once that person renders an opinion, I think both of you, for the sake of all of us out here who are still largely mystified by electricity, should accept that opinion and let it end there and not dispute anyone in the future that ventures the same opinion.

Someone has to be wrong here and someone has to be right, so I'm challenging you both to end this by choosing an impartial 3rd party to say who right. Who's it going to be? Mike Holt? Hansen? Katen? Simmons? Cramer? Cory Friedman? Joe Tedesco? Rex Cauldwell? Who? Comeon, comeon, name your poison, I'll copy the whole damned thread, shoot it to the one person who's opinion you both can accept and get it from the mountaintop so that we can get the definitive answer to this and end the confusion.

Comeon guys, step up.

ONE TEAM - ONE FIGHT!!!

Mike

[Jerry Peck]

Is there anyone in the business that you both (Peck and Chew) would recognize as being THE authority on this subject and who's opinion on this point you both would accept without challenging or endless debate?

Mike,

This isn't "endless", the end is almost in sight. :)



Mike Holt

There's your answer from this end. [ (added with edit) I agree with most of what Mike says, but there are a few things I disagree with him on. ]

But remember, we are almost at the end right now, so hold off until next week (only two more days this week anyway), if we are not in agreement by then, go for it. :cool:

[Brandon Chew]

Hi Mike,

From my point of view I don't see my two posts in a thread 72 posts long as me being locked in an endless debate. I've laid out what my current understanding of the subject is for all to view. I haven't tried to claim I was an expert or insist I'm right. In fact, I'll state flat out that my degree is not in electrical engineering -- but that doesn't mean I don't know anything about the subject.

I do want to get it right. So in those two posts I've climbed out on a very long limb knowing there's plenty of guys eager to start sawing on it. I've got my parachute on and when I hit the ground I hope to dust myself off, look back on where I was, and learn something that I didn't know before I started.

I couldn't name one person who's word I'd accept as gospel without first listening to what they said on the subject and how they said it. That's just the way I am, and why I enjoy the process of learning as much as I do. So, to answer your question, I'd be thrilled if you got all of those guys to weigh in!

I just hope some folks can refrain from the name-calling and personal attacks and we get some new info into the thread instead of recycling the same stuff over and over again.

Brandon

[Brandon Chew]

Brandon,

This is the way most single phase power transformers are wound ( Single Phase Transformers - Power Transformer Co (http://www.powertransformer.us/singlephasetransformers.htm) - see drawings and text).

The current is flowing through the secondary winding in one direction for half of the cycle, then reverses for the other half of the cycle. This creates one current path, in one direction, and one waveform for that current.

Taping off the center give 240 volts to both phase legs or 120 volts between either phase leg and the grounded center point.

When installing the loads as has been discussed, there is no grounded center point current flow (no neutral current), thus, all of the current is going in the same direction at the same time, thus one waveform for the current.

If you were to introduce two opposing waveforms, there would be no current flow. You need the electro-motive force provided by the difference in potential, and two opposing waveforms would cancel each other out, creating no potential difference which is needed to drive the current flow.

Jerry - Thanks. I need some time to think about this and digest the info. Right now my belly says it's time for dinner!

[Michael P. O'Handley]

Hi Brandon,

I know your's was only two posts but I know you're an engineer and I know Jerry is, well, Jerry. In this thread, you two seem to be the.....experts, shall we say.

Nuff said. There's still a need for a definitive answer 'cuz there are others in your corner on this thing. Electrical stuff is important. If we're going to state it on a forum, and it's about the science and not just about how to interpret a rule, folks need to get it right.

ONE TEAM - ONE FIGHT!!!

Mike

[Bruce King]

Here's the real deal, no need for an "expert" but it would be nice to hear from an AC power expert, not a code guy. Keep in mind that engineers are not always right either. I could write a book on that.



Jerry is technically right about the 200 amps being max due to the AC waveform, transformer, neutral, current waveform, series circuit etc. I remember most of this from school but can't explain it very well because that was 30 years ago.

But:

I thought the discussion was primarily about "can you drive four 100 amp loads at 120V?" and the answer is YES, we all agree on that.


Then we started talking about equivalent etc. for the sake of explaining things to those who do not know all of the AC theory in detail. Then Jerry jumped in, which is ok because some of us were trying to simplify it and say the main was passing 400 amps total.


I posted that the main sensed heat/watts as the current monitoring method, this is correct for many smaller breakers but those big ones use a magnetic trip mechanism that is activated by current, so Jerry was right on that.


These things argued previously are actually very correct:
(Jerry will still argue something but will be wrong on these)

120V x 400 amps = 48,000 watts
240V x 200 amps = 48,000 watts

Amps and watts can be easily converted using simple math when the voltage is known. We should have used watts and made this much easier possibly.

50 amp RV's ARE fed with 240/120V up to the interior control panel just like a range circuit although the rigs distribution is wired as two 120V 50 amp busses. You can get 50 amps off each bus (2) in the RV at 120V.

[Carl Eisen]

Actually, what you have is *one* waveform, simultaneously in Phase A and Phase B.

The current flow in Phase A and out Phase B for 1 120/sec, then flows in Phase B and out Phase A for the next 1 120/sec, then it repeats that process for the next 2 120/sec. At this point in time, 1/60 of a sec later, you have completed one cycle, one waveform which rises from 0 to +170 then drops to 0 to -170 then rises back to 0.

Think of the water hoses being switched 60 times a second, what you are describing would be both water hoses being used on at the same time, creating no potential difference and thus no flow (voltage is potential, which is what creates current flow).


I think I have my head just about straight on this. My error is looking at it as two opposing wave forms. Which at having the same amplitude would cancel each other out and thus no flow of current and no horsepower on the business end.

[Jerry Peck]

These things argued previously are actually very correct:
(Jerry will still argue something but will be wrong on these)

120V x 400 amps = 48,000 watts
240V x 200 amps = 48,000 watts

Amps and watts can be easily converted using simple math when the voltage is known. We should have used watts and made this much easier possibly.

50 amp RV's ARE fed with 240/120V up to the interior control panel just like a range circuit although the rigs distribution is wired as two 120V 50 amp busses. You can get 50 amps off each bus (2) in the RV at 120V.

Bruce,

I only argue a point when there is something which needs to be 'argued', i.e., corrected, debated, discussed, resolved to the right answer.

There is nothing to correct in your post above. Nothing which needs to be argued.

It's just that's not what was stated above in the other posts. :)

[Tim Voss]

ONLY IF there are 200 between Phase A and Neutral and 200 Between Phase B and Neutral.




I thought you just said I COULD put 400 1amp 120v lights on a 200amp main. 200 on Phase A and 200 on Phase B. Now you are saying I CAN'T do it?

I'm not quite ready to give up, so let's just start with something we both agree on. On a 200amp 240v service, can I light 400 1amp 120v lights (your basic 120watt regular light bulb)? 200 on each Phase, yes or no?

[Jerry Peck]

I thought you just said I COULD put 400 1amp 120v lights on a 200amp main. 200 on Phase A and 200 on Phase B. Now you are saying I CAN'T do it?

No, what I'm saying, and have been saying, is that you CAN ... BUT ONLY IT you balance them between Phase A and Neutral and between Phase B and Neutral.

Previously you said you could "put 400 1amp 120v lights" ... but did not add the "on a 200amp main" condition. Without that condition you can, but, as I said "You can do what you just said through a 400 amp main, yes, but not through a 200 amp main. All you do is connect each lamp phase leg to neutral - no problem."

AS SOON AS you add that condition (the 200 amp main), the answer changes, and the only way is to balance them out between the neutral and the phases.


I'm not quite ready to give up,

Good, I don't want you to give up, I am hoping you will understand it.

*MIKE* wants you to give up, or someone to give up. Mike does not like disorder, Mike likes to control things, which he offered to do above. Learning involves a lot of disorder in order to be able to work one's way through the thought process.


so let's just start with something we both agree on. On a 200amp 240v service, can I light 400 1amp 120v lights (your basic 120watt regular light bulb)? 200 on each Phase, yes or no?

Yes. As long as each lamp is between a phase leg and neutral, balancing them between Phase A and Phase B.

Yes.

[Tim Voss]

Ok, I agree. I can have 400 lights as long as they are perfectly balanced with 200 on phase A to neutral and 200 on phase B to neutral.

So now I'm trying to determine if I have too much load for a 240v service. Let's say I only have 120v loads - lights, appliances, refrigerator, etc... ALL 120v. Assuming all these 120v loads are perfectly balanced out between the two phases and neutral, can I add up the amps for all my 120v loads up to double the main breaker rating and still not be overloaded?

[Bob White]

Ok, I agree. I can have 400 lights as long as they are perfectly balanced with 200 on phase A to neutral and 200 on phase B to neutral.

So now I'm trying to determine if I have too much load for a 240v service. Let's say I only have 120v loads - lights, appliances, refrigerator, etc... ALL 120v. Assuming all these 120v loads are perfectly balanced out between the two phases and neutral, can I add up the amps for all my 120v loads up to double the main breaker rating and still not be overloaded?


Yeah ---

When can we start adding motor loads and switching electronic loads to the mix???

I'm tired of just using linear loads....;) :p

[Michael P. O'Handley]

*MIKE* wants you to give up, or someone to give up. Mike does not like disorder, Mike likes to control things, which he offered to do above. Learning involves a lot of disorder in order to be able to work one's way through the thought process. Yes. That's pretty funny. Anyone who really knows me would never say that because they realize I live in a constant state of disorder.

No, what I want, and I'm sure others reading this thread want, is to know which side of the argument is accurate. You wear people down Jerry, and if you do it to someone while arguing about electricity, how is anyone reading the thread to know that you've won the argument because you were right or because folks simply got tired and walked away?

It seems to me, that the folks arguing with you here are smart fellows, and that if your arguments were so strong, and accurate, that they wouldn't have stood their ground and continued to argue their points and would have quietly accepted it long ago. So, at the end, when you've completely worn them down, who do the rest of us believe? If we believe your explanation, and you are wrong, what value has all this produced? From where I'm standing, the question has never been answered in such a way that I'd feel comfortable with either side's argument.

It's like a group of students arguing about the solution to a problem where everyone is convinced he knows the right answer but anyone listening to them argue only becomes more unsure. At some point, if they're as smart as they want folks to think they are, they produce documented proof of their positions, or they consult their professor, who is the authority, to settle the argument. In the end, someone obviously has to be wrong but without a professor here to tell the rest of us that he agrees with one side or the other, how can the rest of us, who are not electricians - and the last time I looked you weren't either - have confidence in either side's argument.

At this point, all that's been proven to me is that folks have grown tired arguing and are quitting the discussion. I still don't know who's right and, because electricity is so dangerous, I'd like to know. That's why I suggested taking it to an independent third party.

We need a professor here; that's all I'm saying.

ONE TEAM - ONE FIGHT!!!

Mike

[Bruce King]

We are at the point of just needing another good source of info (in laymans terms) on how the AC voltage and current behaves directionally within the circuit. You have to remember this though, most electrical/electronic instructors start off by asking the class which method the class prefers to use, conventional current flow or electron current flow. This difference alone causes more arguments between engineers and technicians. Automotive guys like conventional current flow, engineer types like electron flow.


The amount of current issue has been settled, its 200A max but you can have balanced series operation with 120V loads that add up to a 48kw load.
Four one-hundred amp loads at 120V equals 48kw.


How many guy's have heard the term ELI the ICE man that is used to help remember the following?

The L is an inductive circuit, the C is a capacitive circuit.
E is voltage and I is current.
Voltage leads current by 90 degress in an inductive circuit.
Current leads voltage by 90 degrees in a capacitive circuit.

It only gets more complicated with different load types.

The 60hz AC wave can be confusing to explain.


Jerry is correct in his ongoing message on this board "getting the terminology right is important to understanding and communicating issues"

[Brandon Chew]

OK, I'll saw my own tree limb. Jerry made a nice start. There are two things I have wrong and one thing I should clarify in my post #70.

First thing wrong and a clarification:

To keep it simple and not go into three phase power, a typical transformer supplying a residential 3-wire single-phase 120/240V system has three windings inside it. On the primary side is one winding on a conductor from the power company's distribution system. On the secondary side are two windings, each one wrapped in opposite directions, that create two 120V wave forms (A & B) that are 180 degrees out of phase with respect to each other.

The typical transformer for a residential 3-wire single-phase 120/240V system is not the one I described with three windings. That transformer is a special transformer that would be used to create two different power sources that you wanted to keep isolated. As Jerry pointed out, the typical transformer for residential systems has two windings, one primary (poco side) and one secondary (house side). The secondary winding has enough turns to create 240V. To get 120V, they tap the secondary winding in the center to create one dual voltage system.

Now the clarification. While it's true that the center tapped secondary results in sine waveforms for voltage and current on the A & B legs that are 180 degrees out of phase with respect to each other when viewed from a common ground, describing that difference with the term "phase" can create confusion among folks in the electrical field, especially when you start talking about the differences between single-phase and three-phase power systems.

To avoid that confusion, it might be better to describe the voltage and current sine waves as having mirror images of each other. If you measured voltage on the A leg and the B leg over time with respect to a common ground and plotted them, you would have a picture like the one attached below.

The secondary coil in the transformer creates a 240V AC power source with a positive pole on one end and a negative pole on the other. When we tap the coil in the center we create two 120V AC power sources linked in series. When we ground that center tap at the transformer and attach 120V loads to A & B, we ensure that power always leaves the transformer at the ends of the coil and returns to the transformer through the grounded conductor at the center.

When loads are pulling current through both A & B, and we have a good system that is keeping the voltage stable, measuring voltage between each leg and the common ground will always result in a plot like the one in the picture -- two sine waves with a common start, equal amplitudes, and that cross the zero line at the same time -- mirror images of each other. What's important to note about the voltage plot is that while each leg is 120V with respect to ground, one is +120V while the other is -120V, depending upon which end of the secondary coil is currently positive or negative. They switch every 1/120 second which is shown as a crossing of the zero line on the plot. With voltage we are concerned about potential difference. So when we measure the voltage across A & B we do not simply add +120 to -120 and get zero and say they cancel each other out, we add the absolute values, 120 + 120, and get 240 volts potential between them.

Now let's look at current flow, or amps. Amps = load/volts. As a practical limitation, the value for load cannot be negative. Something creating a "negative load" would be a power source or a generator. As we saw when we looked at voltage, the voltage on the A & B oscillates between +/- 120 volts. A plot of the amps on A & B will yield two sine waves with a common start and that cross the zero line at the same time. At any given point in time (moving to the right on the x-axis), the amplitudes (distance from the wave to the zero line) of the A & B sine waves will be equal ("mirror images", like they are in the picture) only when equal loads are placed on the A & B -- a/k/a the loads are balanced. When the loads are unbalanced, the amplitude of one sine wave will be greater than the other but the crossover points will be unchanged.

When the current flows through the A & B to their loads and then back to the transformer on a common grounded conductor, the current flowing on that common conductor is arithmetically added together. In this case positive and negative currents subtract from each other, while currents of the same sign (+/+ and -/-) add together. When the loads on A & B are balanced, the magnitudes of the current are the same and the signs are opposite -- the sum is zero and no current is flowing on the grounded conductor. When the loads are unbalanced, magnitudes of the current are now different but the signs are still opposite -- the excess current flows on the grounded conductor at the potential difference between the A & B legs, 240V.

Now the power curve for the system that is derived by taking measurements on the A & B legs with a common ground reference, does not look like the graph in the picture. It would look similar to what would happen if you took that picture and you erased the portions of the waves below the zero line and just had the part above it. The reason is that power = watts = volt amperes = volts X amps. In this system, volts and amps can be either positive or negative, but they are always the same sign at any given point in the system -- when volts are negative, amps are negative, and vice versa. When you multiply two negative numbers together you get a positive number. Power draw is always additive, which makes sense -- more load draws more power.

For the wave forms on the A & B legs, and when measured using a common ground reference, we can say: the voltage waves are 180 degrees out of phase with each other, the current waves are 180 degrees out of phase with each other, but the power waves are additive and are in phase with each other -- as they should be in a single-phase power system.

Now for my second error, and this one was really a "left turn at Albuquerque!" I don't know what the hell I was thinking when I wrote that. If I had thought about it more before I hit the post button I would have caught it. I'm going to blame it on a combination of fatigue due to typing with one hand, pain in my left hand, and pain meds.



Jerry, the analogy between water flowing and electricity flowing is a good one for most things. It breaks down when you transition from discussing DC current to AC current. Thinking in terms of water makes it difficult for people to grasp the concept that electrons can be flowing along a conductor in two directions at the same time. It's the two windings in the transformer generating two 120V currents out-of-phase by 180 degrees that is the key to making this physically possible.

The water analogy (two hoses drawing water from the transformer at the same time) does work to illustrate what happens once the loads on the A & B become unbalanced (which is the common real-world situation in residential panels), or if the A & B are being generated in phase with each other. With A & B 180 degrees out-of-phase with each other, at any given point in time that current is being drawn through the A leg to the house, the capacity of the A leg to return the current that is being drawn from the B leg back to the transformer is limited to an amount that is relative to the current being drawn by the A leg. The excess draw from the B leg has to go back on the neutral. If the A & B are being generated in phase with each other, neither the A or B could carry any flow back to the transformer, and all of the flow from both of them would go back on the neutral at the same time. When A & B "hoses" were both flowing at their maximum rate of flow in phase with each other, the neutral "hose" would need to be twice the size of either one of them.

The current cannot flow on the same conductor in two directions at the same time -- that's gibberish! On the 120V circuits the current flows from the A and B poles on the secondary coil through the main OCPD, then to their respective loads, then back on the grounded conductor. On the 240V circuits the current flows from either the A or the B, whichever pole is positive at the time, through the main OCPD, to the load, and then back to transformer on either the A or the B, whichever pole is negative at the time. On the 120V and 240V circuits the A & B poles are changing polarity every 1/120 seconds.

The reason I said I wasn't going to choose sides when I stepped into this discussion was because I think both sides are right when they speak the same language. You could describe a 200A household system as pulling off the transformer either 200A @ 240V or 400A @ 120V through two 200A legs. The transformer doesn't care how you want to say it as long as you specify the volts along with the amps (a/k/a the power, or watts, or volt amperes are the same either way). My personal preference is 200A @ 240V because in the residential system the ampacity of the service conductors and the rating of the main OCPD are all 200A, and that 400A situation will occur only when the loads on A & B are balanced, which would be unusual to find in a residential system.

[Brandon Chew]

Here's the diagram again.

[Bruce King]

Nice explanation there.

Jerry must be taking the day off.

You are in big trouble for using "400A" in the post.......
Wait till Jerry gets back......

Four 100a loads is the best way to describe that 400a issue.

[Jerry Peck]

You could describe a 200A household system as pulling off the transformer either 200A @ 240V or 400A @ 120V through two 200A legs.

The correct way to address that would be to state that you are drawing 48,000 va at either 120 volts or at 240 volts.

To throw in "400A @ 120V" is a misnomer as there is not "400A" anywhere in that circuit. (Note: I edited this line to change "To through in" to "To throw in".)


The transformer doesn't care how you want to say it as long as you specify the volts along with the amps (a/k/a the power, or watts, or volt amperes are the same either way).

The transformer does not care, because it is not even thinking in volts/amps/va. However, as soon as humans try to describe the limitations of a circuit with a 200 amp main, the limitation is ... 200 amps, not more, and in no way, shape or form is there "400 amps" anywhere in there.

The other correct way would be to state you have 200 amp at 240 volts with 24,000 va on each phase at 120 volts, or, to more simply state it: you have 200 amps at 240 volts with the loads balanced across the phases and the neutral ... (technically, with those lamps, you would not even need to connect to the neutral as their resistance, and resulting current flow, are so close to each other that there would be such little difference at each lamp without a neutral grounding point that the neutral grounding point is not required - with motors, though, the neutral grounding point would absolutely be necessary to maintain the balance across the phase conducts and the neutral conductor and avoid burning motors up).

[Brandon Chew]

The correct way to address that would be to state that you are drawing 48,000 va at either 120 volts or at 240 volts.

To through in "400A @ 120V" is a misnomer as there is not "400A" anywhere in that circuit.

The transformer does not care, because it is not even thinking in volts/amps/va. However, as soon as humans try to describe the limitations of a circuit with a 200 amp main, the limitation is ... 200 amps, not more, and in no way, shape or form is there "400 amps" anywhere in there.

The other correct way would be to state you have 200 amp at 240 volts with 24,000 va on each phase at 120 volts, or, to more simply state it: you have 200 amps at 240 volts with the loads balanced across the phases and the neutral ... (technically, with those lamps, you would not even need to connect to the neutral as their resistance, and resulting current flow, are so close to each other that there would be such little difference at each lamp without a neutral grounding point that the neutral grounding point is not required - with motors, though, the neutral grounding point would absolutely be necessary to maintain the balance across the phase conducts and the neutral conductor and avoid burning motors up).

Ok, I'll buy that. As mentioned I would't attempt to describe it as 400A anyway.

For folks who followed the discussion of sine waves in my last post, nowhere in the system will you measure 400 amps -- the most you will measure is 200 amps. If the system is working right and you are pulling the max 48 kW off the transformer, the most you will see on the A or the B legs is 200 amps. When A & B come together on the common grounded conductor and go back to the transformer, recall that the current waveforms are mirror images of each other, so when you add those currents with opposite signs (+/-) together they want to cancel each other out. As a result the common grounded conductor would have a minimum of zero amps on it from 240V loads or when 120V loads are equally balanced on the A & B, up to a maximum of 200 amps flowing on it when the loads are unbalanced.

:)

[Steve Lowery]

WHEW!

[Tim Voss]

Just one question.

Say I only have 120v loads on my 200a 240v service, and they just happen to be perfectly balanced between the phases. I look at the nameplate on my refrigerator and it says, 120v/60Hz 7.7amps and my vacuum says 120v/60Hz 9.2amps, etc...

If I add up all the amps of my 120v devices I have plugged in and running, can I run 400amps of 120v devices at the same time without blowing the 200a 240v main breaker?

[Richard Moore]

Just one question.

Say I only have 120v loads on my 200a 240v service, and they just happen to be perfectly balanced between the phases. I look at the nameplate on my refrigerator and it says, 120v/60Hz 7.7amps and my vacuum says 120v/60Hz 9.2amps, etc...

If I add up all the amps of my 120v devices I have plugged in and running, can I run 400amps of 120v devices at the same time without blowing the 200a 240v main breaker?

My oh my! I gave up reading this thread as it morphed into something that made my head spin...but they still haven't answered that basic question?

I'll take one final stab at it (without mentioning 400-amps in any way because that seems to get peoples' knickers twisted).

Tim, perhaps a simpler way of putting it would be to ask if you could fully load 20 20-amp 120 volt circuits (assuming they were 10 per leg). And the simple answer is yes.

To make it even simpler, picture a panel with only one double-pole 20-amp breaker, the type with a handle tie. You can run a full 20-amp 240 volt load on that. Remove the handle tie and rewire to create two 120-volt circuits. Hopefully, no one is going to suggest that you couldn’t run 2 20-amp 120 volt devices at the same time. Putting the handle tie back on will not make it trip.

The 200-amp main breaker is no more than a big 200-amp double pole breaker. That is two 200-amp breakers, usually with an internal common trip mechanism, but also sometimes with external handle ties. It works the same as the 20-amp example, just times 10.

I'll let you do the math. :D

[Tim Voss]

Thanks Richard, I did the math and it comes up to the controversial 400amps (at 120v). That's why I was really waiting to see what Jerry had to say.

It's been an interesting discussion, but the home owner doesn't know or care about voltage waveforms and what goes on inside the "black box" of the service panel. They just want to know how much "stuff" can they plug-in.

[Brandon Chew]

Just one question.

Say I only have 120v loads on my 200a 240v service, and they just happen to be perfectly balanced between the phases. I look at the nameplate on my refrigerator and it says, 120v/60Hz 7.7amps and my vacuum says 120v/60Hz 9.2amps, etc...

If I add up all the amps of my 120v devices I have plugged in and running, can I run 400amps of 120v devices at the same time without blowing the 200a 240v main breaker?

Yes. Bruce mentioned something like this a few posts above this one for a situation with four 120V 100A loads when he wrote:

"Four 100a loads is the best way to describe that 400a issue"

The problem comes when you try to say this means the house service is rated for 400A @ 120V. You are drawing 48 kW of power off the transformer by a number of 120 V devices whose nameplates total 400 amps. Because the current (amps) is coming from opposite ends (+/-) of the transformer coil on two different legs to the house and returning on a common neutral, the house service is rated at 200A because that should be the most current you would measure on any of the service conductors.

I entered this discussion thinking you could describe the house service in terms or either 200 or 400 A if you were careful enough about how you did it, but the reality is (and the theory backs it up) that there really isn't a practical reason to try to describe the service in terms of 400A, and wound up agreeing with the 200A side.

Richard - nice description. :)

[Brandon Chew]

It's been an interesting discussion, but the home owner doesn't know or care about voltage waveforms and what goes on inside the "black box" of the service panel. They just want to know how much "stuff" can they plug-in.

Hi Tim,

I wouldn't get into that discussion with a homeowner either. I needed to work through the theory here to get past the "I'm right!", "No, I'm right!" phase that this thread had seemed to be locked into, and to test and clarify my own understanding of the subject. It's been educational.

Brandon

(grits teeth and looks back over shoulder for Jerry's comments on my previous post) ;)

[Jerry Peck]

To make it even simpler, picture a panel with only one double-pole 20-amp breaker, the type with a handle tie. You can run a full 20-amp 240 volt load on that. Remove the handle tie and rewire to create two 120-volt circuits. Hopefully, no one is going to suggest that you couldn’t run 2 20-amp 120 volt devices at the same time. Putting the handle tie back on will not make it trip.

The 200-amp main breaker is no more than a big 200-amp double pole breaker. That is two 200-amp breakers, usually with an internal common trip mechanism, but also sometimes with external handle ties. It works the same as the 20-amp example, just times 10.

(sigh) :D

Richard,

You started out on the right track, keeping everything at 240 volts ...

...but, as soon as you split that double pole 20 amp breaker into 2 separate 20 amp breakers on 120 volt circuit, you changed the entire equation.

How? You ask ...

Here is how:

*IF* (and ONLY *IF*) those two 20 amp breakers remained in the same location (i.e., one on Phase A bus and one on Phase B bus) - then nothing changes - you essentially have (still have) a 240 volt load made up of balances 120 volt loads, with no neutral current.

NOW *HOWEVER*, if you separate those two 2 amp breakers leaving one empty space between them, you have removed them from being on different phase legs and they are now on the same phase leg, making their currents additive *at 120 volts* instead of benefiting from having had them at 240 volts.

The difference between them and the 200 amp main is much more than "just times 10", the difference is that YOU CANNOT SEPARATE the 200 amp main into 120 volt legs ONLY - either BOTH will have the same current and the neutral will have no current (i.e., 200 amps maximum flowing through them), OR BOTH will have the same current as the lower one AND the neutral will have the current difference to the higher one (i.e., *less than* 200 amps flowing through them).

Oh, and, don't forget, "*less than* 200 amps will still flow through a 200 amp main.

[Jerry Peck]

Electricity 001 (it's not even Electricity 101 :) )

You must have a complete circuit for current to flow.

Take a 200 amp main, install a 240 volt breaker, install one conductor from one side of the breaker to a 200 amp 240 volt load, install another conductor from the other side of the breaker to a switch with the switch in the "open" position, then connect the yet another conductor from the other side of the switch to the other side of the 200 amp 240 volt load.

How much current flows?

NONE.

The current MUST be able to get from one side of the main back to the other side of the main. Using one side does nothing.

Throw the switch "closed". Now how much current flows?

200 amps. No more.

But ... but ... but there are *TWO* (count them Jerry) TWO 200 amp breakers there, THAT'S 400 AMPS - DO THE MATH JERRY.

That's what some of you are saying. So I am saying, simply, *SHOW ME THE 400 AMPS*.

I have reduce all the above posts down to this one simple, easy to understand post, with one easy to understand question.

SHOW ME THE 400 AMPS

[Tim Voss]

400 light bulbs x 1 amp each = 400amps

You keeps saying it's the same amp, but it is not exactly the same amp. If you put two 120v light bulbs in series to make a 240v circuit, yes only 1 amp flows through each bulb, but you have to count that amp each time it flows through a load. The first bulb only uses 120volts of the 240volt potiential, the second bulb uses the remaining 120volts of potential.

1amp x 240volts = 2amps x 120volts = 240watts of power

Right there in the math you can see that you have to count the amp each time it passes through the 120volt load. They all add.

The light bulb is stamped 120v and 1amp right on the bulb! If I have 400 of them lit simultaneously, I have 400amps (at 120volts). The light bulb doesn't know what circuit it is in, it only knows that when 120volts is applied it uses 1amp. All 400 could be in series on a 48,000volt circuit with only 1 amp flowing. Doesn't matter, each bulb has a 120volt drop and uses 1 amp. Add them up = 400amps total at 120v or 1amp total at 48,000volts or 200amps total at 240volts.

Since the bulb is 120v, you are running 400amps of 120volt loads. Bottom line is, I can add all the amp ratings of my 120v appliances and put 400amps of load on a 200amp 240v service.

[Richard Moore]

(sigh) :D

Richard,

You started out on the right track, keeping everything at 240 volts ...

...but, as soon as you split that double pole 20 amp breaker into 2 separate 20 amp breakers on 120 volt circuit, you changed the entire equation.

How? You ask ...

Here is how:

*IF* (and ONLY *IF*) those two 20 amp breakers remained in the same location (i.e., one on Phase A bus and one on Phase B bus) - then nothing changes - you essentially have (still have) a 240 volt load made up of balances 120 volt loads, with no neutral current.

NOW *HOWEVER*, if you separate those two 2 amp breakers leaving one empty space between them, you have removed them from being on different phase legs and they are now on the same phase leg, making their currents additive *at 120 volts* instead of benefiting from having had them at 240 volts.

Sigh back at you!
Your ability to complicate simple answers is awsome. All some of the good folk here want to know is if it is possible to have 20 20-amp circuits and run 20 20-amp 120 volt toasters at the same time.

So...just where did I suggest moving the breakers from their original position and putting them on the same leg? Do you really think I don't understand the difference?


The difference between them and the 200 amp main is much more than "just times 10", the difference is that YOU CANNOT SEPARATE the 200 amp main into 120 volt legs ONLY - either BOTH will have the same current and the neutral will have no current (i.e., 200 amps maximum flowing through them), OR BOTH will have the same current as the lower one AND the neutral will have the current difference to the higher one (i.e., *less than* 200 amps flowing through them).

Really? You seem to be saying that I couldn't disconnect one of the SEC's and then just use the one leg for my 10 20-amp ciircuits on that bus bar and power 10 20-amp 120-volt toasters. You're not? That would work? Well, OK...then when I reconnect the SEC to the other bus-bar can I, or can I not run an additional 10 20-amp toasters AT THE SAME TIME!!!


YOU CANNOT SEPARATE the 200 amp main into 120 volt legs ONLY

So...Yes I can! There is no functional difference between a 200-amp residential main and any other amperage double-pole breaker. If you don't believe that, then take a look at a Siemens/ITE quad main and then tell me there aren't separate, but linked, breakers for each side.


Oh, and, don't forget, "*less than* 200 amps will still flow through a 200 amp main.

OMG!!! Thank you for that enlightenment master! :rolleyes: :rolleyes: :rolleyes: :rolleyes:

OK...that's it for me. I gave Tim a simple, understandable answer and I refuse to get sucked into this black hole any more. There was a thread entitled "What do you miss". Me, I miss the old Inspection News where normal HIs could continue normal HI discussions and one person didn't feel the need to dominate the board by disecting each and every damn thread and post for individual statements, usually completely out of context, he could disagree with. Yep....them were the good old days!

I think I'm going to have to take a break from here for a while. It just isn't enjoyable anymore.

[Michael P. O'Handley]

Congratulations Jerry,

You wore another one down. One more victory for the pedant.

Guess I'll copy all of this and send it to Mike Holt in an Email and ask him if he can tell me which one of these "explanations" I should believe.

ONE TEAM - ONE FIGHT!!!

Mike

[Bruce King]

Mike,

I think everyone is in agreement with the theory now but for many its hard to overcome the tendency to total up 120V load amps and use that as the sole supply source. The center tap AC transformer is a really awesome thing when you think about it.


It simply comes down to two ways to discuss it:


If you choose to describe only max 120V loads, total current can be confusing and it is important not to use the wrong terminology.

If you choose to describe the max supply source it is simple, 200A at 240V.


This is the reason power is sold by the kw and not amps, we have two methods of getting power off residential service, at 120V or 240V.


Its the power company not the amp company.

[Brandon Chew]

I have reduce all the above posts down to this one simple, easy to understand post, with one easy to understand question.

SHOW ME THE 400 AMPS

Jerry - This is for my benefit and for those who followed my revised discussion of the volts, amps, and power waves and current flow in post #86. Do you agree with this statement?:

While it is possible to connect a bunch of 120V loads to a 200A 120/240V 1-phase 3-wire residential electrical system at the same time, and have the amp ratings of all of that 120V utilization equipment add up to 400 amps by simple addition of the amps on the nameplates on the equipment, if everything is wired and working properly, you should not measure more than 200A anywhere on the system.

(I realize that instantaneous values of volts & amps fluctuate beyond 120/240 & 200 and those numbers represent "averages"... we are trying to keep it simple.)

If not, how would you change it?

[Bruce King]

Jerry agreed to all that several posts ago.

[Jerry Peck]

400 light bulbs x 1 amp each = 400amps

Nope, not when wired with 400 lamps with 200 sets of 2 lamps in series - that's only *200 amps*.

No matter how you look at it, YOU WILL ONLY HAVE 200 AMPS when wired that way.

You can call a duck a chicken as often as you want to, but to almost everyone else you will look silly when you point to a duck and say, with conviction, 'THAT is a *chicken*, no question about it.'

Well ... they both do lay eggs, and the both do have feathers, and ... BUT, they are NOT one and the same thing.

[Jerry Peck]

(underlining in the following quote is mine)

All some of the good folk here want to know is if it is possible to have 20 20-amp circuits and run 20 20-amp 120 volt toasters at the same time.

Apparently not.

That question has been answered numerous times.

The answer is ALWAYS the same, and ALWAYS WILL BE the same.

To wit: ONLY IF (don't you recall that from earlier posts ... *ONLY IF*) they are equally balanced between the phase legs.

What is so hard to get about *ONLY IF*?

As soon as one starts to talk about "*400 AMPS*" - which is really what this is all about ... a few (1 maybe 2 people) trying to insist that they are getting 400 amps - then everything changes.

THERE ARE NOT 400 AMPS when wired as the question was asked. 200 AMPS, YES. 400 AMPS, NO.

Whenever someone posts about 200 AMP, they are correct, and I state so, and agree ... but ... as soon as someone insists on calling it *400 AMPS*, that is incorrect and I say that too.

Like it or not, the above is exactly what have been going on throughout this thread.

ONE OR TWO PEOPLE INSISTING that there is 400 amps with wired as described - the rest which have posted acknowledge the difference ... those ONE OR TWO PEOPLE have not, and, apparently, will not, think it through or acknowledge the difference.

[Jerry Peck]

Jerry agreed to all that several posts ago.

Bruce, not all, as in "all", there are still one or two who have not yet "got it":


Thanks Richard, I did the math and it comes up to the controversial 400amps (at 120v). That's why I was really waiting to see what Jerry had to say.


400 light bulbs x 1 amp each = 400amps

[Jerry Peck]

Do you agree with this statement?:

While it is possible to connect a bunch of 120V loads to a 200A 120/240V 1-phase 3-wire residential electrical system at the same time, and have the amp ratings of all of that 120V utilization equipment add up to 400 amps by simple addition of the amps on the nameplates on the equipment, if everything is wired and working properly, you should not measure more than 200A anywhere on the system.

( ... we are trying to keep it simple.)

If not, how would you change it?

"If not, how would you change it?"

Brandon,

I would change it to read as follows:

While it is possible to connect a bunch of 120V loads to a 200A 120/240V 1-phase 3-wire residential electrical system at the same time, you cannot simply add up the amp ratings of all of that 120V utilization equipment by simple addition of the amps on the nameplates on the equipment, if everything is wired and working properly, you should not measure more than 200A anywhere on the system. This is just like trying to add up the amp ratings for 6 mains in a service equipment panel - you cannot add up the amp ratings for 6 mains to get the amp rating of the service.

Brandon, I believe that little bit of wording change in the middle of your statement, and the last sentence I added relating it to something which we all understand, makes it understandable for all and makes it work for all.

None of us would walk up to a service equipment panel which has 6 - 100 amp mains and state that it is a 600 amp service, would we? Nor would we spend untold amounts of time trying to convince others that it was 600 amps.

Anyone disagree with that?

Anyone agree with that?

[Jerry Peck]

And, last but not least ... well ... okay, he is least, so I responded last ..


Congratulations Jerry,

You wore another one down. One more victory for the pedant.

Guess I'll copy all of this and send it to Mike Holt in an Email and ask him if he can tell me which one of these "explanations" I should believe.

ONE TEAM - ONE FIGHT!!!

Mike

Mike,

If you got off your high horse (that tall statute horse covered with pigeon droppings) and actually read the posts, you would see that all ... *all* ... except ... yes, except ... except 1, possibly 2, are in agreement of what has been stated.

[Bruce King]

Here is the correct formula for max service entrance load with 120V balanced loads.


Using 400 one amp 120V lamps as the load (200 lamps per leg):


400 x 1A / divided by 2 (since 1/2 of the loads are in series) = 200A.

[Michael P. O'Handley]

Uh huh,

Riiiigggghhhhhhhtt! That's why the thread went over a hundred posts.

OT - OF!!!

M.

[Jerry Peck]

That's why the thread went over a hundred posts.


Mike,

Ummm ... YOUR post put it over 100 ... yours was #101. :)

[Michael P. O'Handley]

Interesting,

Your last response is #112 on my display.

Wonder if this will start another long debate wherein you try and convince me I don't know the difference between the numbers 101 and 112 ? LOL

ONE TEAM - ONE FIGHT!!!

Mike

[Jerry Peck]

Wonder if this will start another long debate wherein you try and convince me I don't know the difference between the numbers 101 and 112 ?

"wherein you try and convince me I don't know the difference between the numbers 101 and 112 ?"

Ummm ... that long debate will only last until you learn to read ...

See attached screen capture of your post, with the post number 101 encircled with an ellipse.

You said: "That's why the thread went over a hundred posts."

I stated: "Ummm ... YOUR post put it over 100 ... yours was #101."

Review the screen capture and you will see who made the thread go over 100 posts - you did, with your post being the 101st post in the thread.

Believe it or not, it's there for all to see.

[Michael P. O'Handley]

LOL,

Well, if I knew how to do a screen capture, I could show you that at this end your response is #114.

Don't ask me why; maybe Brian knows.

OT - OF!!!

M.

[Tim Voss]

I guess all those generator manufacturers are going to have to re-write their manuals, since they can't "really" power double the amount of 120v amps as 240v amps.

[Michael P. O'Handley]

OK,

Sorry for the thread drift, but I finally figured out the screen capture thingy. As you can see, the posts are numbered differently than yours. What's up with that?

OT - OF!!!

M.

[Matt Fellman]

OK,

Sorry for the thread drift, but I finally figured out the screen capture thingy. As you can see, the posts are numbered differently than yours. What's up with that?
OT - OF!!!

M.

Ignore feature?

[Jerry Peck]

As you can see, the posts are numbered differently than yours. What's up with that?

"What's up with that?"

You are looking at DIFFERENT POSTS - go back up and look at post number 100 (bottom of the first page), then look at post number 101 (top of the second page).

WHO's post is post number 101?

Hint: Yours.

Thus, WHO's post caused this thread to go beyond 100 posts (which is what you were commenting on)?

Hint: Yours.

[Michael Larson]

Wow, Jerry can count to a hundred.:p

[Tim Voss]

Ok Jerry, so we disagree on how the amps of the loads may (or may not) add up.

But you DO agree that you can put 120v loads whose "nameplate amps" add up to 400, on a 200a 240v service as long as the loads are perfectly balanced between the phases?

[Michael P. O'Handley]

Jerry,

You're a flaming jackass. Do you realize that? It doesn't matter who's post put it over 100, there wouldn't have been 100 posts about this in the first place, if you hadn't insisted on dissecting everyone's commentary, muddying the waters, and yammering on, and on, and on, and on, and on, and on......

A horse's ass with teeth and a Napolean complex to boot; what a friggin putz!

I need to unsubscribe to this place so I don't keep coming back here. Ta Ta!

Never mind, I just found out how to turn on the ignore feature. That's pretty kewl. I think I'll add that at the crib.

M.

[Jerry Peck]

Jerry,

You're a flaming jackass. Do you realize that?

Ohhh .... Mickie's hot now.

First you try to blame someone else for making the thread go beyond 100 posts, then, when you find out it's you who did that, you try to re-direct your comment so you look innocent.

You sure turned into a jackass ... wait, no ... you always were one, I will correct myself ... you sure exposed the real jackass you are.


It doesn't matter who's post put it over 100, there wouldn't have been 100 posts about this in the first place, if ...

... you, Mike O', had not tried to insist your way was the right way, the only way.


I need to unsubscribe to this place so I don't keep coming back here. Ta Ta!

Now that's not a bad idea :cool: , it will save me from having to read your ignorant, intrusive and obnoxious posts. :rolleyes:

I don't use the ignore feature, I read what jerks like you write, and choose to ignore or to respond - that way I can track what jerks are saying what.

[Jerry Peck]

Ok Jerry, so we disagree on how the amps of the loads may (or may not) add up.

But you DO agree that you can put 120v loads whose "nameplate amps" add up to 400, on a 200a 240v service as long as the loads are perfectly balanced between the phases?

Tim,

That's a poor way to state what you are trying to indicate (unless you are still trying to "add up to 400") because you are still "adding up" the "nameplate amps" as though that will work.

In reality, you will need to "add up and divide out by the number of loads the same amps is flowing through" such the "nameplate amps" are no more than 200 amps are on the 200 amp 240 volt supply.

You are still 'trying to "add up" the amps to get 400', and that does not work.

Going back to my example of 6 mains, let's try this":

Say you have 6 mains as follows"

1 - DP 30 amp (DP stands for "double pole")
2 - DP 60 amp
2 - DP 100 amp
1 - DP 50 amp

Can you say that you can add those up 30 + 60 + 60 + 100 + 100 + 50 = 400 amps service?

No, of course not, not anymore than you can add those 120 volt loads up to get 400 amps on a 200 amp 240 volt supply.

You are close, but still not quite there.

However, if you wanted to change your reference to 48,000 va of 120 volt loads on a 48,000 va 240 volt supply, then, yes.

[Tim Voss]

You may say it's a poor way to say it, but I don't think so.

I'm trying to say how much 120v "stuff" I can plug into a 200amp 240v service. My devices have Nameplate ratings in 120v and amps. I want to make sure I don't overload the service. No division necessary, I just add up the Nameplate amp ratings of my 120v devices and as long as that is equal to or less than 400, everything is good. Right?

[Tim Voss]

This is the part that is really killing me:


However, if you wanted to change your reference to 48,000 va of 120 volt loads on a 48,000 va 240 volt supply, then, yes.


Ok, I'll say 48,000 va of 120 volt loads. That is exactly what I'm saying. Let's just plug the 120 into the v and see what comes out for the a.

[Jerry Peck]

Ok, I'll say 48,000 va of 120 volt loads.

Then you would be correct, yes. :D


That is exactly what I'm saying.

Now, yes, but not before.


Let's just plug the 120 into the v and see what comes out for the a.

Not a problem, as long as you want to keep it all in the same context, however, that then does not go with the 200 amps 240 volts ... until you convert it to volt-amps. You still insist on mixing terms, and that does not work, not unless you want to spell out an equation or a problem. You either need to stick to volts and amps, or, volt-amps (watts).

[Tim Voss]

va is an equation, as in P=va



That's a poor way to state what you are trying to indicate (unless you are still trying to "add up to 400") because you are still "adding up" the "nameplate amps" as though that will work.



Are you saying that I can NOT put 400 Nameplate rated amps of 120v equipment on a 200amp 240v service?

[Jerry Peck]

va is an equation, as in P=va

No, va is only a statement, an equation needs all parts. If you noticed, "va" is simply "va", YOU ADDED the "P=" to make it an equation.


Are you saying that I can NOT put 400 Nameplate rated amps of 120v equipment on a 200amp 240v service?

Tim,

I guess either you are just not paying attention, or I am just very poor at communicating this, but, here it goes for one more time:

No, I am saying you cannot ADD THEM UP TOGETHER, which is what you continue to try to do.

[Tim Voss]

I'm not trying to add them up together. I'm trying to put twenty 20 Nameplate rated amp 120v toasters on the system. Are you saying I can't?

[Tim Voss]

Ok, I re-read you post, and I think you are saying I CAN put 400 Nameplate rated amps of 120v equipment on the 200amp 240v service. Is that correct?

[Tim Voss]

Ok, so now I have all this 120v equipment plugged-in.

I'm not measuring or talking about amps in any wire. I'm reading the rating printed on the equipment. You are saying I can't add? Actually it's quite easy. Toaster 1 says "20 amps", toaster 2 says "20 amps", etc... 20+20+...

When I add that up, it adds to 400.

[Jerry Peck]

When I add that up, it adds to 400.


Tim,

I can go (legally - speed limits):

70 on I-95
70 on I-4
45 on SR 40
185 is what that adds up to.

*I* can add numbers too. Just means nothing.

[Tim Voss]

Actually, means I can plug-in 400 Nameplate rated amps of 120v equipment on my 200a 240v service.

It's a good way to know if it's "Too much for 200A" when I don't have a meter handy.

[Tim Voss]

You see, it actually does mean something. When I plug-in 401, I have overloaded the breaker, but when I plug in 400 I haven't.

[Tim Voss]

There must be something special about that number, 400.

[Jim Zborowski]

Tim, What I'm getting out of your last post is you need 120 volts and 1 amp to light a bulb. Then, if you have 2 bulbs you need another 120 volts and another amp to light that bulb, and so on. So the logic then is that each bulb " consumes " the entire 120v/1 amp to light.
Unfortunately you have to consider Volts, Amps, Watts, and Resistance in a circuit. You also have to consider parallel, series, and series/parallel loads
( resistance ). If you have 2 - 100 ohm resistors in series the resistance is added. If you place them in series, the resistance is divided. This then changes the power ( load ) requirement between a series and parallel circuit.
Also, for your scenario to work, the first bulb would need a filament the size of ooo service wire to keep from burning out.

[Tim Voss]

Hi Jim, did you read the entire thread? We've been through all that.

No, what I'm getting at is quite simple and doesn't require much calculation or detailed knowledge of electric theory.

I have 400 standard 120watt, 120volt, 1amp light bulbs, or 20 standard 2400watt, 120volt, 20amp toasters, doesn't matter. The volt and amp rating is stamped right on the nameplate.

For 120volt loads, I can add up 400amps of Nameplate rating and put that on a 200amp 240v service. But if I try to put 401amps of Nameplate rating, that will overload the breaker.

And actually, the light bulb does "consume" the entire 120v 1amp that goes through it. P = va = 120v * 1a = 120watts

[Bruce King]

Lets use twenty, 20 amp 120V circuit breakers each feeding a 20 amp load device as an example, with all turned on.

With 10 breakers on each leg of a regular panel there are only 10 current paths in the panel, each leg has a load in series with the other leg.

Each load is in series with a load on the other leg so ONLY 10 current paths are present in the panel.

Each current path has 20 amps flowing: 10 paths x 20 amps = 200 amps.

Its simply because you have to figure two loads on each path because that is the way the current travels. Each current path travels through one 20 amp breaker on opposite legs with a total of two breakers per path.

Yes, if you add up the loads by looking at it seperately you get 400amps but you can not use this method with series loads.
You have to use VA or watts as the total load so that it does not appear to be implying 400 amps is flowing in any one wire.

[Tim Voss]

Yes. I have always said there is not 400a in one wire. It is the amps of all the 120v loads added together that give you 400amps.

[Bruce King]

You have to use VA or watts when adding up series loads, never use the term 400A, its just not technically correct when discussing this example.

When you state something in "amps", you should only be referring to a current that can be actually measured.

Its one of the reasons that the term "watts" is available for use, this example we have been tossing around is a perfect example of why using the proper terminology at the right time is important.

[Tim Voss]

Actually, it is technically correct in this example.

When you have 120v loads, it does not matter if they are in a 120v circuit or a 240v circuit. The v in the va will always be 120v, because that is the voltage drop across the load. For example, the 120watt, 120volt, 1amp light bulb. Even when you put 2 in series on 240volts, there is a 120v drop across each bulb and 1amp flowing through each bulb. So the va for each bulb is 120v * 1a = 120watts. The Total va for the circuit is (120v * 1a) + (120v * 1a) = 240watts.

You can use math on that equation to come up with some equivalent equations.

(120v * 1a) + (120v * 1a) = 240v * 1a = 240va
(120v * 1a) + (120v * 1a) = 120v * (1a + 1a) = 120v * 2a = 240va

BOTH are equally correct, and in fact, it is MORE appropriate to express it as 120v * 2a BECAUSE it is a 120v light bulb, NOT a 240v light bulb. You don't want to give the impression that you can put that bulb on a 240v circuit, it will blow up.

[Jerry Peck]

it is MORE appropriate to express it as 120v * 2a BECAUSE it is a 120v light bulb, NOT a 240v light bulb.

Tim,

Now THAT is *COMPLETELY* wrong.

You do not have 2 amps at 120 volts.

Think about what you have: you have 1 amp at 120 volts twice (each lamp has 1 amp through it and each lamp has 120 volts across it).

Thus, the better way to say it would be 1 amps at 240 volts (because this *IS* what you have ... whether or not the center point between the two lamps was, or was not, grounded.


You don't want to give the impression that you can put that bulb on a 240v circuit, it will blow up.

You are giving the impression that you are using a 240 watt lamp at 120 volts, and you are not. You are using 120 watt lamps at 120 volts.

[Tim Voss]

* SIGH *

I guess it's time for Electricity 002: Calculating power in a multi-load circuit

Refer to the included diagram for this discussion.

The total power of a circuit is equal to the sum of the power for each load.

P(total) = P(1) + P(2) + ...

In circuit A, there are 2 loads with an ammeter and voltmeter showing the measured values.
Total power in circuit A
P(total) = P(1) + P(2) = (120x200) + (120x200) = 120x400

In circuit B, there are 2 loads with an ammeter and voltmeter showing the measured values.
Total power in circuit B
P(total) = P(1) + P(2) = (120x200) + (120x200) = 120x400

From "outside the box" you CAN NOT determine any difference between the circuits. All measured values are the same. They both deliver 400amps of power at 120volts. They are both equivalent.

For the total power calculation, it does not matter if the circuit is parallel or series, they are both driving the same exact loads (load 1 and 2).

Another way to calculate the Total Power is to reduce the multiple loads into an equivalent theoretical load. Now if we look inside the box, for circuit A, the total equivalent power is 400amps at 120volts and for circuit B the total equivalent power is 200amps at 240volts, but the actual power is 400amps at 120volts, because that is what is actually measured at the loads.

Jerry, you suggest that you can express the power in terms of amps at 240v, and I agree.

But you are * COMPLETELY * wrong when you say that you cannot also express the power in terms of amps at 120v.

[Jerry Peck]

* SIGH *

I guess it's time for Electricity 002: Calculating power in a multi-load circuit

Let's try Electricity 000, which should have come before Electricity 001 and Electricity 002.

When loads are in series, the current flow is not additive for a total amperage current flow. There is the same current flowing through each load, hence "in series".

There is a voltage across each load with the voltages being additive for the total voltage.


Jerry, you suggest that you can express the power in terms of amps at 240v, and I agree.

Yes, that is what I've been saying.


But you are * COMPLETELY * wrong when you say that you cannot also express the power in terms of amps at 120v.

That is not what I've said anywhere. IN FACT *I* have been telling YOU to refer to those loads in power terms (watts for resistive loads such as lamps, VA for other ac loads) ... it has been YOU who has REFUSED to use those power terms and to instead insist on only using the term "amps" in trying to add up series loads.

You seem so close to getting it, then you back away and insist on adding up "amps" for series loads. Why do you refuse to use the power terms you stated above which should be used?

[Tim Voss]

No Jerry, YOU are wrong. Did you even look at the diagram?

In the series circuit, the voltages don't add. Look at the measured voltage for Load 1, it is 120 volts. Look at the measured voltage for Load 2, it is 120 volts.

You don't even seem close to getting it.

[Tim Voss]

Jerry, what part of P(total) = P(1) + P(2) don't you understand?

[Jerry Peck]

When loads are in series, the current flow is not additive for a total amperage current flow. There is the same current flowing through each load, hence "in series".

There is a voltage across each load with the voltages being additive for the total voltage.


No Jerry, YOU are wrong. Did you even look at the diagram?

In the series circuit, the voltages don't add. Look at the measured voltage for Load 1, it is 120 volts. Look at the measured voltage for Load 2, it is 120 volts.

Tim,

I'll make an "assumption" here: YOU drew the diagram, right?

Now (presuming that my "assumption" is correct) you need to learn to understand what you drew.

Circuit A is a parallel circuit, the current flow is additive, the voltages are the same.

Circuit B is a series circuit, the current flow is the same, the voltages are additive.

Don't believe me when I say :

When loads are in series, ...

There is a voltage across each load with the voltages being additive for the total voltage.?

Look at Circuit B: You are showing Load 1 with 120 volts across it, you are also showing Load 2 with 120 volts across it, you are also showing the additive total of those voltages across each load as being 240 volts.

That is correct.

Thus YOU DREW it right (if you drew it), but apparently you do not understand it (did you copy it from someone else?) because you are now stating:

In the series circuit, the voltages don't add.

and


Look at the measured voltage for Load 1, it is 120 volts. Look at the measured voltage for Load 2, it is 120 volts.

Yeah, I have looked at them, and in Circuit B (the series circuit) Load 1 is connected to Load 2, with the voltage across Load 1 (120 volts) added to the voltage across Load 2 (120 volts) to gives the total voltage (240 volts), which, in Circuit B is 120 volts + 120 volts = 240 volts. Did you catch the "+" sign? That's for "addition".

Now, if that is not additive ... I don't know what kind of "new math" you are using.

*IF* *YOU* are referring to Circuit A, then you are even more confused than I thought as Circuit A is a parallel circuit, not a series circuit.

[Tim Voss]

In the series circuit, what is the power in Load 1?

120w = 120v * 1a

So what is the voltage across Load 1?

[Tim Voss]

Ok Jerry, let me try to explain this another way.

Let's say you cannot see "inside the box" because it's a locked panel and you have no idea what is in there.

What would you say is the total power for loads 1 and 2 on circuit B. Just look at the diagram and give me an honest answer.

[Jerry Peck]

In the series circuit, what is the power in Load 1?

120w = 120v * 1a

So what is the voltage across Load 1?


Ok Jerry, let me try to explain this another way.

Good, another way would be better as you've answered you own question in the above.


Let's say you cannot see "inside the box" because it's a locked panel and you have no idea what is in there.

What would you say is the total power for loads 1 and 2 on circuit B. Just look at the diagram and give me an honest answer.

First thing you do is use a VOM and check the voltage across each load and across both loads, getting 120 volts, 120 volts, and 240 volts (talking about Circuit B here).

The second thing you do is use a clamp-on ammeter to read the current through Load 1 and through Load 2.

Now, though, even doing that would not tell you whether or not they were in series unless you also knew the supply voltage (which we do not know because it is hidden inside the box).

Based on your drawing, BOTH Circuit A and Circuit B would show 120 volts at each load, BOTH Circuit A and Circuit B would show the same 200 amps current through each load.

The only way to find out if one was in series or not would be to disconnect one load, in the series Circuit B (as drawn with no common grounded connection), the other load would also 'shut off', whereas in the parallel Circuit A, the other load would 'stay on' ... HOWEVER ... if Circuit B had a common grounded neutral point, then the other load would also 'stay on' - ruining your ability to determine what was what based on those tests.

Now, back to your question:

What would you say is the total power for loads 1 and 2 on circuit B. Just look at the diagram and give me an honest answer.

For Circuit B *AND FOR CIRCUIT A*, the "total power for loads 1 and 2" would be 24,000 VA + 24,000 VA = 48,000 VA. If the loads were purely resistive loads, that would also be 24,000 watts + 24,000 watts = 48,000 watts ... REGARDLESS which circuit was being referred to, Circuit A or Circuit B - the "total power" (which is what you asked for) would be the same.

I've been saying that all along, so I'm not sure why you are asking that question.

Now it is *my turn* to ask questions (the cover to *the box* has been removed to allow you to see and check what is inside *the box*):

a) In Circuit A, what is total current flow through Load 1? Through Load 2? Through Loads 1 and 2?

b) In Circuit B, what is the total current flow through Load 1? Through Load 2?, Through Loads 1 and 2?

[Corn Walker]

What a fun thread to reply to for my first post.

I think you need to get away from measuring the voltages between each load and common. You will always measure 120V between each leg and common and 240V between the two legs.

What you need to do is follow the electrons, and here is where phase is important and what it means to "balance" the panel.

If you have a 1A load attached from leg A (0 degree phase) and common 1A worth of electrons will travel from leg A to common. If you place a second 1A load on leg A you will now have 2A worth of electrons traveling from leg A to common.

Okay, now lets put a 1A load attached from leg B (180 degree phase) to common. Now what you have is 1A worth of electrons traveling from leg A to common and 1A worth of electrons traveling from leg A to leg B. You do NOT have 3A traveling from A and B onto common.

In a balanced panel all of the electrons are flowing from leg A to leg B and none are flowing back on common. Of course, in an AC system, these voltages flip direction every few milliseconds (so you have electrons occasionally flowing from B back to A) but the principle is not changed because of it.

Corn Walker
(former Physics Teacher)

[Tim Voss]

a) total current flow through load 1 = 200a, total current flow through load 2 = 200a, total current in circuit A = 400a
b) total current flow through load 1 = 200a, total current flow through load 2 = 200a, total current in circuit B = 200a

But you didn't really answer my question.



First thing you do is use a VOM and check the voltage across each load and across both loads, getting 120 volts, 120 volts, and 240 volts (talking about Circuit B here).



I don't think you are really going to try and measure the voltage across both loads, nor start disconnecting them to see if they are in series or parallel. That is completely impractical in a real situation. In my example there are only 2 loads, but if there where 30 circuits in a panel that is 2.6525286 &#215; 10^32 combinations you would have to measure.

No, I think you would use a VOM and check the voltage across each load (only).

And you have measured the current with your clamp-on ammeter.

Then what would you do?

[Jerry Peck]

But you didn't really answer my question.

I did.

You asked: "What would you say is the total power for loads 1 and 2 on circuit B. Just look at the diagram and give me an honest answer."

I answered (I'll add bold and underlining for you): "For Circuit B *AND FOR CIRCUIT A*, the "total power for loads 1 and 2" would be 24,000 VA + 24,000 VA = 48,000 VA. If the loads were purely resistive loads, that would also be 24,000 watts + 24,000 watts = 48,000 watts ... REGARDLESS which circuit was being referred to, Circuit A or Circuit B - the "total power" (which is what you asked for) would be the same."

[Tim Voss]

Ok, that is correct, but show your work please.

How did you calculate 24,000 VA for Load 1 and 2?

[Jerry Peck]

Ok, that is correct,

I know it is correct.


but show your work please.

How did you calculate 24,000 VA for Load 1 and 2?

Your drawing shows 200 amps flowing through a load with 120 volts across it.

120 volts times 200 amps equals 24,000 volt-amps

The rest I already showed you. :)

But here it is AGAIN.

24,000 VA + 24,000 VA = 48,000 VA

I know you are trying to eventually show that you get "400 amps", but you don't.

You can keep playing these games, the outcome will never be "400 amps". :D

Because you never have "400 amps" on that circuit we were discussing. :rolleyes:

Are you trying to use up another 100 posts having me show you that?

[Tim Voss]

Yes, I'm trying to help you "see" the 400amps, as you requested.

Ok, so now you have said:

(120v * 200a) + (120v * 200a) is the total power for circuit B.

So we have two 120v loads on circuit B using 200amps each. Now if someone said, "I have two 120v loads here. How much amps at 120v do I need to power these loads? Hmmm, looks like I need 200amps for the first one and another 200amps for the second one."

Would you not say, "You need 400amps at 120v to power those loads"?

[Corn Walker]

The way I would explain (incorrectly - but it helped to get the point across) power phasing to my students is that you have 120V pushing 200A through the circuit and -120V pulling that same 200A through the circuit. The total amount of Power available in the circuit to perform any work is 48000VA. That is NOT to say you have 400A total current flowing though the circuit - you don't.

If you start with a 120V 200A load placed on one leg the load will be referenced to common (we'll say this leg pushes 200A onto common). If you then add a 120V 200A load to the second leg then this second circuit doesn't push an additional 200A onto common; instead it pulls the same 200A through the second load that was pushed through the first. The combined push/pull force is additive (240V) allowing you to do more work (48000VA) with the same 200A (assuming ideal resistive loads).

You can say you have two 120V 200A loads on the circuit but you can not say you have 400A on the circuit because the amperages on the load side of the circuit can't be added, only the force or the power.

Corn Walker
(former Math teacher too)

[Jerry Peck]

Yes, I'm trying to help you "see" the 400amps, as you requested.

There is nothing to "see", because there are no "400 amps" anywhere there.


Ok, so now you have said:

(120v * 200a) + (120v * 200a) is the total power for circuit B.

So we have two 120v loads on circuit B using 200amps each. Now if someone said, "I have two 120v loads here. How much amps at 120v do I need to power these loads? Hmmm, looks like I need 200amps for the first one and another 200amps for the second one."

That's where you become incorrect, ending up totally incorrect here ...


Would you not say, "You need 400amps at 120v to power those loads"?

No. (This is the part where you "ending up totally incorrect".)

See, first, I would have to know how the circuits were wired (see "That's where you become incorrect" above and your quote above it), then I would ...

... AS I HAVE BEEN for the past 100 posts or so ...

... explain that THERE ARE ONLY 200 amps in that circuit.

Do you understand yet? Or do we need to try for 200 posts?

[Tim Voss]

I'm not saying I have 400amps on the circuit, I'm saying I have two 120v 200amp loads on the circuit and that is equal to 400amps of 120v loads.

[Jerry Peck]

The way I would explain (incorrectly - but it helped to get the point across) power phasing to my students is that you have 120V pushing 200A through the circuit and -120V pulling that same 200A through the circuit. The total amount of Power available in the circuit to perform any work is 48000VA. That is NOT to say you have 400A total current flowing though the circuit - you don't.

If you start with a 120V 200A load placed on one leg the load will be referenced to common (we'll say this leg pushes 200A onto common). If you then add a 120V 200A load to the second leg then this second circuit doesn't push an additional 200A onto common; instead it pulls the same 200A through the second load that was pushed through the first. The combined push/pull force is additive (240V) allowing you to do more work (48000VA) with the same 200A (assuming ideal resistive loads).

You can say you have two 120V 200A loads on the circuit but you can not say you have 400A on the circuit because the amperages on the load side of the circuit can't be added, only the force or the power.

Corn Walker
(former Math teacher too)

Maybe you can do a better job of explaining that to Tim than I have been able to ... I've never been a Physics Teacher nor a Math Teacher - maybe that's my problem? :D

[Carl Eisen]

Jerry,

You must have the patients of a Saint. I would have thrown in the towel a hundred post or so ago.

[Tim Voss]

Jerry,

How do you explain the fact that generator manufacturers agree with me and list the capacity of their generators in BOTH 120v and 240v amps?

Baldor Generator 125amps at 240v OR 250amps at 120v
Honda Generator 20.8amps at 240v OR 41.6amps at 120v
Grainger Generator 41.6amps at 240v OR 83.2amps at 120v

[Jerry Peck]

I'm not saying I have 400amps on the circuit, I'm saying I have two 120v 200amp loads on the circuit and that is equal to 400amps of 120v loads.

You don't, though,

You have two 120 volt loads (you agree with that) at 200 amps (you agree with that), and you thus have two 120 volt loads on a 240 volt system with 200 amps.

As as as ... and this is where you veer off into netherland ... you want to start comparing those two 200 amp 120 volt loads on the 240 volt circuit to two 200 amps 120 volt loads on two 102 volt circuits?

You need to switch to ... and we've been here before, even you have ... power ratings.

I.e., each 200 amp 120 volt load has a power rating of 24,000 VA, two of them have a power rating of 48,000 VA. That does not change regardless if they are connected to two 120 volt circuits or are connected across one 240 volt circuit - THE POWER does not change.

In one case (your parallel circuits) you are drawing 200 amps + 200 amps = 400 amps. You have two circuits with 24,000 VA on each, or 48,000 VA.

In the other case (the one where we started and are still discussing - the series 240 volt circuit) you have 200 amps ... period. You have one circuit with 48,000 VA on it.

The POWER does not change.

[Tim Voss]

Corn Walker,

All I'm trying to say is that the force of the power is equal to 400amps at 120volts. Is that not true?

[Tim Voss]

Jerry,

How do the loads "know" they are on a 240v circuit? They are 120v loads. ( Not to get sidetracked, please answer my question about the generators.)

[Jerry Peck]

Jerry,

How do you explain the fact that generator manufacturers agree with me and list the capacity of their generators in BOTH 120v and 240v amps?

Baldor Generator 125amps at 240v OR 250amps at 120v
Honda Generator 20.8amps at 240v OR 41.6amps at 120v
Grainger Generator 41.6amps at 240v OR 83.2amps at 120v

Tim,

They are not agreeing with you.

They are stating the MAXIMUM CAPACITY of the generator, how you hook it up is up to you.

They are saying that *IF* you hook it up as in your parallel circuit drawing, you will get twice the current at half the voltage than *IF* you hook it up in the series circuit, where you will get twice the voltage and half the current.

You will notice they are talking about the maximum POWER the generator is capable of, and POWER does not change (see my other post above), and then giving you the two options of how you might wire it up (parallel - 120 volt circuits, or series - 240 volt circuit).

[Jerry Peck]

All I'm trying to say is that the force of the power is equal to 400amps at 120volts. Is that not true?


Tim,

Now you are talking POWER again.

Make up your mind. :D

POWER does not change, the voltage doubles, the current drops by half, and vice versa.

You cannot, however, say you are getting 400 amps through two 200 amps loads at 120 volts each on a 240 volt circuit (which is where all of this started).

[Tim Voss]

No Jerry, these are service backup generators that are hooked up as split-phase 240v emergency power backup. And yet, for the Baldor generator you can get 250a of 120v out of that 240v service.

How do you explain that?

[Corn Walker]

All I'm trying to say is that the force of the power is equal to 400amps at 120volts. Is that not true?

When I taught I would always tell my students to check their units.

Volts is a measurement of force
Amperes is a measurement of charge/sec
Power is a measurement of force*charge/sec

If you're talking about abstract numbers outside of the system then you can say 400A at 120V (i.e. 48000VA) is the same as 200A at 240V and is the same as 4800A at 10V which is the same as 48000A at 1V (Quiz: what size Cu wire is required for this assuming a 100m length?). Whenever you say this, however, you aren't talking about force or volume but power (aka work). The force in the first example is 120V and the force in the last example is 1V. The power in both examples is the same: 48000VA.

*Within* a system, however, you can't abstract charge/sec from force. You can't talk about the amperes in the system without also talking about the volts required to move those amps. In the case of a main panel with 200A service, we're talking about 200A at 240V. If you attach a 200A*120V load to one leg and then balance it with another 200A*120V load on the other leg, you are moving 200A*240V. In your circuit when you attach the second load you change the phase angle of the force vector, giving you 240V force for the combined loads. The current stays the same - 200A.

It is convenient when we talk about circuits to refer only to amps because the voltage is implied in a residential panel (not an assumption I'd make in a commercial panel). It is nonetheless incorrect to refer to only amps when doing the math to calculate the total load of a panel. When adding loads you need to consider the phase angle of the load to determine how much current will travel over each leg and common. 130A*120V at 0 deg + 70A*120V at 180 deg results in 70A*240V phase balanced power + 60A*120V unbalanced power.

Theoretically the 4/0 supply cable could have an undersized common conductor (2AWG - much like they do with ground conductors) with the assumption that your panel will never be more than 50&#37; out of balance at full load. In practice, however, it is a safer bet to assume a 100% out of balance panel and size the conductor the same as the mains.

As for the generators, they are talking about the amount of power they can generate (e.g. 4800VA) and explaining it in terms the user will likely understand (40A at 120V or 20A at 240V). That is, if the user connects a 40A*120V load to the generator it will reliably produce the required power. If the user connects a 25A*240V load to the generator they can expect a voltage drop (or the circuit to open if it has an overload protector).

As a practical analogy, let's assume I can push 200 pounds up onto the 1st story roof using a pole and you can pull 200 pounds up onto the second story roof using a rope. If we were both on the ground pushing or both on the second story roof pulling we could move 400 lbs from the ground to the first story or from the first story to the second. But because I'm on the ground and you're on the roof (180 deg phase separation) we can only each move the 200 pounds one story or together move the 200 pounds from the ground to the second story.

Corn Walker

[David Banks]

Corn Walker. What a cool name.

[Tim Voss]

Corn Walker,



As for the generators, they are talking about the amount of power they can generate (e.g. 4800VA) and explaining it in terms the user will likely understand (40A at 120V or 20A at 240V). That is, if the user connects a 40A*120V load to the generator it will reliably produce the required power. If the user connects a 25A*240V load to the generator they can expect a voltage drop (or the circuit to open if it has an overload protector).



That's ALL I've been trying to say. Jerry says you CAN'T say that. Have you read the whole thread from the beginning? Are you saying I can connect 400amps of 120volt devices to a 200amp 240volt split-phase service? (Better watch out what you say, or Jerry will get you.)

[James Duffin]

The power company bills you kw you use per hour (khw). They don't care if the loads are 120 volt or 240 volt. The charge is the same for a 2-pole 20 amp breaker or two 1-pole 20 amp breakers if the breakers are all loaded to the max.

[Jerry Peck]

That's ALL I've been trying to say. Jerry says you CAN'T say that.

If I understand what Corn Walker said, he also said you can't say that.

(underlining is mine)

If you're talking about abstract numbers outside of the system then you can say 400A at 120V (i.e. 48000VA) is the same as 200A at 240V and is the same as 4800A at 10V which is the same as 48000A at 1V (Quiz: what size Cu wire is required for this assuming a 100m length?). Whenever you say this, however, you aren't talking about force or volume but power (aka work). The force in the first example is 120V and the force in the last example is 1V. The power in both examples is the same: 48000VA.

*Within* a system, however, you can't abstract charge/sec from force. You can't talk about the amperes in the system without also talking about the volts required to move those amps. In the case of a main panel with 200A service, we're talking about 200A at 240V. If you attach a 200A*120V load to one leg and then balance it with another 200A*120V load on the other leg, you are moving 200A*240V. In your circuit when you attach the second load you change the phase angle of the force vector, giving you 240V force for the combined loads. The current stays the same - 200A.

[James Duffin]

The volts don't matter in this application. You have the same wattage available at all times. In my area 48KW cost you $4.80 per hour no matter if you have a 200 amp 240 welder connected or 240 two hundred watt 120 volt light bulbs connected.

[Tim Voss]

Actually, I thought Corn Walker was agreeing with me. He said:



As for the generators, they are talking about the amount of power they can generate (e.g. 4800VA) and explaining it in terms the user will likely understand (40A at 120V or 20A at 240V). That is, if the user connects a 40A*120V load to the generator it will reliably produce the required power.


Seems like he is saying it is correct to say it both ways.

So how can Baldor Generators say 250amps at 120volts when they only have a double pole 125amp breaker on their 240v generator?

[Corn Walker]

So how can Baldor Generators say 250amps at 120volts when they only have a double pole 125amp breaker on their 240v generator?

Explaining how auto-switching circuits work is perhaps a bit beyond this topic but suffice it to say they operate differently than the utility feeds in a typical home. The Baldor generators have the ability to supply either 120V or 240V to a standard L1430R receptacle. Once you add voltage switching circuitry to the mix you're no longer comparing apples to apples.

Still they're not exactly being precise here - but they don't need to be. It would be exceedingly difficult to connect a total load of 250A to a single 120V leg.
The two 125A breakers are on independent power legs (fed by independent generator coils) that are kept 180 deg out of phase with each other to provide split-phase power. Baldor is answering a consumer question here: how much stuff can I run with this generator? The consumer will look at the name plates on the table saw, work lights, air compressor, and battery chargers (all of which operate at 120V) and add up the current draw of those devices. They get a number (62A) and see that it's less than the rating of the generator and they're good to go.

Do note, btw, that the primary unit by which generator power output is rated is watts, not VA.

[Tim Voss]

They are not supplying 250a to a single 120v leg, they are supplying 125a to one leg and 125a to the other leg. The unit doesn't have a receptacle, it is hardwired to the service panel and operates just like split-phase utility power.



Baldor is answering a consumer question here: how much stuff can I run with this generator? The consumer will look at the name plates on the table saw, work lights, air compressor, and battery chargers (all of which operate at 120V) and add up the current draw of those devices. They get a number (62A) and see that it's less than the rating of the generator and they're good to go.



That is EXACTLY the question I was answering. You can add up all the amp ratings on the name plates of your 120v equipment. And if it is less than 400A, then you can run that on your 200A 240V service, correct?

Jerry says I can't say it that way.

[Corn Walker]

That is EXACTLY the question I was answering. You can add up all the amp ratings on the name plates of your 120v equipment. And if it is less than 400A, then you can run that on your 200A 240V service, correct?

Jerry says I can't say it that way.

Well... it depends. :)

That's where the whole balancing thing comes into play. Once you reach 1/2 of the rated capacity you need to be concerned about where you're connecting those loads. If the loads were all 240V there'd be no problem since they would automatically balance across the two service legs and you'd have up to 200A of current draw to play with. 120V loads, on the other hand, need to be planned (hence why electrical inspectors require load calculations from the electrician laying out the panel).

If you were to ask me, "can I run 205A of 120V equipment on a 240V 200A service?" I'd still have to say it depends. There's no definite answer you can give to such a question - you need to know where each of the loads is to be connected. Less than the rated capacity (say 198A total of 120V equipment) and you can be unequivocal - above that you need to be cautious.

Now practically speaking, and bringing this back to home inspection, I would not be concerned if I saw 400A worth of breakers on a 200A Mains panel. However I would be concerned if I saw electric space heaters in every room since many homes share a single circuit for two or three bedrooms. In my own house every bedroom has 12/2 home run back to the panel for receptacles (on AFCI breakers at that) and an additional 12/2 home run for lighting serving no more than two rooms. Perhaps it was a overreaction to what was there before - two 20A and two 15A circuits feeding the entire house.

Corn Walker
tentatively unequivocal

[Tim Voss]

Right, I left out "Perfectly Balanced". We covered that before. So, yes, then, as long as it is balanced.

[Jerry Peck]

If you're talking about abstract numbers outside of the system ...

*Within* a system, however, you can't abstract charge/sec from force.

Tim,

You and Corn Walker can sort out the abstract numbers, it is getting to abstract for me, since we were initially talking about within the system.

Let me know if Corn Walker makes you feel good about your position :cool: ... I'm abstracting out of here ... :D

[Brandon Chew]

This thread is like a vampire in a bad horror film. No matter how many times you kill it, it just won't die. :)

[Tim Voss]

Ok, Jerry, but the part about running 400A of nameplate rated 120V equipment on the 200A 240V service is real. I guess since you already agreed that was possible (but to be fair, meaningless) it's a good time for you to "abstract" out.

[Jerry Peck]

it's a good time for you to "abstract" out.

*I* am "abstracting out", as I said, however, *YOU* have already "abstracted out". :D

As soon as you stepped out of talking about what we were talking about, i.e., "within the circuits" as Corn Walker put it, you "abstracted out" ... go for it man.

96,000 volts at 0.50 amps = 48,000 VA.

As does 1 volt at 48,000 amps. :rolleyes:

[Tim Voss]

No, I'm still talking about the same 200A 240V split-phase service we started with. The same service you said you could put 400 name plate amps of 120v devices on.

[Corn Walker]

Only 14 more posts after this one and we can hit 200. Come on team, we can do it! :p

[Bruce King]

I thought you could only hit 200 posts on the subject of "how to test a garage door"

[Jim Zborowski]

#188
Corn, I like your annalogy.

[James Duffin]

I'll help reach the 200 mark.

I still say that the volts and amps don't matter. The PC transformer determines how much power is available. On a 200 amp service it is 48KW at 120 volt, 240 volt, or a combination of both.