An electric shower circuit would work well for what you are doing threephase. 
AndyGuinness
AndyGuinness
Called calcs and volt drop confusion?
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threephase123
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Thanks guys. I always want to push myself farther. I want to push my mind.
three,
That is fine, but get to grips fully with the basics first!
Get a full grasp of the fundamental science.
Once you have ask Kirchoff, to see what I am getting at. ; )
That is fine, but get to grips fully with the basics first!
Get a full grasp of the fundamental science.
Once you have ask Kirchoff, to see what I am getting at.
; ) 
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Without an engineering degree most people will be stumbled when you mention Kirchoff, I do think Sidewinder is the best one on this forum to progress you further if you are willing to learn, but be warned, he will make you work for the answer.
I hope you do not mind me saying that Sidewinder?
I hope you do not mind me saying that Sidewinder?
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Rings are parallel circuits...
Just try and get your head around a parallel circuit..
Ignore the actual cables used or real loads..
just get your head back around current traveling along parallel circuits...
Then you can try to apply that logic to the real ring circuit....
e.g. Hypothetical circuit..
i)
Two parallel conductors of equal CSA & equal length supplying 12A load
(this would be the mid point of a ring)
6A down each leg.
ii)
Two parallel conductors equal CSA & but one 3x the length of the other supplying 12A load
(this would be a quarter point around a ring)
3A down the long length
& 9A down the short length
3:1 ratio of length & current.
iii)
Say our ring is in exact quarters call each section A, B, C & D.
If we have TWO 12A loads on the mid point and the quarter point between A & B..
A&B will both carry 6A from the mid point load
C&D will both carry 6A from the mid point load
A will carry 9A from the load between A & B
B,C & D will carry 3A from the load between A & B
So overall
A will carry 6+9
B will carry 6+3
C will carry 6+3
D will carry 6+3
As volt drop is related to current there will be more volt drop down section 'A'...
Now try and work out some numbers for unequal loads at unequally spaced positions around the ring....??
bad day explode
After your head has exploded you will then see that.....
there is NO easy calculation you can do for a true volt drop on an unbalanced ring final circuit!
:coffee
Just try and get your head around a parallel circuit..
Ignore the actual cables used or real loads..
just get your head back around current traveling along parallel circuits...
Then you can try to apply that logic to the real ring circuit....
e.g. Hypothetical circuit..
i)
Two parallel conductors of equal CSA & equal length supplying 12A load
(this would be the mid point of a ring)
6A down each leg.
ii)
Two parallel conductors equal CSA & but one 3x the length of the other supplying 12A load
(this would be a quarter point around a ring)
3A down the long length
& 9A down the short length
3:1 ratio of length & current.
iii)
Say our ring is in exact quarters call each section A, B, C & D.
If we have TWO 12A loads on the mid point and the quarter point between A & B..
A&B will both carry 6A from the mid point load
C&D will both carry 6A from the mid point load
A will carry 9A from the load between A & B
B,C & D will carry 3A from the load between A & B
So overall
A will carry 6+9
B will carry 6+3
C will carry 6+3
D will carry 6+3
As volt drop is related to current there will be more volt drop down section 'A'...
Now try and work out some numbers for unequal loads at unequally spaced positions around the ring....??
bad day explode
After your head has exploded you will then see that.....
there is NO easy calculation you can do for a true volt drop on an unbalanced ring final circuit!
:coffee
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