Two Voltages in same Box

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Keri_WW

Keri_WW

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I have a double gang junction box which will house two switches, one controlling main lights which are at 277V and the other controlling decorative lighting at 120V. I know NEC 300.3 (C) (1) says that they can occupy the same enclosure, but do I need to add any kind of identification to the switchplate letting people know that there are two different sources?

Thanks,
Keri :grin::grin:
 
The way I read that is that I am ok without doing anything special since the voltage difference isn't up to 300V. Is it common practice to use barriers within the junction box anyway?
 
Keri_WW said:
The way I read that is that I am ok without doing anything special since the voltage difference isn't up to 300V. Is it common practice to use barriers within the junction box anyway?
Click to expand...

Keri if you take a volt meter between the 277v and the 120v you have exceeded the 300v potential difference so a barrier is required per 404.8(B)
 
If you measured the hot legs of the two circuits what do you think you will read?

You can buy a metal box and matching mud ring, the box has indents to hold the isolation blade that slot into the mud ring. Even plastic boxes have the same application!

Glad you stated 404.8(B), the OP can also look at really all off 210.5 but really210.5(D) as a refresher cause I was thinking it / knowing it existed but not all the way back @ 404.8(B). :roll:
 
Agreed.

voltagebarrier.jpg
 
I think it's sort of obvious that you place a barrier. Id hate to see someone touch those two terminals together. What a bang that would make.
 
I'm stumped

I'm stumped

Geez, just when you think you've got a handle on voltage readings, calculations and theory this comes up. How does one arrive at 381v from reading across 277v and 120v?
 
mark32 said:
Geez, just when you think you've got a handle on voltage readings, calculations and theory this comes up. How does one arrive at 381v from reading across 277v and 120v?
Click to expand...

See this thread for some help.

Roger
 
Keri_WW said:
The way I read that is that I am ok without doing anything special since the voltage difference isn't up to 300V.
Click to expand...

dcspector said:
Keri if you take a volt meter between the 277v and the 120v you have exceeded the 300v potential difference . . .
Click to expand...
The voltage between conductors of two different grounded sources like this depends on which phase of each supply each conductor is on. You can actually change this difference by swapping phases on either or both circuits.
 
Wow, vectors bite me in the tail once again!

Thanks for all of the replies. The barrier is definitely the way to go!

Keri:grin::grin:
 
Thank you Roger for the link to the older thread.

To Quote Don in post #17


"Note my above post is not correct, the voltage between one phase of the 480/277 volt system and a different phase of the 208/120 volt system would be ~356 volts if there is a 120 degree difference. In most cases the 208/120 volt system would be supplied by a delta/wye transformer that would have its primary connected to the 480 volt supply. In this case there would be an additional 30 degree phase shift from the delta wye transformer and if this phase shift is positive, there will be a 150 degree difference between A phase of one system and B phase of the other system. This would result in ~385 volts between the ungrounded conductors. If the phase shift was negative, there would be ~301 volts. The resultant voltage is found by using the rule of cosines to add the voltage vectors.
c^2 =a^2 + b^2 -2abCOS(angle)
c^2= c squared
c is the resultant voltage
a is 120 volts
b is 277 volts
Don"

Then in post #19 you posted a graph from Smart. I understand the 277v and 120v lines being drawn the way they are (Being 120 degrees apart) but it's been awhile since school and I'm a little rusty with my math. Could you or anyone work Don's formula posted above to get at least one of the numbers presented in smarts graphic? Here's a link to the thread,

http://forums.mikeholt.com/showthread.php?t=86550&highlight=404.8(B)&page=2


Thanks, Mark
 
mark32 said:
Thank you Roger for the link to the older thread.

To Quote Don in post #17


"Note my above post is not correct, the voltage between one phase of the 480/277 volt system and a different phase of the 208/120 volt system would be ~356 volts if there is a 120 degree difference. In most cases the 208/120 volt system would be supplied by a delta/wye transformer that would have its primary connected to the 480 volt supply. In this case there would be an additional 30 degree phase shift from the delta wye transformer and if this phase shift is positive, there will be a 150 degree difference between A phase of one system and B phase of the other system. This would result in ~385 volts between the ungrounded conductors. If the phase shift was negative, there would be ~301 volts. The resultant voltage is found by using the rule of cosines to add the voltage vectors.
c^2 =a^2 + b^2 -2abCOS(angle)
c^2= c squared
c is the resultant voltage
a is 120 volts
b is 277 volts
Don"

Then in post #19 you posted a graph from Smart. I understand the 277v and 120v lines being drawn the way they are (Being 120 degrees apart) but it's been awhile since school and I'm a little rusty with my math. Could you or anyone work Don's formula posted above to get at least one of the numbers presented in smarts graphic? Here's a link to the thread,

http://forums.mikeholt.com/showthread.php?t=86550&highlight=404.8(B)&page=2


Thanks, Mark
Click to expand...
The 150? displacement in my text is the same as the 210? in Smart's graphic.
voltage a=120 volts
voltage b=277 volts
c^2=120^2 + 277^2 -2(120)(277)(COS 210)
c^2=14400 +76729 - 2(33240)(-0.866)
c^2 =91129 + 57571.68
c^2 = 140700.68
c = 385.62
 
As chairman of the stupid question department, it is my duty to ask... I suppose you need a barrier if the 277 v lights are on different phases? ... which would possibly lead you to group switches of the same phases in different boxes. Would that cause any inductive heating issues?

my duty is done.
 
realolman said:
As chairman of the stupid question department, it is my duty to ask... I suppose you need a barrier if the 277 v lights are on different phases? ... which would possibly lead you to group switches of the same phases in different boxes. Would that cause any inductive heating issues?

my duty is done.
Click to expand...

Yes


What some who have not done this before should be aware of, is the entry of the cables into the enclosure. When a barrier would be necessary, the arrangement of the conductors is restricted by the placement of the barrier in the box. This takes planning before the cables are installed to the lights and from to the supply.
 
don_resqcapt19 said:
The 150? displacement in my text is the same as the 210? in Smart's graphic.
voltage a=120 volts
voltage b=277 volts
c^2=120^2 + 277^2 -2(120)(277)(COS 210)
c^2=14400 +76729 - 2(33240)(-0.866)
c^2 =91129 + 57571.68
c^2 = 140700.68
c = 385.62
Click to expand...


Thank you very much Don for your help, just another question if you don't mind. You used 150 degrees in your example by adding 30 to 120, I am guessing smart used 180 and added 30 to that to arrive at 210. It was my understanding that the phase shift is 90 degrees when a voltage is induced, did I overlook the 30 degree shift somewhere in my studies?


ps, I'm sure you are aware the # above would be 148700.68 instead of 140700.68

Thanks again,
Mark
 
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