Return Path of Current

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For a 1 phase circuit, current flows through hot and comes back through neutral.

For a 3 phase system ( delta-star) we have three hot wires and 1 neutral. so current flows
through the three hot wires and comes back through neutral.

How does current find its return path in delta system since there are only three hot wires?
 

Smart $

Esteemed Member
Location
Ohio
For a 1 phase circuit, current flows through hot and comes back through neutral.

For a 3 phase system ( delta-star) we have three hot wires and 1 neutral. so current flows
through the three hot wires and comes back through neutral.

How does current find its return path in delta system since there are only three hot wires?
You have to know how to apply Kirchoff's Current Law (Google it) with respect to 3?, essentially for each point in time. It essentially states the sum of currents entering a junctioin (node) is equal to the sum of currents leaving the node. When it comes to 3?, there are methods to not have to do the math for each point in time, assuming the current's waveform is sinusoidal.

Anyway, your assessment isn't quite correct...

For a 2-wire circuit, current travels to load on one conductor and returns on the other, reversing 120 times per second (60Hz)

For a 3- or 4-wire circuit, current travels to the load on one or two conductors and returns on the others, depending on where the 3- and 4-wire nodes are locating and Kirchoff's Current Law at that point in time.

More later if required....
 

big john

Senior Member
Location
Portland, ME
For a 1 phase circuit, current flows through hot and comes back through neutral...
I know it's common to think of the neutral as the return path, but it makes it confusing. A neutral wire is just a transformer lead that we marked white. The same current flows over both the hot and neutral conductors, and in the case of alternating current, it flows both directions.

When you've got a 1? piece of equipment, you're connecting it between two transformer leads. When you've got a 3? piece of equipment you're connecting it between three transformer leads. If it's 3? 4W then it's using four. In any of the cases all that's happening is that voltage across any two leads is being applied to the load, but there are simply different combinations of current depending on what the load is between the wires. It's no different in a delta circuit: In fact you can make a delta transformer out of three 1? transformers by connecting the secondaries to each other.
 

jcormack

Member
Location
Pennsylvania
For a 1 phase circuit, current flows through hot and comes back through neutral.

For a 3 phase system ( delta-star) we have three hot wires and 1 neutral. so current flows
through the three hot wires and comes back through neutral.

How does current find its return path in delta system since there are only three hot wires?

Often 3-phase loads are Delta loads, with no connection to neutral, and hence no neutral current flow- all currents exist in the phase leads. 3-phase motors are typically Delta loads with zero neutral current flow.
In a balanced 3-phase system there will be a zero net neutral current, even with neutral connected loads.

Actually, if you had a 240/120 circuit utilizing a shared neutral for two 120 loads that were fully equal (magnitude & angle) - you would have a net zero neutral current.
 

Smart $

Esteemed Member
Location
Ohio
How does current find its return path in delta system since there are only three hot wires?
Okay, let's say you have a 240V 1? heater load. There no neutral, current travels back and forth on the two conductors.

Now let's say we have three of them on a 240/120V 3? 4W system and we connect them to 2-pole breakers, say 1) ckt# 1-3, 2) ckt# 5-7, and 3) ckt# 9-11. As such they would be connected tot he following phase: 1) AB, 2) CA, and 3) BC. Note in the preceding there are two of each letter. The panel that supplies these has 4 wires, but there is no neutral current from these three heaters. All the current is on Lines A, B, and C.

What if we removed the panel and wired these three heaters directly to the supply wires. Would the heaters work? Let's say each heater draws 10A. How much current would there be on each Line?

BTW, in the preceding example with the panel, the junctions/nodes are located on the panel buses.
 

Besoeker

Senior Member
Location
UK
For a 1 phase circuit, current flows through hot and comes back through neutral.

For a 3 phase system ( delta-star) we have three hot wires and 1 neutral. so current flows
through the three hot wires and comes back through neutral.

How does current find its return path in delta system since there are only three hot wires?

I think I can understand why you have asked the question.
I suppose we all came at the subject first from either DC or single phase AC. A two wire circuit. So we have the notion of a go and return.
So think of a 4-wire star. Say a 208/120V system. Use labels A, B, C, and N for the three phases and neutral.

Connect a 208V load between A and B. The current path is between A and B with no neutral involved.
Same if you do a B to C load and a C to A.

That's how delta works.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
Just to be a little more explicit about this, when you have a single phase load on a delta system, the current flows from one phase to the other. There is NO valid connection from one phase to ground on a true delta system, hence no neutral.

The one exception that screws people up sometimes is the "High Leg" 240V Delta transformer system. There IS an neutral, but it is created by center tapping (splitting) only ONE winding of the transformer and grounding that center tap connection. So you can get 120V from A or C to that grounded center tap neutral, but not from B phase, because it has no direct ground reference.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
For a 1 phase circuit, current flows through hot and comes back through neutral.

For a 3 phase system ( delta-star) we have three hot wires and 1 neutral. so current flows
through the three hot wires and comes back through neutral.

How does current find its return path in delta system since there are only three hot wires?

If you look at a time plot of the current in all three phases of a balanced three phase load, you will see that the values of all three phases at any point in time sum to zero. For example, whenever the current on any one phase is at its maximum positive value, the other two phases are at 1/2 their maximum negative value.
 

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Okay, let's say you have a 240V 1? heater load. There no neutral, current travels back and forth on the two conductors.

Now let's say we have three of them on a 240/120V 3? 4W system and we connect them to 2-pole breakers, say 1) ckt# 1-3, 2) ckt# 5-7, and 3) ckt# 9-11. As such they would be connected tot he following phase: 1) AB, 2) CA, and 3) BC. Note in the preceding there are two of each letter. The panel that supplies these has 4 wires, but there is no neutral current from these three heaters. All the current is on Lines A, B, and C.

What if we removed the panel and wired these three heaters directly to the supply wires. Would the heaters work? Let's say each heater draws 10A. How much current would there be on each Line?

BTW, in the preceding example with the panel, the junctions/nodes are located on the panel buses.

1. Well if u r removing the panel then how will the cable from panel to heater be protected. There needs to be protection for the heater cables. but basically answering your question that whether heaters will work or not, my answer is YES.
2. Each individual heater will be connected across two phases i.e hence will see line voltage. each heater will draw line current
of 10A

Thanks
 

Smart $

Esteemed Member
Location
Ohio
1. Well if u r removing the panel then how will the cable from panel to heater be protected. There needs to be protection for the heater cables. but basically answering your question that whether heaters will work or not, my answer is YES.
The point wasn't about protecting the conductors. It was simply an exercise for which you concluded correctly.

But just to add a bit of innuendo, if all electrical equipment worked perfectly, forever, there'd be no need for conductor protection. Just wire it up like the branches of a tree. :p

2. Each individual heater will be connected across two phases i.e hence will see line voltage. each heater will draw line current of 10A
Technically, each heater is connected across a phase. As such, they are connected between two lines. (Yes, there are a lot of tradesmen who do not use the correct terminology.) In all conventional applications of wiring, you connect to a line, typically an ungrounded conductor (but not always), or you connect to a neutral, typically a grounded conductor (but not always).

With that said, each heater draws 10A. Your answer of 10A is incorrect. Each line's current will be powering two heaters. What is the line current?

Hint (actually, an outright gimme :blink:): Determine total load in kVA (apparent power), then determine line current using 3? power formula.
 

mahaney03

Member
Location
DMV area
The point wasn't about protecting the conductors. It was simply an exercise for which you concluded correctly.

But just to add a bit of innuendo, if all electrical equipment worked perfectly, forever, there'd be no need for conductor protection. Just wire it up like the branches of a tree. :p


Technically, each heater is connected across a phase. As such, they are connected between two lines. (Yes, there are a lot of tradesmen who do not use the correct terminology.) In all conventional applications of wiring, you connect to a line, typically an ungrounded conductor (but not always), or you connect to a neutral, typically a grounded conductor (but not always).

With that said, each heater draws 10A. Your answer of 10A is incorrect. Each line's current will be powering two heaters. What is the line current?

Hint (actually, an outright gimme :blink:): Determine total load in kVA (apparent power), then determine line current using 3? power formula.


New Junior Engineer here trying to learn as much as I can! So this means you have:

240V * 10A = 2400VA

2400VA / (240*1.73) = 5.78 A on each phase?

If this is correct, why wouldn't it be 5 A on each phase? What happens to the extra current?
 

david luchini

Moderator
Staff member
Location
Connecticut
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Engineer
New Junior Engineer here trying to learn as much as I can! So this means you have:

240V * 10A = 2400VA

2400VA / (240*1.73) = 5.78 A on each phase?

If this is correct, why wouldn't it be 5 A on each phase? What happens to the extra current?

There are three heaters, not one.

7200VA/(240*1.73) = 17.32A on each line.
 

GoldDigger

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Location
Placerville, CA, USA
Occupation
Retired PV System Designer
New Junior Engineer here trying to learn as much as I can! So this means you have:

240V * 10A = 2400VA

2400VA / (240*1.73) = 5.78 A on each phase?

If this is correct, why wouldn't it be 5 A on each phase? What happens to the extra current?

If each heater is 2400VA, (actually 2400W since the power factor of a heater is going to be darned close to 1), then you will measure a current of 10A at each terminal of each heater, assuming that each is a single phase heater. Now look at what happens when we connect three such heaters in a delta configuration to three phase source (either 240 delta or 240Y/139):
Each line will supply current to both heaters which are connected to that line.
But since the current from the two heater terminals connected to one line (not to one phase!) are vectors which are 120 degrees apart their sum is neither zero nor 20A. Instead the line current will be 10A times sqrt(3). **
Not at all sure where the 5A came in unless you were trying to calculate the line current for a unit 2400W three-phase heater. In that case you do not start out with 2400 = 240 x 10 in the first place.

**: The component of the current that is out of phase between the two terminal currents is 10 sin(30), which is just 5. The 5 from one heater cancels out the 5 from the other heater and does not result in any line current. The component of each which adds up will be 10 cos(30) which is 10 sqrt(3)/2. The sum of those two components is twice that, namely 10 sqrt(3)

At the risk of confusing the issue even more, I cannot resist taking this a little further.

The total current in each line will be 10 x sqrt3, and that net current will be in phase with the 139V from that line to neutral if it is a wye source. You can therefore calculate the total power for all three lines as (3) times (10 x sqrt3) times (139). But sqrt3 times 139 is just 240, so the power is 2400 times 3.

Now if you take the total wattage, 7200, and apply the three phase power rule, you also get that the line current is 10 x sqrt(3). (Making use of the fact that sqrt3 divided by 3 is just 1/sqrt3.)

It all works out logically together, but if you are not comfortable with doing vector math in your head your best bet is just to memorize and apply the three phase power rule and not try to "simplify" it in any way. :)

 
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dereckbc

Moderator
Staff member
Location
Plano, TX
For a 3 phase system ( delta-star) we have three hot wires and 1 neutral. so current flows
through the three hot wires and comes back through neutral.

What makes you think that? How is the load connected? L-L or L-N?

How does a single phase 240/120 system work when you connect a motor L-L for 240 volt operation. No neutral is there because the load is L-L, not L-N. Only reason the neutral is there if for 120 volt loads.
 
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JRW 70

Senior Member
Location
Eastern Central Missouri
Occupation
Testing and Engineer
Configuration

Configuration

The three phase lines are just connected
in a Y arrangement is what I got out of the
post above. The transformer is just D:Y.

e.t.a. : if this is a Y config. there should be no
return on the "N" unless one or two of the elements
fail.

JR
 
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GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
The three phase lines are just connected
in a Y arrangement is what I got out of the
post above. The transformer is just D:Y.

e.t.a. : if this is a Y config. there should be no
return on the "N" unless one or two of the elements
fail.

JR
Given that the actual source is a wye, then all that really matters is whether it is a 240Y/139 with the heaters connected line to line or a 415Y/240 (not bloody likely, but possible, especially outside the US) with the heaters connected line to neutral.
And as you say, in either case there should be no neutral current with the three heaters connected line to line, even if they are not perfectly balanced and small neutral current related to mismatch only if they are connected line to neutral.
 

mahaney03

Member
Location
DMV area
If each heater is 2400VA, (actually 2400W since the power factor of a heater is going to be darned close to 1), then you will measure a current of 10A at each terminal of each heater, assuming that each is a single phase heater. Now look at what happens when we connect three such heaters in a delta configuration to three phase source (either 240 delta or 240Y/139):
Each line will supply current to both heaters which are connected to that line.
But since the current from the two heater terminals connected to one line (not to one phase!) are vectors which are 120 degrees apart their sum is neither zero nor 20A. Instead the line current will be 10A times sqrt(3). **
Not at all sure where the 5A came in unless you were trying to calculate the line current for a unit 2400W three-phase heater. In that case you do not start out with 2400 = 240 x 10 in the first place.
**: The component of the current that is out of phase between the two terminal currents is 10 sin(30), which is just 5. The 5 from one heater cancels out the 5 from the other heater and does not result in any line current. The component of each which adds up will be 10 cos(30) which is 10 sqrt(3)/2. The sum of those two components is twice that, namely 10 sqrt(3)

At the risk of confusing the issue even more, I cannot resist taking this a little further.

The total current in each line will be 10 x sqrt3, and that net current will be in phase with the 139V from that line to neutral if it is a wye source. You can therefore calculate the total power for all three lines as (3) times (10 x sqrt3) times (139). But sqrt3 times 139 is just 240, so the power is 2400 times 3.

Now if you take the total wattage, 7200, and apply the three phase power rule, you also get that the line current is 10 x sqrt(3). (Making use of the fact that sqrt3 divided by 3 is just 1/sqrt3.)

It all works out logically together, but if you are not comfortable with doing vector math in your head your best bet is just to memorize and apply the three phase power rule and not try to "simplify" it in any way. :)


But I would rather understand than just plug and chug numbers into equations. Your explanation definitely helped me out immensely so thank you :)
 

under8ed

Senior Member
I have actually ran a 3 phase water heater that called for a neutral without it. I realized the elements were tied to a common point, and that the neutral was really only needed to protect the system in the case of a failure. It was new, and we were in the process of installing a delta/wye transformer to accommodate the load; but until then it operated just fine.. I have also seen 3 balanced rows of 277v. light fixtures operate just fine without the neutral connected, (we were trying to determine if the had been wired correctly), they will operate as a single 3 phase load seeing the normal voltage if balanced.
 
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