multiwire branch circuts / follow up.

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domnic

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ON a 208 v four wire if i have two hots and one neutral and i run two 10 amp loads one on each hot wire what is the amp draw of the neutral ? and what is the math for your answer ?
 
Zero

I?A + I?B + I?C - (IA x IB) - (IB x IC) - (IC x IA)

300 - 300 = zero

Square root of zero = 0
 
Bob, that formula is incorrect. The first set of parentheses should have A squared, B squared, C squared not A times 2, B times 2, and C times 2.
 
infinity said:
Bob, that formula is incorrect.
Click to expand...

Thanks.


I believe this one is on the money


NeutralCurrentFormula.gif
 
Seems to me It should be 10

10< 0 + 10< 120 = 10 < 60


sqr (10 ^2 + 10 ^2 )-(10 x 10))

sqr (100 +100 )- 100

sqr ((200)-100)

sqr (100)= 10

Where did I screw up?
 
realolman said:
Seems to me It should be 10

10< 0 + 10< 120 = 10 < 60


sqr (10 ^2 + 10 ^2 )-(10 x 10))

sqr (100 +100 )- 100

sqr ((200)-100)

sqr (100)= 10

Where did I screw up?
Click to expand...

Why do you think you screwed up. The answer is not zero since the op is talking about a 3 phase system with the mwbc using 2 phases.
 
Actually I didn't think I did... but one of the things I've learned on this forum is that it's best to make your words soft and sweet, because you don't know which ones you might have to eat.:)

And while I 'm on cliches... I thought I saw where someone posted that there was no current on the neutral, and that did not seem correct to me.. so I thought I'd see what it was....

and the cliche is...

there's a fine line between a hero and a zero.:smile:
 
multiwire branch circuts

multiwire branch circuts

10 x 10 = 100/10 =10 10/1.732 = 5.77 amps ?????
 
I say 10a, and here's why:

If it were a balanced 4-wire MWBC, the neutral current would be zero. Subtract the 1a from one phase, and the neutral current becomes 1a.

Remove all 10a from that phase, and the neutral becomes 10a. The neutral current equals what is missing from a fully-balanced 4-wire MWBC.


"Formula? We don't need no stinkin' formula!"
 
I am trying to reason the difference if any. Assuming the above ckts on a 1 ph. Edison 3 wire, the net current thru the neutral would be 0. What would the net be if the 2 legs are taken from a 208 3 ph where the phases are 120 deg. apart?
 
I was viewing this from an "imbalance perspective". where if connected to a 1 ph multiwire branch ckt, vectorly the neutral amps would be 0, would it not? I.E the neutral could be clipped downstream of the load connections w/o any current change thru the loads. On the 208 3 ph supply, is the 5.56 amp value an imbalance current?
 
Neutral current formula's are system specific. Meaning that single phase 120/240 system will find neutral current differently than a 208Y/120 volt system. In the single phase system just subtract the smaller current from the larger and you'll find the neutral current. The WYE system has a much more complex computation.
 
automate0042 said:
I was viewing this from an "imbalance perspective". where if connected to a 1 ph multiwire branch ckt, vectorly the neutral amps would be 0, would it not? I.E the neutral could be clipped downstream of the load connections w/o any current change thru the loads. On the 208 3 ph supply, is the 5.56 amp value an imbalance current?
Click to expand...
I am not sure I understand. If you have a single phase system with current flowing the same in cir. A and Cir B then the neutral current will be zero. If it is imbalanced then how could it be zero.

Let's say you have 10 amps on "A" and 5 amps on "B" then the neutral current will be 10-5 or 5 amps.
 
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