Neutral calclation

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Arkansas74

Arkansas74

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Ozark,arkansas
Hey guys if I have a 3 phase pull a roundhouse for 120 drops and each one is drawing 10 amps. If I turn one breaker off and have 10 amps on each phase and 10 amps on a an 10 on b and c is off my neutral will have 10 amps on it then. the way my calculations come out is that correct? 10x10+10x10+0x0=200. 10x10+10x0+0x10=100
200-100=100
sqrt 100=10
 
Your calculation is correct. :thumbsup: However, your installation is a code violation. :slaphead: If you are sharing a neutral, then the act of turning one breaker off is required to also turn off the other two breakers. Let's pose the question by saying that you turn off one of the loads at its individual switch, rather than by opening a breaker. :happyyes:
 
Arkansas74 said:
Hey guys if I have a 3 phase pull a roundhouse for 120 drops and each one is drawing 10 amps. If I turn one breaker off and have 10 amps on each phase and 10 amps on a an 10 on b and c is off my neutral will have 10 amps on it then. the way my calculations come out is that correct? 10x10+10x10+0x0=200. 10x10+10x0+0x10=100
200-100=100
sqrt 100=10
Click to expand...
You have found a way to manipulate the numbers which happens to work for one special case. The actual formula, which has been posted in another thread, involves more than just the squares of the individual line currents, but that formula happens to give the same answer as your method when one of the line currents is zero.
 
This is a formula for neutral current


ry%3D400
 
charlie b said:
Your calculation is correct. :thumbsup: However, your installation is a code violation. :slaphead: If you are sharing a neutral, then the act of turning one breaker off is required to also turn off the other two breakers. Let's pose the question by saying that you turn off one of the loads at its individual switch, rather than by opening a breaker. :happyyes:
Click to expand...

Depends on the code cycle.;)
 
Arkansas74 said:
Hey guys if I have a 3 phase pull a roundhouse for 120 drops and each one is drawing 10 amps.
Click to expand...

When you use the term 3 phase pull, are you referring to 3 separate phases as in a 120/208 volt system? if so, you
would not have 10 amps in the neutral.
 
bob said:
When you use the term 3 phase pull, are you referring to 3 separate phases as in a 120/208 volt system? if so, you
would not have 10 amps in the neutral.
Click to expand...
bob, the OP knew that there would be no neutral current for a balanced load. He asked what would be the neutral current if he then opened the breaker on only one of the three lines.
In fact, his answer was correct under the assumption that all of the loads were connected line-to-neutral only. (Which would be required for an MWBC unless fed by a multi-pole breaker with common trip instead of just handle ties.)
 
GoldDigger said:
You have found a way to manipulate the numbers which happens to work for one special case.
Click to expand...
Uh, well, actually his method is correct and would apply in general. He did the math in four steps that exactly match the formula that Dennis posted. First he calculated AA + BB + CC. Then he calculated AB + AC + BC. Then he subtracted the second result from the first. Then he took the square root.

 
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