Single Phase vs Three Phase flow of electrons with no neutral/grounded path?

[EEC]

Are there differences in the flow of electrons in a single plase 230V circuit to a 208V 3-phase circuit uses only two of the 3 phases?
1. Take the single phase 115V part of the 230V circuit. (Does the current flow at a different time then the other 115v part of the 230V circuit?)
2. What is the path that the electrons flow on for the 115V part of the 230V circuit [since there is no return path for each leg of the 230V circuit, no neutral/grounded path]?
3. Whats the difference in a 208V circuit between two phases for 208V feeding a load. There does seem to be a time difference between phases.
4. In both of the above cases do the electrons flow back to the source or what happens to them.
5. Can you provide a reference source to back up your conclusions, if so what are they?

 

[david luchini]

Are there differences in the flow of electrons in a single plase 230V circuit to a 208V 3-phase circuit uses only two of the 3 phases?

I've long forgotten the physics of electron flow, but I'll give it a shot...No, there is not a difference in electron flow in a 230V single phase circuit compared to a 208V 3phase circuit using only two phases.

1. Take the single phase 115V part of the 230V circuit. (Does the current flow at a different time then the other 115v part of the 230V circuit?)

In a 230V single phase circuit, there isn't really two 115V parts of the circuit. The current flows in the complete 230V circuit at the same time.

2. What is the path that the electrons flow on for the 115V part of the 230V circuit [since there is no return path for each leg of the 230V circuit, no neutral/grounded path]?

The electrons don't really flow anywhere in an AC circuit. They move back and forth, first one direction then the opposite.

3. Whats the difference in a 208V circuit between two phases for 208V feeding a load. There does seem to be a time difference between phases.

I'm not sure what you mean by "difference." If you had a 208/120V wye system, then the 208V between A-B would be at a different phase angle from the 208V between B-C, etc. If you plotted the three 208V sine waves together, the 120 degree phase shift could be looked at as a time difference.

4. In both of the above cases do the electrons flow back to the source or what happens to them.

Again, electrons don't really flow anywhere in AC current, they move back and forth in an alternating fashion.

5. Can you provide a reference source to back up your conclusions, if so what are they?

I'm sure the forum members who still remember the physics of electron flow in current could expound.

 

[EEC]

In a 230V single phase circuit, there isn't really two 115V parts of the circuit.

In a 230V single phase circuit, there isn't really two 115V parts of the circuit.

The post before this post says "In a 230V single phase circuit, there isn't really two 115V parts of the circuit." Can you explain in more detail in order for me to understand that statement?

 

[chris kennedy]

If you had a 208/120V wye system, then the 208V between A-B would be at a different phase angle from the 208V between B-C, etc. If you plotted the three 208V sine waves together, the 120 degree phase shift could be looked at as a time difference.




I'm sure the forum members who still remember the physics of electron flow in current could expound.

I'll take the over on 314 posts on this one.

 

[gar]

110710-1305 EDT

EEC:

If I have a voltage source, could be a battery, wires from the source to a load, could be a simple resistor, then I have a single loop. When that circuit is closed, wires connected to the voltage source and the load, then a current flows in that loop.

Connect one terminal of a second battery to one terminal of the first battery. No current flows in the second battery and you have not changed the current flow in the first battery.

Your question 1. The answer to your first question is no because there is no current flow in the other half of the 230/115 supply. But I do not believe you asked the question you wanted to ask.

Your question 2. I do not understand the question.

It seems that in some way you are asking about the instantaneous relationship of voltage and current in several types of circuits.

If you consider a single loop of a voltage and load resistance, then the current is always exactly proportional to voltage. When the voltage is zero the current is zero. When the voltage is peak positive, then the current is peak positive, and for peak minus voltage the current is peak minus.

In a single phase supply with a center tapped transformer you have two hot output voltages relative to the center tap and these are of opposite polarity, but of the same magnitude, at any instant of time. Whether the center tap is grounded or at 1,000,000 V DC does not change the statement because the statement is relative to the center tap.

Change the reference point to one end of the secondary, then the two voltages are in phase with each other, but one is 1/2 the value of the other.

To have current flow you need a closed loop. You can have multiple loops overlap each other as in the case of the center tapped secondary from the power company. Here you have a feeder neutral that is common to both loops.


Your question 3. If you pick two hot wires of a three phase circuit, Y or delta, these are the equivalent of a single phase voltage source and you don't care about the other phases unless you are concerned about voltage to ground, or load current from the other phases. The other phases could affect your phase if they had loads, but no load on the others and your phase just looks like a single phase supply.

Your question 4. You can speak of current flow, but no single electron starts at one end of the voltage source and instantaneously flows to the other voltage source terminal. Electrons bounce around in the wire and there is an energy flow that in many cables is about 0.7 times the velocity of light.

In a CAT-5 cable it takes time for a signal to travel from one end to the other. I show this time delay in photos at my web site
http://www.beta-a2.com/cat-5e_photo.html
If you do a calculation using data from Photo P3, then 1.5 microseconds at 1000 ft produces a velocity of 126,262 miles per hour. This is 0.678 time the velocity of light.

If I look at something like a cathode ray tube, then I do have a direct flow of electrons from the emitting cathode to the CRT face. These do have a transit time much slower than the velocity of light. The velocity of light is 186,000 miles per second. The transit time is a function of the accelerating voltage, and problems with this transit time in vacuum tubes occurs in the region of 1000 MHz.

Suggested references would be books on basic electrical circuit analysis. Start with DC.

.

 

[david luchini]

The post before this post says "In a 230V single phase circuit, there isn't really two 115V parts of the circuit." Can you explain in more detail in order for me to understand that statement?

I can try...I assume with your single phase vs. two phases question you are talking about single phase vs. three phases systems.

Lets assume our single phase system to be the secondary of a transformer, and we'll make it a 230/115V secondary to match your question. If you draw your transformer secondary, you would have one winding from a terminal L1 to terminal N, lets say, with 115V between L1 and N. A second winding is connected between N and L2, also with 115V between them. We can see that there are 230V between L1 and L2.

Now if we draw a line out from L1 and another line out from L2, and we connect some simple load (such as a resistor) between these 2 lines. We know have a completed 230V circuit with a source (the transformer) and a load (the resistor.)

You can see that current will flow thru the load between L1 and L2, and the voltage drop across the load will be 230V. Current doesn't flow halfway through the load and somehow back to the neutral point. That is to say, there are not two 115V halves of the circuit, just one 230V part.

 

[EEC]

From the reply to questions this is my understanding thus far

From the reply to questions this is my understanding thus far

Single phase 230/115V, each 115V leg is opposite polarity at the same instant of time. The current through the load is also traveling through the load in opposite directions at the same instant of time. Therefore the voltage drop will be 115V in each polarity to total 230V. Is this correct?

 

[gar]

110710-1731 EDT

EEC:

Suppose you have a 230/115 supply. This is most likely from a center tapped transformer.

Connect two identical 115 ohm resistors in series across L1 to L2. There is no connection from neutral to the mid-point of the resistors. How much current is flowing thru the two resistors? 1 ampere.

What is the direction of current in each resistor?

What is the voltage difference between the neutral and the resistor mid-point?

.

 

[david luchini]

Single phase 230/115V, each 115V leg is opposite polarity at the same instant of time. The current through the load is also traveling through the load in opposite directions at the same instant of time. Therefore the voltage drop will be 115V in each polarity to total 230V. Is this correct?

Personally, I don't like the term "polarity" in an AC circuit. To me it denotes "positive" and "negative." But to answer your question...no, current is not flowing through the load in opposite directions at the same.

It is better to think of phase angle rather than polarity with AC circuits. In your 230/115V AC source (such as the transformer secondary) you have one winding that is 115V at an angle of 0, say from L1 to neutral. The second winding is 115V at an angle of 180, from L2 to neutral. This can be written as VL1=115<0 and VL2=115<180. The voltage across both windings at the same time is VL1-VL2 = 115<0-115<180=230<0.

If at one instant in time, current is flowing "out" of L1 toward the load, at the same instant it is flowing in the same direction thru the load, from the load back to L2. At the source (the transformer secondary) current is flowing (still in the same direction) from V2 to N, and from N to V1. The circuit is a loop in which current is flowing in the same direction at the same time.

 

[EEC]

110710-1731 EDT


What is the voltage difference between the neutral and the resistor mid-point?

.

My answer would be 115V

 

[EEC]

It is better to think of phase angle rather than polarity with AC circuits. In your 230/115V AC source (such as the transformer secondary) you have one winding that is 115V at an angle of 0, say from L1 to neutral. The second winding is 115V at an angle of 180, from L2 to neutral. This can be written as VL1=115<0 and VL2=115<180. The voltage across both windings at the same time is VL1-VL2 = 115<0-115<180=230<0.

0 to 180 degree phase angle would mean each 115V will peak at the same time in opposte directions. Therefore it seems that current will peak the same as voltage unless its a LCR circuit. Correct?

 

[david luchini]

0 to 180 degree phase angle would mean each 115V will peak at the same time in opposte directions. Therefore it seems that current will peak the same as voltage unless its a LCR circuit. Correct?

Yes, the two 115V (to neutral) will peak at the same time in opposite direction. But don't forget that they are also connected together in opposite directions. So the voltage across both 115V sources peak at the same time in the same direction. This is why we have 230V instead of 0V.

If our source voltages are VL1-N (Voltage L1 to neutral)=115<0 and VL2-N=115<180, then the voltage from neutral to line of the two sources is 180 degrees opposite: VN-L1=115<180 and VN-L2=115<0.

So if you had two 115 Ohm resistors for loads, and connect the first from L1 to N. The load current is 1A at 0 degrees. If the second resistor is connected from L2 to N, the load current in that resistor will be 1A at 180 degrees. The current flowing in the neutral, figured by KCL will be In=IL1+IL2, which will be zero amps. There will be 1A flowing thru both resistors in the same direction at the same time (and nothing flowing on the neutral.)

 

[gar]

110710-2003 EDT


EEC:

Your answer of 115 is not correct. The correct answer is 0 volts in an ideal circuit and very close to 0 in an experimental circuit you could setup.

Find two same wattage incandescent bulbs from the same batch. These will be close to the same resistance with the same voltage applied. You could test them in parallel and measure the current to each with an accurate meter. If within 1/2% of the same current then put them in series and connect across your 230 V supply.

Accurately measure each of the 115 V supplies, also the voltage across each bulb. Then measure the voltage from the bulbs mid-point to the 230 V neutral. You should see a small voltage, and maybe you can see this voltage difference in relation to the four separate voltages you measured.

.

 

[Smart $]

Single phase 230/115V, each 115V leg is opposite polarity at the same instant of time. The current through the load is also traveling through the load in opposite directions at the same instant of time. Therefore the voltage drop will be 115V in each polarity to total 230V. Is this correct?

When discussing AC systems, at any single instance, DC theory applies. In this sense, single phase 230/115 supply is comparable to two batteries connected in series. That means the positive of one battery is connected to the negative of the other. The voltage across both batteries is double that of either single battery. The current flows in the same direction whether you connect a circuit to either or across both batteries.

If you connected the negative of both batteries, your voltage across the positive terminals would be zero.

Therefore, in order for there to be double the voltage across L1 and L2, the source has to be in series.

Does that help any?

 

[charlie b]

This entire discussion has me confused, mostly because I don?t think the question is very clear. You speak of a single phase 230 volt circuit, you mention the absence of a neutral path, and somehow the notion of 230/115 comes into play. I am going to ignore everything said by everyone so far. Then I am going to present a question of my own, and answer that question. If my question has anything to do with what you originally tried to ask, then perhaps I will have helped.

Circuit 1: Consider a common household power panel. It gets 120/240 volt, single phase power from the utility. Insert a two-pole breaker into positions 2 and 4. Attach a black wire to the part of the breaker in position 2, and a red wire to the part in position 4, and run them to some machine in the garage. For this discussion, we will disregard the ground wire.

Circuit 2: Consider a common industrial power panel. It gets 120/208 volt, three phase power from the utility. Insert a two-pole breaker into positions 2 and 4. Attach a black wire to the part of the breaker in position 2, and a red wire to the part in position 4, and run them to some machine in the garage. For this discussion, we will disregard the ground wire.

Question: How does the current flow in Circuits 1 and 2, and is it different?

Warning: If anyone brings up the concept of ?electron drift,? you will be sentenced to 300 lashes with a wet noodle. We don?t need to go there, not today anyway. I am well aware that what I am about to say is not strictly true. But it is true enough, for practical purposes, and it can help a person understand current flow.

Answer: For starters, it is not different. In both cases, electrons leave the source (we will consider the panels to be the source), and electrons return to the source. In both cases, during one half of the cycle (a cycle occurring 60 times each second), electrons leave the source via the black wire, proceed to the garage, flow through the machine, and return to the source via the red wire. In the other half cycle, electrons leave the source via the red wire, proceed to the garage, flow through the machine, and return to the source via the black wire. During that first half cycle, any single electron would be moving so fast that it will be able to make it from the source, along the black wire to the load, and along the red wire back to the source, and complete the trip in such a short time, that it will start the round trip again, and again, and again, about a bazillion times, before the end of that half cycle. On the other half cycle, it will make the round trip, in the other direction, another bazillion times or so, then once again reverse direction. Please note that you do not need a neutral wire, in order for the electrons to have a path that returns them to the source.

 

[Smart $]

...
Warning: If anyone brings up the concept of ?electron drift,? you will be sentenced to 300 lashes with a wet noodle. ....

Please report for sentence to be carried out

 

[IamJonscranium]

Great explanation Charlie!

I like to say it like this: DC current is like a chainsaw, AC is like a hacksaw.


Now for more fun. I was talking to a cable guy and he gave me this story:

He went to a house where a lady was having problems. He looked at the main coax cable coming in, saw that it was melting, and cut it. All the electric equipment in the house began smoking.

The theory is that the house ground went bad, and the coax (which had it's shield connected to house ground) became the ground.

Now an electrician speculated that losing the ground caused all the 120V circuits to be 240V. The only way I can think this would happen is if there was a short from neutral to one of the hot legs. Thoughts?

I figured this thread was a semi-appropriate place to put this, since it might be a good exercise in learning about the topics discussed so far.

 

[Besoeker]

Are there differences in the flow of electrons in a single plase 230V circuit to a 208V 3-phase circuit uses only two of the 3 phases?

In an AC circuit, the electrons don't really flow. They just wiggle back a little bit. I think it's simpler and more appropriate to talk about currents rather than electron flow.

1. Take the single phase 115V part of the 230V circuit. (Does the current flow at a different time then the other 115v part of the 230V circuit?)

Assuming you mean a 115-0-115 supply with no neutral (per the threat title), you have a single 230V source and thus one current path and just one current so there are no different times.

2. What is the path that the electrons flow on for the 115V part of the 230V circuit [since there is no return path for each leg of the 230V circuit, no neutral/grounded path]?

Same answer as above. With no neutral, you just have a 230V circuit.

3. Whats the difference in a 208V circuit between two phases for 208V feeding a load. There does seem to be a time difference between phases.

If it's 208V load connected between two phases of a 208V 3-phase system with no connection to the neutral, then it's a single phase circuit.

4. In both of the above cases do the electrons flow back to the source or what happens to them.

Se initial comment.

5. Can you provide a reference source to back up your conclusions, if so what are they?

I didn't think I needed one. Maybe Shepherd, Morton & Spence Basic Electrical Engineering.

One final point that has a fundamental on the answers above.
Did you really mean no neutral connection?

 

[charlie b]

In an AC circuit, the electrons don't really flow. They just wiggle back a little bit.

I had asked that we not go there. :happysad: That will cost you 300 lashes with a wet noodle!

 

[John120/240]

Electron drift any chance we can get an explanation how it works..................