How does current flow in a circuit?

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bluescotty

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I am a new member and cannot find where to post a new thread so hopefully someone can instruct me on this?

In the meantime I have a query which is:

When connecting a single phase load across 2 phases of a 208/120 3 phase 4 wire system what is the actual current drawn in each of the 2 phases?
If for example I had an appliance that had a full load current of 20A would the current be 10A on each of the the two phases connected. If it was phase to neutral I know that theoretically it would be 20A on the phase conductor with no current on the neutral, but what about phase to phase?
 
LarryFine

LarryFine

Master Electrician Electric Contractor Richmond VA
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Scotty, welcome to the forum! :)

In any 2-wire circuit, regardless of whether the source is line-to-neutral or line-to-line, the current is the same throughout the circuit. If the load is 20a, the current anywhere in the circuit is the same 20a.

More power at a given current results from the source being a higher voltage. Remember that voltage is measured between two points, and current flows through a conductive pathway between those two points.
 
chris kennedy

chris kennedy

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60 yr old tool twisting electrician
bluescotty said:
If it was phase to neutral I know that theoretically it would be 20A on the phase conductor with no current on the neutral
Click to expand...

This is incorrect and I believe Larry's post explained that very clearly. FWIW, here is the formula for calculating neutral current in a 3? system.


3neutralcurrent.jpg


And welcome to the Forum.
 
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Besoeker

Besoeker

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chris kennedy said:
This is incorrect and I believe Larry's post explained that very clearly. FWIW, here is the formula for calculating neutral current in a 3? system.


3neutralcurrent.jpg
Click to expand...
That's true only for the specific case of all three phases having the same power factor.
 
steve66

steve66

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This formula has given me deja-vu every since I first saw it posted on this fourm. I 'm sure I knew where it came from at one time. And I think it does only work for all three phases having the same power factor (all the currents are 120 degrees out of phase.) But I can't be sure without seeing where this formula came from.

I'll give kudos to anyone that can derive this formula. And you get double kudos if you can do it with a matrix equation and determinates.

Steve
 
mivey

mivey

Senior Member
steve66 said:
This formula has given me deja-vu every since I first saw it posted on this fourm. I 'm sure I knew where it came from at one time. And I think it does only work for all three phases having the same power factor (all the currents are 120 degrees out of phase.) But I can't be sure without seeing where this formula came from.

I'll give kudos to anyone that can derive this formula. And you get double kudos if you can do it with a matrix equation and determinates.

Steve
Click to expand...
Already been done. See:
http://forums.mikeholt.com/showthread.php?t=53520
 
Besoeker

Besoeker

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steve66 said:
And I think it does only work for all three phases having the same power factor (all the currents are 120 degrees out of phase.)
Click to expand...
You are right.
This, like other forums, isn't really set up to present mathematical expressions.

So I'll just give a couple of examples of results.
Suppose Ia, Ib, and Ic are 10A, 20A, and 30A respectively.
If all have the same power factor this would result in a neutral current of 17A which accords with the formula given by Chris K.

OK. Give Ic a 0.8 pf lag and In would be 13A.
However, make Ib 20A at 0.8 pf lag and In would be closer to 30A.
 
mivey

mivey

Senior Member
chris kennedy said:
Where could one find this formula?
Click to expand...
In = Ia<A + Ib<B +Ic<C

Ia = |Ia|*cos(A) +j |Ia|*sin(A)
Ib = |Ib|*cos(B) +j |Ib|*sin(B)
Ic = |Ic|*cos(C) +j |Ic|*sin(C)

where A, B, & C are the 0, 240, 120 angles of the voltages

cos(angle between voltage and current)=power factor

so for 0.8 pf: arccos(0.8) = 36.87deg.

If Ic is 0.8 lagging, Ic= Ic<120deg-36.87deg = Ic<83.13deg = 30<83.13
In = 10<0 + 20<240 + 30<83.13 = 12.97<73.94

If Ib is 0.8 lagging, Ib= Ib<240deg-36.87deg = Ib<203.13deg = 20<203.13
In = 10<0 + 20<203.13 + 30<120 = 29.59<142.23
 
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Mayimbe

Senior Member
Location
Horsham, UK
bluescotty said:
When connecting a single phase load across 2 phases of a 208/120 3 phase 4 wire system what is the actual current drawn in each of the 2 phases?
If for example I had an appliance that had a full load current of 20A would the current be 10A on each of the the two phases connected.
Click to expand...


You will have an unbalance system. You can analyze that with symmetrical components.
say, Z+= 120/20 = 6 ohms = Z-

Z0 = 3*Z+

I+ = V(posistive secuence)/Z+
I- = I+
I0 = I+/3

a=1<120?


It wont be 10 A on each phase.

see this link. chapter four ;)

http://books.google.co.ve/books?id=R7GSIXxQMvAC&dq=Protective+Relaying:+Principles+and+Applications,+by+J+Lewis+Blackburn&printsec=frontcover&source=bn&hl=es&ei=bGLKSuvOI5ivtgekz5TYDg&sa=X&oi=book_result&ct=result&resnum=4#v=onepage&q=&f=false
 
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