MWBC= more heat or Less heat?

Status
Not open for further replies.

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
If you want the current from the rest of the circuit to flow through the wire in question, you must add enough additional potential force to overcome the wire resistance or some of the current is going to flow through a different channel.
Mivey, this is not true, nor is it relevant to the point being made. The point being made is that the voltage drop along the circuit does not cause the current to move in the circuit, but that current moving through the circuit causes the voltage drop.

All of the current from the "rest of the circuit" will flow through the wire in question without adding any additional force. The force that moves the current in the circuit is the potential source (such as the battery.)

If I connect a 12v battery to a 12 ohm resistor with "superconductors," 1 Amp will flow from the battery, through the super conductors and the resistor. There will be a 12 volt drop across the resistor, and 0 volt drop along each conductor. If I then replace the "superconductors" with "real" conductors that each add 0.5 ohms resistance to the circuit, 0.923 Amps will flow through the circuit. There will be a 0.0462 Volt drop along each of the conductors and a 11.076 volt drop across the resistor.

Current is flowing over the wire in question, but I have not added any additional potential to the circuit (and none of the current is flowing through any different channel.) So your first statement is wrong.

If it takes 40 volts to push 10 amps of current through a circuit with superconductor wire, you are going to have to add more voltage force to push the same circuit current through a wire with some resistance.
This statement is true, if we want the "same current" to flow through the circuit, but it is not "necessary" to add potential to CAUSE current to flow the circuit. Consider a 120V circuit with 1200 watt of incandescent lighting on it. We know that there will be a voltage drop on the circuit conductors feeding the lighting. In an ideal world, 1200 watt on a 120v circuit would create a 10 Amp current, but lets assume the circuit conductors add 0.25 ohms each. The current flowing the the circuit would the be 9.6 amps and my 1200 watts of light only output 1106 watts. But in this real world case, I'm not going to retap my transformer to increase the voltage to make sure I get 10 Amps, and therefore, 1200 Watts out of the circuit.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
Ideal components are simply a tool used to simplify real world problems.

The implications of Ohm's Law break down under certain ideal conditions. For example: Considering a fixed resistance connected to a voltage source - if you increase the voltage, the current increases. With ideal components, there are cases where this relationship is not true.

There are times when the use of ideal components is misleading. The case of the current in the neutral wire is one of those cases. I firmly believe that the current flow in the neutral is a result of the potential difference from one end of the conductor to the other.
Can you give me an example where the relationship of increased voltage/increased current is not true with ideal components? This statement is not true.

You can follow, if you desire, the erroneous belief that the current flow in the neutral is a result of the potential difference from one end of the conductor to the other. However, you will be doing yourself no favors by following this belief. We have clearly demonstrated herein that the current flow in the entire circuit is CREATED by the potential source in the circuit, and that the potential difference from one end to the other end of the neutral conductor is a RESULT of the current that is flowing in that conductor. To reject this is to reject the basic tenets of Electrical Engineering.
 

crossman gary

Senior Member
Can you give me an example where the relationship of increased voltage/increased current is not true with ideal components? This statement is not true.
Of course.

Example 1: Take an ideal 12 volt source with no internal resistance and connect it to each end of an ideal conductor with zero impedance. What does Ohm's Law predict for the current in the conductor?

Example 2: Exact same scenario, but increase the source voltage to 480 volts. What does Ohm's Law predict for the current flow in the conductor?

I can give more examples where Ohm's Law will fail in its predictions when we are working at the extremes, such as conductors with zero resistance.
 

mivey

Senior Member
If you want the current from the rest of the circuit to flow through the wire in question, you must add enough additional potential force to overcome the wire resistance or some of the current is going to flow through a different channel.
Mivey, this is not true...So your first statement is wrong.
If it takes 40 volts to push 10 amps of current through a circuit with superconductor wire, you are going to have to add more voltage force to push the same circuit current through a wire with some resistance.
This statement is true, if we want the "same current" to flow through the circuit...
You disagree with my first statement in the first 1/2 of your post. Then you agree with my second statement in the second 1/2 of your post. Both of my statements are saying the same thing, my second one just threw some numbers in there.

"current from the rest of the circuit" = "40 volts to push 10 amps of current through a circuit with superconductor wire"

"you must add enough additional potential force to overcome the wire resistance" = "you are going to have to add more voltage force to push the same circuit current through a wire with some resistance"

In strictly comparing the ideal conductor or superconductor with the conductor, I did not change the power requirements of the load in the rest of the circuit.

Why are you changing the power requirements of the load? The conductor then becomes parasitic to the other load instead of just being a conductor.

You must add a force to be able to push the current through the lattice of the conductor and to compensate for the collisions that take place. The superconductor allows current to pass through without the normal collisions that take place in a conductor.

Think of the conductor as a parking lot full of randomly cars that a semi-truck barrels into. Eventually a car is going to be pushed out the other end but a lot of energy will be expended.

Now think of the superconductor as a lot full of cars but with a narrow path straight through with a car on the far end. The semi-truck can cruise through with practically no resistance, other than the pressure wave bouncing off the cars and pop the car out the other end with very little extra effort.

This is a crude analogy but works. The superconductor is not a perpetual motion machine but the cost to move the current through is of such magnitudes difference that we call it lossless. It does not have resistance in the traditional sense of the conductor because the electrons travel differently. But there still is a cost.

In the random parking lot, you have to add enough extra energy to the truck at the start so the car will pop out the end. This extra energy gets left in the parking lot full of smashed cars.

The extra energy for the conductor gets radiated out of the conductor as heat. You added more potential energy to deliver the same current to the load elsewhere in the circuit.
To reject this is to reject the basic tenets of Electrical Engineering.
And you claim these are: what?
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
The second half of the second sentence is false. Electrons exist and move only in the real world. In your theoretical contrivance of ideal zero impedance conductors and zero impedance infinite sources, there is nothing that moves at all. It is all pencil and paper and math. Show me your theoretical circuit and put a volt meter and amp meter on it and show me that real electrons are moving on the neutral without a potential difference from one end of the neutral to the other. You can't.
Crossman gary, you have got to be kidding me. No, the second half of the second sentence in not false. If you cannot make the correlation between electrical theory (pencil, paper and math) and the real world, then what is the point. We should just give up electrical engineering. Let's not teach it in school, let's not offer it as a profession, let's just get rid of it all together.

Your theoretical circuits are only an approximation of what is actually happening. There are real world MWBC neutrals all over the world with potential differences from one end to the other, with electrons moving because of these differences.
NO, NO, NO. The electrons moving in these neutrals are NOT moving BECAUSE of the potential difference from one end to the other. The electrons are moving BECAUSE we have applied a POTENTIAL SOURCE to the circuit. The potential source and the potential source alone is what moves the electrons in the circuit.

The potential difference that you measure from one end to the other of the neutral wire is a RESULT of the current that flows in the circuit. It does not matter if you look at a theoretical circuit with "ideal" circuit conductors or a theoretical circuit with "real world" circuit conductors, or at an actual real world circuit. The current will (theoretically - hope this makes you happy) flow in the neutral of the theoretical circuit with ideal circuit conductors. There will be no voltage drop along that conductor. Current will also (theoretically) flow in the the theoretical circuit with conductors with impedance. There will be a (theoretical) measure voltage drop along this neutral conductor. And of course in a real circuit, there will be a (real) current flow in the neutral with a (real) measureable voltage drop along that conductor.

And, not a single real electron ever ever moved in a theoretically perfect circuit in all the history of the earth, and therefore we cannot use the perfect circuit to make blanket statements about what is really happening.
This is irrelevant. Electrical theory is used to demonstrate with paper and pencil how a circuit will perform in the real world. It does not matter if a single real electron never moved in a theoretically perfect circuit in all the history of the earth. What matters is that the electrical theory works the same when modelling an "ideal" circuit or when modelling a "real world" circuit. It is simple enough to model an ideal circuit, and then add impedances to the circuit conductors to model a "real world" circuit.

The crux of the matter is that electrical theory, that is Ohm's Law, Kirchoff's Current Law, Kirchoff's Voltage Law, etc., can be applied to "ideal" modelled circuits and "real world" modeled circuits equally, therefore, we can model both "ideal" circuits and "real world" circuits to demonstrate how circuits will perform under different conditions.
 

crossman gary

Senior Member
You can follow, if you desire, the erroneous belief that the current flow in the neutral is a result of the potential difference from one end of the conductor to the other. However, you will be doing yourself no favors by following this belief.
David, if you can convince me that I am wrong, I will be the first to admit the mistake. I take these discussions as a learning opportunity and my mind is open. However, I am not yet convinced. Because Ohm's Law fails when we are at the extreme edge of zeros and infinities, we cannot use it to mathamatically predict what would happen under these circumstances.

We have clearly demonstrated herein that the current flow in the entire circuit is CREATED by the potential source in the circuit
Agreed

and that the potential difference from one end to the other end of the neutral conductor is a RESULT of the current that is flowing in that conductor. To reject this is to reject the basic tenets of Electrical Engineering.
Voltage drops are, nevertheless, indicative of a potential difference between two points, and, regardless of the cause of the potential difference, current will flow between these potential differences if a load is connected between them. How does a resistive voltage divider work? How does a potentiometer work?
 

crossman gary

Senior Member
Crossman gary, you have got to be kidding me. No, the second half of the second sentence in not false. If you cannot make the correlation between electrical theory (pencil, paper and math) and the real world, then what is the point. We should just give up electrical engineering. Let's not teach it in school, let's not offer it as a profession, let's just get rid of it all together.
David: To be an engineer, a person should understand where the laws apply and where the laws fail. One cannot assume that they are accurate under every condition which could possibly or theoretically exist in the universe.

Take your zero impedance conductor and close it up into a loop. There is no source voltage connected and there is zero resistance. What does math and Ohm's Law predict about the current flow?

E = I x R
0 = I x 0

What is I?
 

mivey

Senior Member
How about another try?

Suppose we have a charge train flowing down an ideal conductor. This charge train was created by a force in a galaxy far, far, away. Call the charge train something created by some source somewhere to serve the destination load. The current is a measure of the time rate of change of charges going through a point.

When this charge train hits the resistive conductor, we must add energy to push it through the conductor or part or all of it will be absorbed.

If we let the train sacrifice some of its charge to get through, the original source supplied the needed energy. The voltage across the conductor is a measure of the sacrifice.

If we want to make a complete delivery of what was ordered by the load, we must add additional power to that originally carried by the train. The voltage across the conductor is a measure of what we had to add to deliver the full load. The load will not be happy if you lose part of their order.

The extra power does not necessarily have to be added by the original source. It can be supplied by a compensating source that is an integral part of the conductor. This extra power is not part of the original charge train.

If we are going to let the original charge train supply the power, then the folks at the delivery point have already added extra to their order to compensate for these losses. They have included the extra power needed in the original order. The voltage across the conductor is a measure of those losses. They could have ordered off-setting losses from another supplier. It is a separate calculation from what they will actually need at the plant.
 

mivey

Senior Member
David: To be an engineer, a person should understand where the laws apply and where the laws fail. One cannot assume that they are accurate under every condition which could possibly or theoretically exist in the universe.

Take your zero impedance conductor and close it up into a loop. There is no source voltage connected and there is zero resistance. What does math and Ohm's Law predict about the current flow?

E = I x R
0 = I x 0

What is I?
Ohm's Law becomes a limit and the current will remain in whatever state it is already in.

If the current is already flowing, it will continue to flow unimpeded. If it is not flowing, it will not start.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
You disagree with my first statement in the first 1/2 of your post. Then you agree with my second statement in the second 1/2 of your post. Both of my statements are saying the same thing, my second one just threw some numbers in there.
Mivey, no, both of your statements are NOT saying the same thing. The first statement says that if we add impedance to the circuit, we must increase the circuit voltage to push the current through this additional impedance. This is not true, as I demonstrated.

Your second statement says that if we add impedance to the circuit, AND we want to maintain the SAME current level flowing in the circuit, then we must increase the circuit voltage. This being true, I agreed with your statement. But, the two statements are clearly NOT saying the same thing.

Why are you changing the power requirements of the load? The conductor then becomes parasitic to the other load instead of just being a conductor.
I am not "changing" the power requirements of the load. I am not even addressing the power "requirements" of the load. I gave an example of connecting a 12 ohm resistor to a 12v battery with "superconductors." the 12 ohm resistor is the circuit load, and the 12v battery is the potential source. By Ohm's law, 1 Amp will flow through this circuit. I then changed to "superconductors" to "real world" conductors, adding 0.5 ohms for each conductor to the circuit. This changes the circuit impedance to 13 ohms, but I still have my 12 ohm load resistor. I have not "changed" the power requirements of the load. In this new circuit, only 0.923 amps will flow because of the increased impedance in the circuit.

Let look at this in another fashion. Lets say we take an incandescent light bulb which is rated for 250W at 120V. This light bulb is essentially a resistance load. The resistance for this bulb is 57.6 Ohms (R=V^2/P.) If i connect this light bulb to a 120V source, 100 feet away, using #10 Awg wire. The resistance of the #10 wire is approx. 0.124 ohms over the hundred feet, so the resistance of the entire load circuit is 0.124+57.6+0.124 ohms = 57.848 Ohms. The current flowing through this circuit will be (I=V/R) 2.0744 Amps. The output of the 250W rated light bulb will be 247.86 Watts. I haven't changed the load requirement, or changed the load device, but I am getting a reduced load output because of the losses due to the impedance of the conductors.

Now lets say we take this same circuit and replace the #10 Awg conductors with #14 Awg conductors. The resistance for each of the #14 wires will be 0.314 ohms. The resistance of the load circuit is now 0.314+57.6+0.314 = 58.228 Ohms. The current flowing through circuit is now 2.0608 Amps. The output of the light bulb is now 244.6 Watts. Again, I haven't changed the load device, but the output of the lamp has been reduced because of the increase in the impedance of the circuit.

You must add a force to be able to push the current through the lattice of the conductor and to compensate for the collisions that take place. The superconductor allows current to pass through without the normal collisions that take place in a conductor.
I'm not going to get into the collisions of electrons in the circuits, but clearly the circuit still operates without an increase of voltage.

And you claim these are: what?
The tenets to which I refer are Ohm's law, KCL, KVL, etc.
 

crossman gary

Senior Member
Ohm's Law becomes a limit and the current will remain in whatever state it is already in.

If the current is already flowing, it will continue to flow unimpeded. If it is not flowing, it will not start.
Mivey, first, let me say that you are one of the forum members who I respect the most for your clear thinking. However, I cannot see that the simple E = I x R predicts what you just stated. I'm not sure what you mean when you say Ohm's Law becomes a limit. Are you talking about a mathamatical limit such as used in calculus?
 

crossman gary

Senior Member
Example 1: Take an ideal 12 volt source with no internal resistance and connect it to each end of an ideal conductor with zero impedance. What does Ohm's Law predict for the current in the conductor?

Example 2: Exact same scenario, but increase the source voltage to 480 volts. What does Ohm's Law predict for the current flow in the conductor?
David: Still waiting on an answer to these.
 

mivey

Senior Member
Mivey, no, both of your statements are NOT saying the same thing.
Pardon you. I'm the one that wrote the statements so I'm the one who knows what I said.

How you are reading it is your problem. I have made multiple attempts to clarify but you are not hearing me.
I am not "changing" the power requirements of the load....but I am getting a reduced load output because of the losses due to the impedance of the conductors.
Then you have changed the power delivered to the load. Six of one...
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
David: To be an engineer, a person should understand where the laws apply and where the laws fail. One cannot assume that they are accurate under every condition which could possibly or theoretically exist in the universe.

Take your zero impedance conductor and close it up into a loop. There is no source voltage connected and there is zero resistance. What does math and Ohm's Law predict about the current flow?

E = I x R
0 = I x 0

What is I?
I is zero. 0 x 0 = 0. Seems to me Ohm's Law works fine in this instance. There is no potential source, therefore no force to create the current.

Voltage drops are, nevertheless, indicative of a potential difference between two points, and, regardless of the cause of the potential difference, current will flow between these potential differences if a load is connected between them.
Just above this point, you agreed with me that it is the potential source that causes the current to flow in the circuit, yet here you resort to your previous belief. Voltage drops are not an indicator of a potential difference between two points, but instead a potential difference between two points is an indication of a voltage drop. (Sorry, don't mean to get all "chicken and the egg" here, but it is an important point.)

You say that regardless of the cause of the potential difference current will flow between the potential difference. Again, I would say NO. Not regardless of the cause, but because of the cause. This may be overly simplified, but try this: You must start at the beginning - you must have a potential source for current to flow in a circuit. This is step 1. Next you must have a load device, and you must complete a closed circuit to start the current flowing in the circuit. This is step two. If we complete a circuit with a switch in it and leave the switch open, then current will not flow. The potential exists on the circuit, but current will flow. I can measure a potential difference between the line side of the switch and the "neutral" side of the load, but no current flows.

With the switch closed, and current flowing in the circuit, there will be a voltage drop along the impedance elements of the circuit. The load (resistor or light bulb, or whatever) will have a voltage drop across it because current is flowing through it. The conductors, which offer impedance to the circuit, will have a voltage drop across them because of the current flowing through them. The voltage drops of each of the elements in the load circuit must add up to the voltage of the potential source, according to Kirchoff's voltage law (KVL.) If I connect a lamp to a 120v circuit, and measure a 1 volt drop on one supply conductor, and a 1 volt drop on the second supply conductor, then the voltage drop across the lamp must be 118 Volts.

So from beginning to end, you must have a potential source to get current to flow, you must complete a closed load circuit to get current to flow, and once that current is flowing, you will achieve a voltage drop across the impedances in the circuit. The voltage drops on the circuit are only a Consequence of the current flowing in the circuit, and have nothing to do with the current actually flowing in the circuit.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
Pardon you. I'm the one that wrote the statements so I'm the one who knows what I said.

How you are reading it is your problem. I have made multiple attempts to clarify but you are not hearing me.Then you have changed the power delivered to the load. Six of one...
Yes, I understand that you wrote the statements, however, sometimes when writing a statement, what you write conveys a different meaning that what you intend. That is why I always have someone proofread letters before I send them out. I believe that is the case here.

Your first statement said that if you increase the impedance in a circuit (or replace a superconductor with a real conductor) that you will need to increase the voltage to get current to flow through the circuit. This is not true, surely you see that. Specifically, you said that one must add enough potential source to overcome the resistance in the wire. Even if I took an existing circuit with #10 Awg wires and a light bulb and replaced the #10 Awg wires with #14 Awg wires, current will still flow through the circuit without increase the voltage of the potential source.

Your second statement said that if a circuit is pushing 10 amps through a circuit with 40 Volts, then by adding more resistance to the circuit wires, you will have to increase the voltage in order to maintain 10 amps flowing in the circuit. This is undoubtedly true, but it is not the same statement as the first statement.
 

crossman gary

Senior Member
I is zero. 0 x 0 = 0. Seems to me Ohm's Law works fine in this instance. There is no potential source, therefore no force to create the current.
Mathamatically, your conclusion is false. Dealing on theory and math only, and not the real world, the current can be any real number of amps and the solution still works. For example:

E = I x R
0 = 6 x 0

Why can't the current be 6 amps?

The proper mathamatical answer to

0 = I x 0

is not I = 0

I in this case is undefined, arbitrary.

For you to say that the current is zero is an admission that ohm's law and math has limitations when dealing with this subject.
 

mivey

Senior Member
Mivey, first, let me say that you are one of the forum members who I respect the most for your clear thinking. However, I cannot see that the simple E = I x R predicts what you just stated. I'm not sure what you mean when you say Ohm's Law becomes a limit. Are you talking about a mathamatical limit such as used in calculus?
Intuition. There might be a proof, but that would take more thought.

Look at the two limits of I = E/R. As the resistance approaches zero, the current approaches infinity. As the voltage approaches zero, the current approaches zero.

One is pushing the current to infinity while the other is pushing it to zero. The net is an equal and opposite action on the current so the current state will be unchanged.

If we approach zero by cranking down the voltage or approach infinity by cooling down the super conductor, a difference in the rate of change will make a change in the current. But in our ideal case, we have warped instantly to zero resistance and voltage.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
David: Still waiting on an answer to these.
Crossman gary, you are discussing a theoretical ideal short circuit (with even zero impedance in the potential source.) The current that will flow in these circuit, according to Ohm's law will theoretically be Infinity. I do not see why Ohm's law would be a problem in this theoretical case.

Let's look at it this way. Suppose we put a theoretical variable resistance into this circuit would could be adjusted down to zero ohms. According to Ohm's law I=V/R, as the resistance decreases, the current increases. As resistance reaches zero, the current goes to infinity.

This is fine in my eyes, because "Infinty" is a theoretical concept. You cannot divide by zero in the real world, because "infinity" does not exist in the real world. But in theory, Ohm's law works even in this extreme situation.

As an aside, for another theoritcal concept that is used in electrical engineering that does not exist in the real world, consider to multiple ways to represent voltage and current in power systems. A voltage of 120V at a phase angle of 30 can be represented in a polar form: 120V<0, or in a phasor form: 120*sin(21660*t+30)Volts, or in a rectangular form: 103.9 + j60 Volts. There constant "j" in the rectangular form is a theoretical value of the square root of negative one. This constant cannot exist in the real world, but is used every day in electrical analysis.
 

mivey

Senior Member
...Voltage drops are not an indicator of a potential difference between two points, but instead a potential difference between two points is an indication of a voltage drop...
OK, I'm lumping a voltage across something and a potential across something as the same thing. Maybe there is a point you are making that I'm missing.
...So from beginning to end, you must have a potential source to get current to flow, you must complete a closed load circuit to get current to flow, and once that current is flowing, you will achieve a voltage drop across the impedances in the circuit. The voltage drops on the circuit are only a Consequence of the current flowing in the circuit, and have nothing to do with the current actually flowing in the circuit.
Are you saying there is no current flow if you have a series capacitor or are you saying there is no potential across a capacitor unless current is flowing?
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
Mathamatically, your conclusion is false. Dealing on theory and math only, and not the real world, the current can be any real number of amps and the solution still works. For example:

E = I x R
0 = 6 x 0

Why can't the current be 6 amps?

The proper mathamatical answer to

0 = I x 0

is not I = 0

I in this case is undefined, arbitrary.

For you to say that the current is zero is an admission that ohm's law and math has limitations when dealing with this subject.
Crossman gary, I'm afraid you are wrong on this matter. You are looking at the Ohm's Law equation of V=I*R and saying - aha, if V = 0 and R = 0, then I = anything I want it to.

You are not following Ohm's law in this case. Ohm's law says "the current through a conductor between two points is directly proportional to the potential difference or voltage across the two points, and inversely proportional to the resistance between them. You have clearly stated that there is no potential source in your circuit, therefore there is no potential difference and therefore no current. The current is zero.
 
Status
Not open for further replies.
Top