Even loads on branch circuits

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crtemp

Senior Member
Location
Wa state
I got tagged the other day and this made no sense to me. I had 3 bedrooms in a house and the total square footage totaled 493 sq ft for the 3 bedrooms. Here in Washington we are only required to arc fault bedrooms and bedroom closets. I only installed one arc fault circuit. The inspector wrote me up for this..

Where the load is calculated on the basis of volt-amperes per square meter or per square foot, the wiring system up to and including the branch-circuit panelboard(s) shall be provided to serve not less than the calculated load. This load shall be evenly proportioned among multioutlet branch circuits within the panelboard(s). Branch-circuit overcurrent devices and circuits shall be required to be installed only to serve the connected load.


He said I was only allowed 2 bedrooms per arc fault (which is not in the NEC nor in the WAC (Washington administrative code)). He said it is just something that the inspectors came up with on their own. Not sure how I am suppose to know about it then. He then told me that I did not have my circuit loads evenly proportioned. The house total square footage was 1192 sq ft. I had installed three 15 amp lighting circuits for the house and one 15 amp circuit for the garage lights and plugs.

I had read another post a few weeks back that had mentioned that the small appliance circuits had to feed at least 2 counter top plugs. Basically just said that I could have one plug on the counter top on one circuit and then the remaining plugs on the other circuit and still be code compliant. As long as the counter top plugs were served by at least 2 circuits it did not matter how they were split up. Is this true?

Also why does the inspector not have an issue with my garage circuit not having a load similar as the rest of the house? It only has 1 keyless light fixture and 1 gfci plug on it. I don't even understand why it matters if all the circuits in the panel are evenly proportioned. What is the big deal?
 

jxofaltrds

Inspector Mike®
Location
Mike P. Columbus Ohio
Occupation
ESI, PI, RBO
I got tagged the other day and this made no sense to me. I had 3 bedrooms in a house and the total square footage totaled 493 sq ft for the 3 bedrooms. Here in Washington we are only required to arc fault bedrooms and bedroom closets. I only installed one arc fault circuit. The inspector wrote me up for this..

Where the load is calculated on the basis of volt-amperes per square meter or per square foot, the wiring system up to and including the branch-circuit panelboard(s) shall be provided to serve not less than the calculated load. This load shall be evenly proportioned among multioutlet branch circuits within the panelboard(s). Branch-circuit overcurrent devices and circuits shall be required to be installed only to serve the connected load.


He said I was only allowed 2 bedrooms per arc fault (which is not in the NEC nor in the WAC (Washington administrative code)). He said it is just something that the inspectors came up with on their own. Not sure how I am suppose to know about it then. He then told me that I did not have my circuit loads evenly proportioned. The house total square footage was 1192 sq ft. I had installed three 15 amp lighting circuits for the house and one 15 amp circuit for the garage lights and plugs.

I had read another post a few weeks back that had mentioned that the small appliance circuits had to feed at least 2 counter top plugs. Basically just said that I could have one plug on the counter top on one circuit and then the remaining plugs on the other circuit and still be code compliant. As long as the counter top plugs were served by at least 2 circuits it did not matter how they were split up. Is this true?

Also why does the inspector not have an issue with my garage circuit not having a load similar as the rest of the house? It only has 1 keyless light fixture and 1 gfci plug on it. I don't even understand why it matters if all the circuits in the panel are evenly proportioned. What is the big deal?

Let's see:

2 bedrooms at 100SF each = 200 X 3VA = 600VA ----- 2 AFCI limit - OK

2 bedrooms at 200SF each = 400 X 3VA = 800VA ----- 2 AFCIs? - OK or NOT OK

Does he have a chart that you can use?

1 AFCI = XXX SF

or

1 AFCI = XXX VA
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
The inspector is flat out wrong on all counts. There is no limit on the number of rooms any circuit can supply, and inspectors cannot make up rules on their own. I would look at the total number of lights and plugs that are connected on the one circuit. So long as it is within the rating of the AFCI breaker, you would be compliant. That said, as a homeowner I would prefer to have a separate circuit. I would not like to have to turn off every bedroom in the house, in order to do work on one circuit. But that is not a code requirement.

Also, I think there is a rule about balancing the loads on the two (or three) phases within any given panel. But I just took a look and could not find it. However, that rule has nothing to do with having one circuit be loaded differently than any other circuit. If the 15 amp circuit (let us say on phase A) is lightly loaded, but if you arrange the loading on the other circuits to provide an overall balance, then you are compliant.
 

iceworm

Curmudgeon still using printed IEEE Color Books
Location
North of the 65 parallel
Occupation
EE (Field - as little design as possible)
... He said it is just something that the inspectors came up with on their own. Not sure how I am suppose to know about it then. ...
JMO:
Generally speaking - you're fubared. Once he puffs up and pulls his badge and gun, it is cost prohibitive to back them down - not to mention all the hate and discontent you get on your next job. You being right doesn't change the costs.

Just curious - did he play the, "It's for the safety of the children." card.

ice
 

suemarkp

Senior Member
Location
Kent, WA
Occupation
Retired Engineer
I could argue this a couple of ways. First, with 1192 sq ft of space, you need to supply 3576 VA. That is 29.8A at 120V, so you'd need 2 general lighting circuits. With 2 circuits, one of them could serve all the bedroom space (up to 596 sq ft of space per circuit) and that would be the AFCI breaker. However, you put in 4 circuits. Does that mean you can only serve 298 sq ft per circuit? Using that mentality, I could see how he'd say you need 2 circuits to cover the bedroom space.

I've never heard of anyone making sure the circuits are evenly apportioned, but its another nebulous portion of the code.

You could ask him if you remove two 15A circuits and merge them onto the remaining two 15A circuits if he'd be happier? That would be a case of code compliant and consumer getting a worse design.
 

Smart $

Esteemed Member
Location
Ohio
The inspector is wrong.

For dwelling units, the load of general-use receptacles are included in general lighting.

1192ft? ? 3VA/ft? = 3576VA

3576VA ? 120V = 29.8A

29.8A ? 15A/ckt = 2 ckt

As long as you 2 or more lighting and receptacle circuits (and none are overloaded), you have met the requirement.

Code does require the calculated load be evenly apportioned among the branch circuits. The rub is Code says nothing about calculating receptacle load vs lighting load. So technically, dwelling unit general-use receptacles have no load in apportioning branch circuit loads evenly. Where lighting circuits are separate from general-use receptacle circuits, the lighting circuits alone must meet the minimum number and the load be evenly apportioned.
 
The inspector is flat out wrong on all counts. There is no limit on the number of rooms any circuit can supply, and inspectors cannot make up rules on their own. I would look at the total number of lights and plugs that are connected on the one circuit. So long as it is within the rating of the AFCI breaker, you would be compliant. That said, as a homeowner I would prefer to have a separate circuit. I would not like to have to turn off every bedroom in the house, in order to do work on one circuit. But that is not a code requirement.

Also, I think there is a rule about balancing the loads on the two (or three) phases within any given panel. But I just took a look and could not find it. However, that rule has nothing to do with having one circuit be loaded differently than any other circuit. If the 15 amp circuit (let us say on phase A) is lightly loaded, but if you arrange the loading on the other circuits to provide an overall balance, then you are compliant.

Being the devil's advocate here, just want to probe this a bit more as I have always been curious about this: You calculate the load per square foot. If the load must be evenly proportioned among the branch circuits, than each branch circuit should supply (square feet/number of circuits) square feet right? Sure, how picky and ridiculous can the inspector be, but within reason it doesnt seem out of line for him to say the load is not evenly divided right? Ok unleash the hounds, tell me what I am missing.
 

Smart $

Esteemed Member
Location
Ohio
Being the devil's advocate here, just want to probe this a bit more as I have always been curious about this: You calculate the load per square foot. If the load must be evenly proportioned among the branch circuits, than each branch circuit should supply (square feet/number of circuits) square feet right? Sure, how picky and ridiculous can the inspector be, but within reason it doesnt seem out of line for him to say the load is not evenly divided right? Ok unleash the hounds, tell me what I am missing.
I believe it to be a common misconception that the load be apportioned by ft?/circuits. Fact is, it is apportioned by (total general lighting load)/(general lighting circuits*). Nothing says one portion of the total area can't have a higher VA/ft? than another portion (which would end up being a lower VA/ft? to average out the total).

*includes combination general lighting and general-use receptacle circuits, but not general-use receptacle only circuits... IMO
 
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jxofaltrds

Inspector Mike®
Location
Mike P. Columbus Ohio
Occupation
ESI, PI, RBO
210.11 Branch Circuits Required.

(B) Load Evenly Proportioned Among Branch Circuits.
Where the load is calculated on the basis of volt-amperes per square meter or per square foot, the wiring system up to and including the branch-circuit panelboard(s) shall be provided to serve not less than the calculated load. This load shall be evenly proportioned among multioutlet branch circuits within the panelboard(s). Branch-circuit overcurrent devices and circuits shall be required to be installed only to serve the connected load.

The "Evenly Proportioned" ends at the panelboard?
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
That means, IMHO, that you do not have to proportion two branch circuits originating in different sub-panels.

Sent from my XT1080 using Tapatalk
 

Smart $

Esteemed Member
Location
Ohio
210.11 Branch Circuits Required.

(B) Load Evenly Proportioned Among Branch Circuits.
Where the load is calculated on the basis of volt-amperes per square meter or per square foot, the wiring system up to and including the branch-circuit panelboard(s) shall be provided to serve not less than the calculated load. This load shall be evenly proportioned among multioutlet branch circuits within the panelboard(s). Branch-circuit overcurrent devices and circuits shall be required to be installed only to serve the connected load.

The "Evenly Proportioned" ends at the panelboard?

That means, IMHO, that you do not have to proportion two branch circuits originating in different sub-panels.

Sent from my XT1080 using Tapatalk
Correct.

Note a "portion" of the total area has to be assigned a panelboard to perform Article 220 calculations. However, that does not mean the total area must be equally apportioned among multiple panelboards.
 
Correct.v

Note a "portion" of the total area has to be assigned a panelboard to perform Article 220 calculations. However, that does not mean the total area must be equally apportioned among multiple panelboards.

Ok I reread that a bunch of times and I amend my interpretation a little. It seems to me that - technically - you would find the square footage served by each panelboard, get your VA number from that, and divide that evenly among the number of circuits you are providing in that panelboard. Thus in each panelboard, each circuit corresponds to X square feet or 3X VA. Panelboard "B" could have a different square footage per circuit than panelboard "A". I think the ambiguity comes in when you have two branch circuits serving the same room - then it breaks down because there is no way to define the square footage served be each circuit.
 

Smart $

Esteemed Member
Location
Ohio
... I think the ambiguity comes in when you have two branch circuits serving the same room - then it breaks down because there is no way to define the square footage served be each circuit.
Where supplied by the same panel, it doesn't matter. It's only an issue if the room's general lighting is being supplied by two panels. In that case, you divvy up the area of the room by the percentage of connected load. :blink::D
 
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