MWBC= more heat or Less heat?

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crossman gary

Senior Member
Hi, just popping my head up when I really should not be spending time on the intarwebs :)
Thanks for popping in. As always,I highly respect your thoughts and clear-headedness from past discussions, and say "thank you" for the insights. I hope you can shed more light on the discussion.

Strictly speaking, Ohm's law applies to DC systems, not AC systems. The concept of Ohm's law has been extended to AC, by adding terms for things such as inductance and capacitance, but the original concept put forth by Ohm himself was DC.
I am perfectly fine with limiting the issues I am discussing to DC. No problem there.

For what its worth, the 'continuous loop of superconducting wire' is regularly used in industry, and to the best ability to measure, V and R are zero, and I has a finite non-zero value. The practical application is the superconducting magnet, for example used in NMR and particle accelerators.

With the switch open, a power supply applies a voltage to the terminals. Even though this is a loop of superconductor, the applied voltage results in a finite current; the loop of wire (usually many turns) is an inductor, thus an applied voltage results a finite and continuously changing current flow. This is a very low frequency AC effect. Once the current has ramped up to the desired value, the shorting switch is closed, and the power supply simply disconnected.

Once the switch is closed, the system is DC, with no voltage drop anywhere, no resistance anywhere, a finite current flow, and a constant magnetic field.
I can't dispute that. Ohm's law would say that the current is undefined.

There are dissipation mechanisms (resistance) for AC current flow, and theories suggest some dissipation mechanisms for DC, which would mean very low resistance rather than true zero resistance...but the resistance is so low that it would take tens thousands of years for the current to decay, if not much longer.
Seems to me that if the superconductor magnetic field is being used to do work, then there would be an "impedance" acting on the electrons in the loop. Energy would have to be added to the circuit to maintain whatever work the magnetic field is doing. So, if we are deriving "work" from the magnetic field, we would have to be adding energy into the system, if not continuously, then at least on occasion. The amount of work being done could possibly be very small considering that we are pushing around individual atomic particles. But, when we take into account that the speeds are relativistic, the energy required would be massive.

Obviously I don't know anything about the above. It all seems very strange. I'm going to go out on a limb and say that massive amounts of energy are being routed into the super-collider system to make the atomic particles move at relativistic speeds. The superconductor magnet is not some kind of magic panacea to create energy out of nothing. My guess is that the superconductors are being used to increase the efficiency from the massive amount of power used.... not that current can flow in the loop with no voltage applied. I guess some research is in order.

Thanks winnie.
 

crossman gary

Senior Member
Rattus had posted that we can figure out what is happening at zero resistance by taking limits as the resistance approaches zero. This may very well be the case for current flow. However, other issues arise. Everyone would understand that a non-zero voltage applied to a resistance is going to cause current flow. As we lower the resistance, the current goes up. (assume an infinite, ideal voltage source)

The closer we get to zero ohms, the higher the current is. We can get arbitrarily close to zero, and the current is huge, but still finite.

Once the resistance actually equals zero, then by intuition, we can say the current is infinite.

All well and good, everything is acting in a linear fashion, with no huge gaps or discontinuities.

But let's look at power.

With the voltage and the resistor, current flows, and we have a certain amount of power, equal to current squared times resistance.

As the resistance decreases, current increases and power increases.

As we get arbitrarily close to zero ohms, power is tremendously high.

Once we reach zero ohms however, we have a huge discontinuity because the power goes from massive to zero. Current squared x zero ohms = zero.

Just an absurdity I see with zero resistance.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Gary,

Thanks for the kind words.

winnie said:
Strictly speaking, Ohm's law applies to DC systems, not AC systems. The concept of Ohm's law has been extended to AC, by adding terms for things such as inductance and capacitance, but the original concept put forth by Ohm himself was DC.
I am perfectly fine with limiting the issues I am discussing to DC. No problem there.
To emphasize this point, so that I make perfectly clear what you are agreeing to, consider a very simple circuit consisting of a 1 ohm resistor, a switch, and a 1 volt source. Ohms law doesn't tell you the currint flow after you close the switch. Rather it tells you the _steady_state_ current after the AC effects of the switching event have died down.

winnie said:
Once the switch is closed, the system is DC, with no voltage drop anywhere, no resistance anywhere, a finite current flow, and a constant magnetic field.
I can't dispute that. Ohm's law would say that the current is undefined.
I would say instead that Ohm's law is not capable of telling us what the current is, or that Ohm's law does not define the current in this case. In these systems, the current is known and quite precisely defined by the AC history of the applied voltages to the loop terminals. The current flow in the superconducting loop remains constant at the level it was at when the switch was closed.

Seems to me that if the superconductor magnetic field is being used to do work, then there would be an "impedance" acting on the electrons in the loop. Energy would have to be added to the circuit to maintain whatever work the magnetic field is doing. So, if we are deriving "work" from the magnetic field, we would have to be adding energy into the system, if not continuously, then at least on occasion. The amount of work being done could possibly be very small considering that we are pushing around individual atomic particles. But, when we take into account that the speeds are relativistic, the energy required would be massive.
Remember that magnets do no work on charged particles. The magnetic force that the particle experiences is always perpendicular to the motion of the particle, so force*distance is always zero. There is considerable energy _stored_ in the magnetic field, equal to the integral of V*I during the 'ramp up' period when the magnet was being turned on; but this energy doesn't get used up running the accelerator.

These particle accelerators have to expend lots of energy in the changing electric fields used to speed up the particles, and lots of energy is spent running the refrigerators to keep the magnet coils cold enough to be superconductors; so we are not talking about any sort of free lunch. As far as the accelerator is concerned, the superconducting magnets are simply very powerful permanent magnets.

-Jon
 

Rick Christopherson

Senior Member
.....(let's call it Question 2) of whether or not Ohm's Law holds for a zero resistance conductor.
Yes, Ohm's Law is still valid with zero resistance. You're expecting it to give you a concrete answer, but it is giving you exactly what it is supposed to be giving you. It isn't the equation that is failing, but your expectation from the equation that is faulty.

Ohm's Law maintains that with zero resistance, there will be no voltage regardless what the current may be. Rearanging the equation to solve for current, I=V/R, this becomes 0/0 which is undetermined, and from mathematics, an undertermined value can be anything from zero to infinity.

Ohm's Law does what it is supposed to do.

By the way, the circuit you drew back in post #209 is not a failure of Ohm's Law, but a failure of your ideal circuit components.
Hence my getting in the "trap."
As you have now come to realize, the trap is not believing that a voltage source and voltage drop are the same. They are the same, and that is what I pointed out a long time ago. The individual electrons don't care where or why the potential difference exists, only that they need to see one in order to move through a reistive path.

Saying that a voltage source is different from a voltage drop is nothing more than semantics, as I have already said before. The term "voltage source" carries an "implication" (whether actually true or not) of a proper power supply (like Radio Shack). However, most people would not consider the induced voltage from an adjacent circuit as a power source, yet it will still result in a small current to flow. This small current is why there is a difference between reading a "phantom voltage" (I do not like that term, personally) using a high impedance DVM versus a lower impedance AVM.

The electrons do not care where/why a potential difference exists, they know they need to move. Furthermore, the labels "active" and "passive" circuit elements is also a matter of semantics, as an active element can become passive, and a passive element can become active (such as the induced voltage in a wire). Putting a label on a device does not change its characteristics. We could label a wire as being a passive element, but our assumption would be shown to be incorrect if subsequent circuit analysis reveals that a magnet stuck to a nearby spinning driveshaft turned this otherwise passive element into an active element.
And this is where the issue arises. Really, my interest in this topic came from viewing your discussion of the zero resistance neutral in the unbalanced MWBC. Your point was that current could flow through the conductor even without a voltage drop on it.
Here is where the problem arose in the very beginning. You can't blindly replace a real conductor with an ideal conductor, when it is the behavior of the conductor itself that you are examining.

The original discussion was examining the behavior of the voltage between two nodes in a circuit that were separated by a real conductor. When David replaced that real conductor with an ideal conductor and no resistance, he changed the two separate nodes into a single node. This would have been acceptable if the sole examination of the circuit was limited to load balance, but it was also looking at the behavior of the conductor itself.
 

Rick Christopherson

Senior Member
As we get arbitrarily close to zero ohms, power is tremendously high.

Once we reach zero ohms however, we have a huge discontinuity because the power goes from massive to zero. Current squared x zero ohms = zero.

Just an absurdity I see with zero resistance.
Just because it seems absurd, does not mean it is wrong. I don't remember how to define an equation as it approaches an indeterminate value, but I suspect that the limit of the equation (is that called the asymptote?) would be commensurate with the voltage of the source, and not running away toward infinity before dropping to zero. I am just guessing at this, but something tells me that as R approaches zero, the power function is going to be something similar to the voltage.

However, notice in my previous posting that the circuit you are trying to examine is not a failure of Ohm's Law, but that you are approaching a limitation of using Ideal components. When you reach R=0, you encounter a paradox, in that your conductor cannot have a voltage across it, and your power supply will always maintain a voltage.

It is your Ideal components that are in conflict, not the equations.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
If you pile on enough 'ideal' situations, you will reach a point where your equations fall apart. These ideal components of which we speak are generally only approximations to reality; by using 'ideals' you get a useful idea of what is happening, but only by throwing out a bunch of messy complex reality.

The question of 'what happens when you apply a zero resistance to a voltage source' is actually answered pretty simply. The source has an internal resistance, and so the output terminal voltage of the source will drop as the current increases. In the limit of zero resistance, the output terminal voltage is _zero_, the current is set by the internal resistance of the source, and all the power is dissipated in the source itself.

Now pile on another level of 'ideal', and make it an ideal voltage source, one that has zero internal resistance and that maintains its output terminals at some constant finite value. If you connect such an ideal voltage source to a zero resistance, then Ohm's law tells us that the current will be infinite. _However_ that is in the DC case, after your switching transients have settled out. IMHO you _must_ think AC for this case.

I would claim that the current is still limited by the AC inductance of the system, and rather than instantly getting infinite current, the current will ramp up in a steady fashion. The rate of this ramp will depend upon the applied voltage and the inductance of the system. After any finite amount of time, the current flowing will be finite, and the energy stored in the inductance will be finite. Only after infinite time will the current flow become infinite.

Now if you pile on the third 'ideal' of zero inductance, and your system has diverged so far from reality that you don't learn anything about the universe by trying to make a prediction.

IMHO it is quite a wonderful thing that we have something that actually approaches one of these 'ideals'. You can actually buy a piece of material that really has zero resistance. You can't buy a perfect voltage source (though with amplifiers you can approximate one), and you will _always_ have some inductance.

But you really can have zero resistance, with some limitations. Superconductors can only carry so much current; push too much current through them, and they will cease to be superconductors. Put them in too strong a magnetic field and they will cease to be superconductors. Too warm? You guessed it....

-Jon
 

crossman gary

Senior Member
Ohm's Law maintains that with zero resistance, there will be no voltage regardless what the current may be. Rearanging the equation to solve for current, I=V/R, this becomes 0/0 which is undetermined, and from mathematics, an undertermined value can be anything from zero to infinity. Ohm's Law does what it is supposed to do.
Considering normal voltage sources, there is no doubt.

By the way, the circuit you drew back in post #209 is not a failure of Ohm's Law, but a failure of your ideal circuit components.
That is certainly one way to look at it.

As you have now come to realize, the trap is not believing that a voltage source and voltage drop are the same. They are the same, and that is what I pointed out a long time ago. The individual electrons don't care where or why the potential difference exists, only that they need to see one in order to move through a reistive path.
You have been correct all along on this one. For those who had or still have doubts, sometimes more discussion is needed to convince them of the facts. Simply stating the fact, no matter how obvious, is sometimes not enough. I certainly appreciate the members here on the forum who expend a considerable amount of energy and time discussing the opinions, and giving explanations or proofs of WHY things are like they are, instead of just saying "this is how it is" but failing to explain. I understand that ultimately it is up to each member to decide what is worth their time and effort.

I feel that David is still going to disagree on the VD vs PD issue. I do not mind spending time trying to convince him otherwise, in whatever manner I can think of.

You can't blindly replace a real conductor with an ideal conductor, when it is the behavior of the conductor itself that you are examining.
Agreed.

The original discussion was examining the behavior of the voltage between two nodes in a circuit that were separated by a real conductor. When David replaced that real conductor with an ideal conductor and no resistance, he changed the two separate nodes into a single node.
Excellent point, and something that was in the back of my mind earlier in the thread. I even mentioned that using the zero resistance neutral was akin to connecting the loads directly to the source, thereby eliminating the MWBC. The fact that this zero resistance neutral becomes a single node certainly seems obvious now.
 

crossman gary

Senior Member
Ohms law doesn't tell you the currint flow after you close the switch. Rather it tells you the _steady_state_ current after the AC effects of the switching event have died down.
Yes. It is all too easy to fall into the habit of dismissing those things. Now, under most circumstances, especially those dealt with by us typical electricians, forgetting about the increasing magnetic field and inductance and self induction isn't a big deal. But at the extremes, they certainly warrant examination.

I would say instead that Ohm's law is not capable of telling us what the current is, or that Ohm's law does not define the current in this case. In these systems, the current is known and quite precisely defined by the AC history of the applied voltages to the loop terminals. The current flow in the superconducting loop remains constant at the level it was at when the switch was closed.
Thanks for that insight. Time for me to do some research!

Remember that magnets do no work on charged particles. The magnetic force that the particle experiences is always perpendicular to the motion of the particle, so force*distance is always zero. There is considerable energy _stored_ in the magnetic field, equal to the integral of V*I during the 'ramp up' period when the magnet was being turned on; but this energy doesn't get used up running the accelerator.
I'm certainly not up on the aspects of what you just said. However, now that I think of it, a permanent magnet doesn't get weaker everytime we use it to move something. So I think I get what you are saying.

These particle accelerators have to expend lots of energy in the changing electric fields used to speed up the particles, and lots of energy is spent running the refrigerators to keep the magnet coils cold enough to be superconductors; so we are not talking about any sort of free lunch. As far as the accelerator is concerned, the superconducting magnets are simply very powerful permanent magnets.
I can see that there would be a huge energy requirement to turn on/off the superconductor magnets for the timing needed to speed up the particle.

The same could be done with permanent magnets - except we would have to move the magnets into place and then back in a precise sequence. This of course would require energy.

Anyway, I am going way off topic. I need to do some more studying of magnetic fields.
 

crossman gary

Senior Member
Thank you Rick and Winnie.

I am satisfied with my understanding of the topics at hand, so I won't be saying much more - unless David still wants to spend more time with it. I'm always happy to discuss a topic, even when I don't know much about it.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
Hi David,

At the time I authored that post, my thinking was wishy-washy on the "voltage drop" and "potential difference" issue. Intuitively I was thinking that they are the same thing, but your line of reasoning had me contemplating the possibility that they aren't. Hence my getting in the "trap."

Yesterday, while doing yard work and mowing the grass, I was thinking about this. I have come to the conclusion that potential difference and voltage drop are the same thing, or at the very least, they accomplish the same exact thing. Remember the Thevinin equivalent circuit? For any given voltage drop across a load, we can substitute the Thevinin equivalent voltage source and you cannot tell the difference between the voltage drop or the potential difference. Functionally, voltage drop and potential difference are exactly the same thing.
Crossman gary, maybe this is just too much of a philosophical concept (you know, if a tree falls in a forest) or maybe I'm not making my point very well. You say that you have concluded that "potential difference" and "voltage drop" accomplish the same exact same thing. I take interest in your use of the word "accomplish," because it goes to the heart of the point I have been trying to make. I do not believe that "voltage drop" ACCOMPLISHES anything. I believe that "voltage drop" is a RESULT of current moving in a circuit. If you think about the discussion we had of "active" and "passive" elements in a circuit (and by the way, I have borrowed these concepts from Engineering Circuit Analysis, Hayt & Kemmerly, 4th Edition,) then we could perhaps think of "potential difference" as being "active" and "voltage drop" as being "passive." Part of the problem, of course, is that there is not a defined convention for these two terms, and "potential" & "voltage" or "difference" & "drop" are used interchangably all the time.

Let me use a mechanical analogy to demonstrate my point (though I'm sure most engineers in here would like to avoid those type of analogies,) since I have the pleasure of working with mechanical engineers everyday. Lets imagine a closed loop hot water pumping system that delivers water to heating coils in fan coil units. The pump would equate to the "potential source," as it provides the pressure in the system which causes the water to flow. The water flowing through the pipes equates to the current, with the friction in the pipe being being the resistance in the conductor. The heating coil equates to the load (resistor, etc.) As water flows through the heating coil, there is a pressure drop across the coil (like voltage drop.) As water moves through the pipes, the friction in the pipe creates a pressure drop along the pipe (like voltage drop.) But would anyone suggest that it is the pressure drops in the system that is causing water to flow? Or is it only the pump that causes water to flow? (Sorry for going there, and yes, I know the mechanical analogy doesn't provide an accurate analogy - its just a thought experiment.)

Lets try another thought experiment based on your idea of connecting a resistor across an existing voltage drop. Imagine a 12V battery with leads brought out and labeled A & B. The open circuit potential between A& B will be 12V. So if we follow your idea and connect a 1 Ohm resistor across that "voltage drop," then current will flow through it. We can predict that the current flow will be 12 Amps, which it will be. Now consider if we disconnect that resistor only at the B terminal, and label the other end of the resistor as terminal C. The open circuit potential between C & B will be 12V. If I add another 1 Ohm resistor across the 12V "voltage drop" between C & B, then we would expect the current through the resistor will be 12 Amps (V=IR: we have an existing "voltage drop" of 12V and we are adding a resistor of 1 Ohm, so current must be 12 Amps.) Of course the current through the resistor will only be 6 Amps. Now you might say, isn't that a failure of Ohm's law, but, no it is not. It is a failure of reasoning in believing that "potential difference" and "voltage drop" are the same thing.


And this is where the issue arises. Really, my interest in this topic came from viewing your discussion of the zero resistance neutral in the unbalanced MWBC. Your point was that current could flow through the conductor even without a voltage drop on it. This leads to two questions in my mind: The question (let's call it "Question 1) of potential difference versus voltage drop, and the question (let's call it Question 2) of whether or not Ohm's Law holds for a zero resistance conductor. Both of these questions are interelated.

On Question 1, I now believe that Thevinin shows that voltage drop and potential difference are the same thing.
I haven't looked at Thevenin, but I hope I have made clear above why you must look at "potential difference" and "voltage drop" as being different things. Potential difference is created by the potential source (battery, generator, etc.) and is the only force in the circuit which causes current to flow. Voltage drop is created by the flow of current through the resistance of the circuit. Voltage drop is not active, it is created by current flow. It does not cause current to flow.

On Question 2, I am attempting to show the absurd results that one may get by applying ohm's law to a zero resistance. This is my Black Box experiment. I haven't read all the latest posts yet, but I hope you or someone will/has commented on the expected results so I can finish my line of thought.
I haven't has a chance to review your Black Box experiment after you provided some clarifications, but I will try to do so. I'm sure it won't be something that I'd be able to just breeze through though.

In my mind, we cannot simply assume that the zero volt, zero ohm situation results in zero current. If we are looking at ohm's law at face value, and we stick to the accepted principles of mathamatics, then ohm's law fails to provide insight into zero ohm, zero volt situations. If a mathamatician without preconceived notions of electricity were to apply ohm's law to this situation, he would contend that the current is undefined, not zero. He would say that ohm's law cannot give us an exact level of current flow.

Ohm's law also says:
E = I x R
0 = 1 x 0
0 = 2 x 0
0 = 3 x 0
0 = 500 x 0
0 = 100000000000000000 x 0
and so on....

If Ohm's Law is valid for the zero volts, zero ohms situation, why would we preconclude that the current is zero? Just apply ohm's law and let the math fall where it may.
Again, you are missing the point. You say that "if Ohm's law is valid for the zero volts, zero ohms situation, why would we "preconclude" that the current is zero?" I honestly think you are trying to over analyze this situation. Think more rationally than theoretically. You defined a circuit with NO potential source, and a loop of zero resistance conductor. If there is no potential source, then what is going to cause current to flow in the circuit?:confused: We KNOW that the current is zero, because the parameters that you have set up have told us that the current will be zero. We would we need to calculate what we already know?

If we use our intuition to decide that it is pointless to apply ohm's law when we "know" that the current is zero, then we are not being mathamatically rigorous, and we are not being intellectually honest. And if zero ohms and zero volts is automatically zero amps, then ohm's law fails to predict this.
It is not "intuition" that is at force here, it is rationality. We haven't assumed that the current is zero, you have told us that it is when you defined your circuit. Zero ohms and zero volts is NOT automatically zero amps, but it IS zero amps in the example circuit with no potential source and a loop of zero resistance wire for the reasons listed above.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
Yes, Ohm's Law is still valid with zero resistance. You're expecting it to give you a concrete answer, but it is giving you exactly what it is supposed to be giving you. It isn't the equation that is failing, but your expectation from the equation that is faulty.

Ohm's Law maintains that with zero resistance, there will be no voltage regardless what the current may be. Rearanging the equation to solve for current, I=V/R, this becomes 0/0 which is undetermined, and from mathematics, an undertermined value can be anything from zero to infinity.

Ohm's Law does what it is supposed to do.
Rick, brilliantly said!:wink: Crossman gary has been hung up on trying to find a situation where Ohm's Law fails (and I do not mean that in a negative fashion - I think this thread has been great for stimulating the old gray matter (and its getting older every day.)) So, last night I was trying to think of a circuit that would get the type of failure that crossman gary was looking for that would be in the same vein in as his closed loop of zero resistance wire. What I came up with was, instead of having a circuit with no volts (and therefore by definition, no amps) and no ohms, and trying to figure out the the current would be, why not connect two resistors in a loop (or one, or three, etc.) with no potential source (and therefore no current) and try to use Ohm's law to try to determine resistance in the circuit, for clearly the resistance is not zero. It's almost the same as the original circuit, but in the original circuit 0 Volts, 0 Ohms AND 0 Amps were all KNOWN values. But lets assume that we take two resistors and don't read what their values are. Applying Ohm's law R=V/I yields R=0/0 which is undeterminable.

The thought that occured to me when pondering this is the exact same thought that you expressed above. I suppose one could argue that Ohm's law has failed in this case, because it didn't tell us what the resistance in the circuit is, but really, Ohm's law has done exactly what it should do. It has told us that the resistance in the circuit cannot be determined. And logically it cannot be determined because there is not enough information in the circuit to provide an answer. What has failed, as you noted above, is the expectation that Ohm's law must give us a concrete answer, when an indeterminent answer is just as valid.

By the way, the circuit you drew back in post #209 is not a failure of Ohm's Law, but a failure of your ideal circuit components. As you have now come to realize, the trap is not believing that a voltage source and voltage drop are the same. They are the same, and that is what I pointed out a long time ago. The individual electrons don't care where or why the potential difference exists, only that they need to see one in order to move through a reistive path.
I must still disagree with you here. Voltage source and voltage drop are not the same. May last post to crossman gary further expounded on this. Hopefully I have made my point more clear. It doesn't matter whether the electrons care where or why the potential difference exits, the fact still remains that the potential source creates the force that moves the electrons, and that the movement of the electrons through the resistive elements in the circuit creates the voltage drops. The model of the resistor connected to the battery with the zero resistance conductors shows that there will be no voltage drop across the conductors even when there is current flowing through those conductors. If there is no voltage drop along the conductor how is there current flowing through the conductor? You say "the individual electrons don't care where or why the potential difference exists, only that they need to see one in order to move through a RESISTIVE path. This is fine, but then how do the electrons move through a path with zero resistance? There must be some force that is moving them. But by your description, they are moving with no force applied to them.

Saying that a voltage source is different from a voltage drop is nothing more than semantics, as I have already said before. The term "voltage source" carries an "implication" (whether actually true or not) of a proper power supply (like Radio Shack). However, most people would not consider the induced voltage from an adjacent circuit as a power source, yet it will still result in a small current to flow. This small current is why there is a difference between reading a "phantom voltage" (I do not like that term, personally) using a high impedance DVM versus a lower impedance AVM.
It is not semantics for the reasons I have outlined above. I make no implication of a "proper power supply," in fact, I do not care what the source of the potential is, only that it exits. But it must exist before current can flow. I think it is pretty logical that there is a causal relationship here. If we have a circuit with various resistive elements, but no potential source, then we have no voltage drops in the circuit. But if we add a potential source, the potential source causes current to flow, and current flowing through the resistive elements creates voltage drops along them.


The electrons do not care where/why a potential difference exists, they know they need to move. Furthermore, the labels "active" and "passive" circuit elements is also a matter of semantics, as an active element can become passive, and a passive element can become active (such as the induced voltage in a wire). Putting a label on a device does not change its characteristics. We could label a wire as being a passive element, but our assumption would be shown to be incorrect if subsequent circuit analysis reveals that a magnet stuck to a nearby spinning driveshaft turned this otherwise passive element into an active element. Here is where the problem arose in the very beginning. You can't blindly replace a real conductor with an ideal conductor, when it is the behavior of the conductor itself that you are examining.
The concept of "active" and 'passive," as I have noted comes from Engineering Circuit Analysis, Hayt & Kemmerly, 4th Edition, and I think are valid concepts. You mention a "passive" element becoming "active" such as the induced voltage in a wire. But the wire has not become "active" by itself. You have introduced a potential source to create the induction. I believe part of the decription of "active" and "passive" that I posted earlier included the idea that an "active" element will deliver power into a circuit (or receive power from another "active" element in a circuit) while a "passive" source is only capable of receiving power (or certain "passive" elements may receive power and then return it to other elements, but they cannot deliver power on their own.) This is the active/passive concept that Hayt and Kimmerly describe. (If I didn't provide this description earlier, I should have.) The wire with the induced voltage is receiving power from a separate source and then delivering it to other elements in the circuit. This is consistent with Hayt & Kimmerly's concept.

The original discussion was examining the behavior of the voltage between two nodes in a circuit that were separated by a real conductor. When David replaced that real conductor with an ideal conductor and no resistance, he changed the two separate nodes into a single node. This would have been acceptable if the sole examination of the circuit was limited to load balance, but it was also looking at the behavior of the conductor itself.
I'd have to look back to the beginning of the post, but I don't believe I ever described a situation of replacing the real conductor between two nodes in the balanced MWBC circuit with an ideal conductor. I think I only introduced the concept of the ideal conductor to create a model where current can flow along a conductor with no voltage drop along that conductor. If I get the time, I'll go back and take a look.

Again, great post Rick. As I was thinking last night trying to find a circuit where Ohm's law would fail in the way that crossman gary was trying to demonstrate, (and coming up with two resistors in a loop which led me to the idea that, although Ohm's Law can't tell us what the resistance in the circuit will be, it is not Ohm's Law which has failed but the "expectation" that Ohm's law must provide a concrete result that has failed) I couldn't not wait to login today to provide crossman gary with my example. Imagine my surprise, when I did login and review the posts since last night that you had posted the same concept about the "failed expectations" regarding Ohm' Law.

Keep up the great work.
 

Rick Christopherson

Senior Member
I must still disagree with you here. Voltage source and voltage drop are not the same. May last post to crossman gary further expounded on this. Hopefully I have made my point more clear.
No. To the contrary, you have not made any points. You have simply repeated the statements and your only supporting argument has been that everyone is wrong except for David.

Which comes first (i.e. chicken or egg :cool:):
a) Does the voltage across a resistor cause the current through the resistor?

OR

b) Does the current through the resistor cause the voltage across the resistor.

Whether you realize it or not, when you claim that voltage drop and voltage source are not the same, you have unwittingly, undermined Norton and Thevenin. Even though you would like to prove yourself to be correct, to do so would disprove Norton and Thevenin. Since these two are just theorems at this time, they are neither proven nor disproven, and as such, your argument can neither be proven or disproven--hence, you are arguing semantics.
It is not semantics for the reasons I have outlined above. I make no implication of a "proper power supply," in fact, I do not care what the source of the potential is, only that it exits.
To the contrary, you have just demonstrated that you are arguing semantics. If it doesn't matter how/why a voltage difference exists, you have contradicted youself that a voltage source is different from a voltage drop.

There is nothing wrong with thinking the way that you are, but when you state that other views are wrong, well, that is when you are wrong.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Again, you are missing the point. You say that "if Ohm's law is valid for the zero volts, zero ohms situation, why would we "preconclude" that the current is zero?" I honestly think you are trying to over analyze this situation. Think more rationally than theoretically. You defined a circuit with NO potential source, and a loop of zero resistance conductor. If there is no potential source, then what is going to cause current to flow in the circuit?:confused: We KNOW that the current is zero, because the parameters that you have set up have told us that the current will be zero. We would we need to calculate what we already know?
This is exactly the reason that I keep bringing up the issue of 'thinking AC'.

The superconducting magnets that I've been describing truly are loops with zero resistance and zero voltage, yet with non-zero current flow. The potential source that started the current flowing was there at some point in the past, but once the magnet is charged the potential source is taken away. Using Ohm's law (DC) doesn't tell us that the current is zero; it tells us that the current can have any value. Only by considering the way voltage was applied over time can you determine what current actually ends up flowing.

-Jon
 

ohmhead

Senior Member
Location
ORLANDO FLA
Well think of this do you think one can have a resistance meaning a conductor alone and produce a voltage and current from that ?

R = E/I E = ZERO I = ZERO but with a resistance & heat you get voltage and current .


With heat i can make voltage and current . There is more than voltage and current and resistance alone . Heres why take a flame apply it to a single strand of copper conductor measure voltage a current flow thur that single copper conductor.

It says to find the unknown you need two but i just found two values from not using ohms law no formula needed.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
This is exactly the reason that I keep bringing up the issue of 'thinking AC'.

The superconducting magnets that I've been describing truly are loops with zero resistance and zero voltage, yet with non-zero current flow. The potential source that started the current flowing was there at some point in the past, but once the magnet is charged the potential source is taken away. Using Ohm's law (DC) doesn't tell us that the current is zero; it tells us that the current can have any value. Only by considering the way voltage was applied over time can you determine what current actually ends up flowing.

-Jon
Is there really any evidence that the resistance is actually zero in a superconducting magnet circuit, or is it just a resistance that is approaching zero?
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Is there really any evidence that the resistance is actually zero in a superconducting magnet circuit, or is it just a resistance that is approaching zero?
I believe the question is still open, and that to the best of our measurement ability the resistance really is zero, but there is enough room for error in the measurements that the conductivity might 'only' be 10^15 times better than copper.

I like the description on page 4 of this document:
http://ocw.mit.edu/NR/rdonlyres/Ele...B7-5407-44DB-87D0-D63A05C8F78E/0/lecture2.pdf

'Superconductivity' means zero resistivity. 'Perfect conductivity' means resistivity that is so low as to be essentially zero (not discernable from zero) over the duration of your experiment.

Page 3 makes the claim that _experimentally_ the resistivity of superconductors is better than 10^-25 ohm-meters (copper is 2*10^-8 ohm-meters), meaning that superconductors are at least 10^17 times better conductors than copper.

This is a brochure from a company that makes high field superconducting magnets for spectroscopic use. This document gives an equation which suggests that superconductors do have some resistance, see the equation in figure 8 (page 17). This equation says that the voltage drop across a length of superconducting wire is an exponential function of the current. For a constant resistance, V is proportional to I; in a superconductor V drops off as I^n where n is between 20 and 40. This suggests that the resistance gets very very very low but remains finite, except right at zero current (and what good is zero resistance if you only have it when no current is flowing...sounds like the guy who is invisible, but only when no-one is looking...).

On page 17 they claim to design their magnets for 0.01uV of voltage drop across the entire magnet. Going back to page 1 where they give the magnet parameters (200A current, 5.0MJ of stored energy) you can calculate the time constant for this inductor. 0.01uV/200A means a resistance of 5*10^-11 ohms, and 5.0MJ=1/2L*200^2 gives an inductance of 250H. This gives a time constant of about 160000 years.

Zero resistance, or very very low resistance? Dunno, you decide :)

-Jon
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
No. To the contrary, you have not made any points. You have simply repeated the statements and your only supporting argument has been that everyone is wrong except for David.
I beg your pardon? I HAVE made several points. However, unlike you, I have provided descriptions of the reasoning that I have applied to the points that I have been making. You have said several time that "electrons don't care where where the potential comes from" in arguing that the potential source and voltage drop are the same thing. "Electrons don't care where where the potential comes from" is not a reasoned argument. Tell us why you believe that. Provide some type of evidence for your assertions (I'm still waiting to hear where I am missing that minus sign.) Provide an example to backup your assertions. I have tried to do that in each of my post, but you seem to make statements and expect us to accept them as fact.

And as far as the argument "that everyone is wrong except for David." Please, PLEASE show me one example in all of these posts where I have made such an argument. You should provide evidence for this statement, just as you should provide evidence for your technical statements! In post #291, I said "Rick, brilliantly said!" and "The thought that occurred to me when pondering this is the exact same thought that you expressed above." How you can read that and then make that statement that I think "everyone is wrong except for David" is beyond my comprehension. Your statement, quite honestly, is petty and mean-spirited. You, sir, are a cur.

Which comes first (i.e. chicken or egg :cool:):
a) Does the voltage across a resistor cause the current through the resistor?

OR

b) Does the current through the resistor cause the voltage across the resistor.
You again, are not understanding my argument. I have argued that a potential source is required for current to flow in a circuit. It does not matter what that source is, but there must be a source. The potential source IS what CAUSES the current to flow in the circuit. Let me try another description of this and another example. I will enlist the help of the Navy (Basic Electricity: Prepared by the Bureau of Naval Personnel,) since I read something in it's definition of KVL which caught my eye. They stated KVL as follows: The algebraic sum of the instantaneous emf's and voltage drops around any closed circuit loop is zero. (In many cases, you see KVL expressed as the sum of the "voltages" around a close loop = zero.) I found their use of "emf" interesting. They define "emf" - electromotive force, as the "force that causes free electrons to move in a conductor as an electric current." This is the same reasoning that I have been arguing. It is the "potential source" or the "electromotive force" that causes the current to flow.

Let me try one more example to explain my point. Imagine a 12V potential source with terminals brought out and labeled as A & B. The "open circuit" potential difference from A-B would be 12V: Voc@ab=12V. The "potential" for current to flow between A and B presently exists in this network, we just need to complete the circuit. If we add a 12 ohm resistor between A-B, we complete the circuit and 1 amp will flow through the resistor (and there will be a 12 volt "drop" across the resistor.) By inserting the resistor and completing the circuit, have we ADDED or REMOVED any more "potential" into the circuit? No. The "potential" for current to flow between A and B was provided by the potential source. The "potential" for current flow existed in the network prior to the resistor being added. Now lets consider adding a second 12 ohm resistor in series with the first 12 ohm resistor between A and B. There will now be 0.5 amps flowing between A-B and there will be of 6 volt "drop" across each resistor. Have we ADDED or REMOVED any more "potential" into the circuit? No, we have not. The "potential" for current to flow that existed between A and B in the "open circuit" is the same "potential" that causes current to flow in the circuit with one resistor, and the same "potential" that causes the current to flow in the circuit with two resistors. And the "potential source" alone has provided that potential.

Whether you realize it or not, when you claim that voltage drop and voltage source are not the same, you have unwittingly, undermined Norton and Thevenin. Even though you would like to prove yourself to be correct, to do so would disprove Norton and Thevenin. Since these two are just theorems at this time, they are neither proven nor disproven, and as such, your argument can neither be proven or disproven--hence, you are arguing semantics. To the contrary, you have just demonstrated that you are arguing semantics. If it doesn't matter how/why a voltage difference exists, you have contradicted youself that a voltage source is different from a voltage drop.
Again, you make a statement, but you offer no evidence to back up your statement. WHY have I undermined Norton and Thevenin? Provide at least SOME explanation as to why you believe this to be true. Thevenin says "Given any linear circuit, rearrange it in the form of two networks A and B that are connected together by two resistanceless conductors. (If either network contains a dependent source, its control variable must be in the same network. Define a voltage Voc as the open-circuit voltage which would appear across the terminals of A if B were disconnected so that no current is drawn from A. Then all the currents and voltages in B will remain unchanged if A is killed (all independant voltage sources and independent current sources in A replaced by short circuits and open circuits, respectively) and an independent voltage source Voc is connected, with proper polarity, in series with the dead (inactive) A network." (per Hayt and Kemmerly - if you have a different statement of Thevenin's theorem, take it up with H&K, not me.)

So how has my claim that voltage drop and voltage source are not the same undermined Thevenin? It seems to me that in the example above with the 12V potential source and 12 ohm resistor, if I divide the circuit into two networks (network A will contain the potential source, and network B will contain the 12 ohm resistor) then the Voc per Thevenin will be 12 Volts. If I then kill network A (short out the 12V potential source) and insert a new voltage source Voc in series with network A, then the currents and voltages in network B HAVE remain unchanged. So how have I undermined Thevenin?

And if I have argued that a "potential source" provides the potential for current to flow in a circuit, and that "voltage drop" is the result of current flow in a circuit, and that it does not matter where the "potential difference" (created by the "potential source," and not to be confused with "voltage drop") comes from (it could come from a battery, generator, transformer secondary with the primary connected to an inverter sourced by storage batteries - it does not matter,) then I have NOT contradicted myself. I have been entirely consistent.

There is nothing wrong with thinking the way that you are, but when you state that other views are wrong, well, that is when you are wrong.
Well that's a heck of an intelligent statement. You've just got done telling me that I was "wrong" when I said that I hope I have made my point more clear. You told me I was "wrong" because I have unwittingly "undermined" Thevenin (even though I haven't.) You told me I was "wrong" because I have "contradicted" myself regarding the difference between "potential source" and "voltage drop" (although I have clearly NOT contradicted myself.) Why is it wrong to state that others views are wrong? That is known as "debate." If we cannot debate opposing views, then we may as well just give up trying to learn, fall in line and follow unquestioningly what Big Brother tells us.

If someone tells me that the sun rises in the west and sets in the east, then it is NOT WRONG of me to tell him/her that he/she is wrong, and to try to explain to that person why he/she is wrong. I CANNOT and WILL NOT subscribe to this point of view.
 

mivey

Senior Member
mivey said:
But my point was that Ohm's Law could not be used to say that there will be zero current on the closed superconductor loop.
crossman gary said:
errrr.... that was MY point!!:grin:
Let me jump in, because I think you are missing the bus here. You are trying to use Ohm's Law to demonstrate the value of current that will be flowing in a passive circuit...
?????:confused:?????
mivey said:
But my point was that Ohm's Law could not be used to say that there will be zero current on the closed superconductor loop.
 
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