# MWBC= more heat or Less heat?

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#### david luchini

##### Moderator
Staff member
Originally Posted by david luchini 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.
No, it will show that it takes zero voltage for a current to flow through a zero resistance. Not that the current has to be zero.
Mivey, again you are missing the point - there seems be quite a bit of that going around. My post above was a response to crossman gary's post #167. He asked in post #167, if you take a zero impedance conductor and close it into a loop with no potential source included, "What does math and Ohm's Law predict about the current flow? E = I x R, 0 = I x 0, What is I?"

You can use Ohm's law to predict the voltage drop across two points in a circuit (V=IR) if you know the current and resistance across the two points, or to predict the current if you know the voltage drop and resistance (I=V/R) or to predict the resistance if you know the voltage drop and current across the two points (R=V/I.)

You responded that Ohm's Law "will show that it takes zero voltage for a current to flow through a zero resistance. Not that the current has to be zero." You are misapplying Ohm's law the same way crossman gary did.

Crossman gary described an "inactive" circuit. He created a loop of wire with no potential source added. In his circuit, there is no potential source, and therefore, no current flowing in the circuit. The must be an "active" source in the circuit to create current flow. My point to crossman gary was that Ohm's law CANNOT be applied to the circuit he described. The circuit resistance in the example is a known and constant value (0 Ohms,) but two predict the voltage drop between two points you must know what the current is between those two points, and to predict the current flow between two points you must know what the voltage drop between the two points.

Since the circuit in question is an "inactive" circuit, we already know that the voltage drop between any two points wil be zero, and that the current flow between any two points will be zero. I suppose that we could mathematically apply Ohm's law as 0=0x0, but what is the point. Since the purpose of Ohm's law is predict voltage drop, or current, or resistance between two points in a circuit, I would suggest that it should only be applied to "active" circuits.

#### crossman gary

##### Senior Member
Let me try to explain why the voltage measured across all points in your illustration will be zero.
You are saying that the voltage measured from N to A across the perfect, ideal, infinite 12 volt source is zero.

Does this mean that all 12 volts was dropped internal in the source?

If so, this is a contradiction of the very definition of an infinite, ideal, no internal resistance source. Such a source can supply infinite current and still maintain its voltage at the full level.

If the voltage is not dropped inside the source, then the voltage had to be dropped external of the source. (KVL was not an explanation and was not even a proof of what happened to the voltage. You arbitrarily inserted zero at N-A and N-G. We could just as easily inserted -12 at N-A and +12 at N-G and it still works)

David, I posed this example in an effort to show you that Ohm's Law has issues when dealing with extreme cases. I hope that you will see the contradictions involved. The explanation you gave basically says that ohm's law for a zero resistance conductor takes precedence over ohm's law for an ideal, infinite source. It can just as easily be assumed that ohm's law for the source takes precedence over the conductor.

An ideal infinite voltage source will maintain the potential difference across its terminals regardless of the applied load, from zero amps to infinite amps.

An ideal zero resistance conductor will maintain the exact same polarity and voltage across any two points on the conductor, from zero amps to infinite amps.

These two "facts" which are inherent in the definitions are contradictory when taken together in a circuit. And ohm's law cannot solve the contradiction.

#### crossman gary

##### Senior Member
Since the purpose of Ohm's law is predict voltage drop, or current, or resistance between two points in a circuit, I would suggest that it should only be applied to "active" circuits.
Now we are getting somewhere. As I have been saying since post 130, Ohm's law will fail under certain extreme circumstances.

Now, what would be your definition of an "active" circuit?

Based on your post, it seems that the definition of an active circuit would be that the circuit contains a non-zero potential difference. Correct?

So if we stick rigorously to the laws and concepts of mathamatics, you are saying that ohm's law will fail to predict the parameters of a circuit with a zero potential?

Even though it sure looks like you just discovered and admitted that ohm's law has limitations, and it feels sort of good to me, I am going to say that the logic is wrong..... Ohm's law does not fail under zero voltage conditions. Ohms law works just fine with zero voltage.

Ohm's law fails under zero resistance conditions. And that applies to any circuit or portion of a circuit with zero impedance.

#### crossman gary

##### Senior Member
If you have a disagreement with one of my posts, then provide a counter argument with examples to back up your argument.
David, in all honesty, we have been doing this very thing all along.

By doing this, this forum will be an educational opportunity for all of us.
Okay. Please allow me some rope to hang myself by using the logical device of reductio ad absurdum to disprove one of your statements. Please endulge me for a moment...

Thought experiment...

I have two externally identical "black boxes", Box A and Box B.

Each box has two ideal, zero impedance conductive terminals protruding from the top. There is a voltmeter mounted on top of each box, with the leads attached to the terminals in a manner to measure the potential difference between the two terminals. The meters are perfect, ideal, infinite impedance, and infinitely precise.

Both voltmeters read exactly 0 volts.

You have two perfect, ideal, zero impedance conductors, each with a built-in, perfect, ideal, zero impedance ammeter. You take one of these conductor/ammeter combinations and attach across the terminals of Box A. You take the other and attach across the terminals of Box B.

What predictions might you conclude about the ammeter readings in each of the conductors?

#### david luchini

##### Moderator
Staff member
You are saying that the voltage measured from N to A across the perfect, ideal, infinite 12 volt source is zero.

Does this mean that all 12 volts was dropped internal in the source?

If so, this is a contradiction of the very definition of an infinite, ideal, no internal resistance source. Such a source can supply infinite current and still maintain its voltage at the full level.

If the voltage is not dropped inside the source, then the voltage had to be dropped external of the source. (KVL was not an explanation and was not even a proof of what happened to the voltage. You arbitrarily inserted zero at N-A and N-G. We could just as easily inserted -12 at N-A and +12 at N-G and it still works)

David, I posed this example in an effort to show you that Ohm's Law has issues when dealing with extreme cases. I hope that you will see the contradictions involved. The explanation you gave basically says that ohm's law for a zero resistance conductor takes precedence over ohm's law for an ideal, infinite source. It can just as easily be assumed that ohm's law for the source takes precedence over the conductor.

An ideal infinite voltage source will maintain the potential difference across its terminals regardless of the applied load, from zero amps to infinite amps.

An ideal zero resistance conductor will maintain the exact same polarity and voltage across any two points on the conductor, from zero amps to infinite amps.

These two "facts" which are inherent in the definitions are contradictory when taken together in a circuit. And ohm's law cannot solve the contradiction.
Crossman gary, you seem to have ignored the part of my post detailing that the potential will drop within a circuit during a short circuit, even in real world situations. You seem to be expecting that if you put two potential sources in series, such as the two 12v batteries that you drew in your example, that during a short circuit that these two sources must still have 12V drops across them. This is not so. If that were so, then in the example that I gave of the 75kVA transformer, the voltage in the secondary should still be 208V during a short circuit. However, we can measure (in the real world) that the voltage drops during a short circuit.

This is the flaw in your reasoning. Your expectation that the voltage drop across the potential source will remain at 12V is wrong. I did not arbitrarily insert 0V across the potential sources, I arrived at 0V across the sources because that is what Ohm's law tells me they should be. It seems to me that ohms law is performing exactly as expected in this case. If Ohm's law tells us that the voltage drop across two points in a circuit is equal to the current flowing in the circuit between the same two points multiplied by the resistance in the circuit between the same two points (V=IR,) then since the resistance between ANY two points in your circuit is zero ohms, then the voltage drop across ANY two points in your circuit is also zero.

I do not see any contradiction in this.

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#### david luchini

##### Moderator
Staff member
Now we are getting somewhere. As I have been saying since post 130, Ohm's law will fail under certain extreme circumstances.
I know that you have been saying since post 130 that Ohm's law will fail under certain situations, but I still do not agree with that.

Now, what would be your definition of an "active" circuit?
I have defined an "active circuit" within these posts as a circuit with an "active" element such as a potential source. The other elements in the circuit, such as conductors, light bulbs, resistors, etc., are "passive." (And further, a "circuit" involves a closed loop, so an "open circuit" is not a "circuit" but an electrical "network.") Active circuit elements can deliver power into a circuit, while passive circuit element can absorb power in the circuit.

Based on your post, it seems that the definition of an active circuit would be that the circuit contains a non-zero potential difference. Correct?
No, that is not precisely correct. I say that because there seems to be a confusion on the terms being used (I am guilty of that myself.) The term "potential difference" is also being used to mean "voltage drop" (eg, "there is a potential difference across that resistor" and "there is a voltage drop across that resistor",) so I would not use that term in this definition. An active circuit will be on that has and "active circuit element" such as a potential source in it.

So if we stick rigorously to the laws and concepts of mathamatics, you are saying that ohm's law will fail to predict the parameters of a circuit with a zero potential?
NO, I am not saying that at all, and this is an important point. You described earlier a circuit (inactive circuit) made by forming a loop out of an ideal conductor. You wanted to know what Ohm's law would say about current flowing in that circuit, since obviously the resistance in the circuit would be zero ohms, and, since there is not potential source in the circuit, the would be no voltage drops across any two points in the circuit.

But if you think about Ohm's law logically, it is used to predict either voltage drop, current flow, or resistance across any two points in a circuit when two of these components are known values. That is to say, if I know the current and resistance across two points in the circuit, I can determine the voltage drop, or if I know the voltage drop and current across two points, then I can determine the resistance, etc. But you are trying to apply Ohm's law to an "inactive" circuit, with (to be redundant) no potential source in the circuit.

My question is "why"? You have described a circuit (your loop of zero resistance conductor) where we already know that the voltage drop across any two points in the circuit will be 0 Volts, the resistance between any two points in the circuit will be 0 Ohms, and the current between any two points in the circuit will be 0 Amps. If you know that V=0, R=0, and I=0 across any two points in the circuit, why do you need to try to apply Ohm's Law? That is why I say that Ohm's law should only be applied to active circuit. Have I explained this concept in a clear manner? I hope that I have.

Even though it sure looks like you just discovered and admitted that ohm's law has limitations, and it feels sort of good to me, I am going to say that the logic is wrong..... Ohm's law does not fail under zero voltage conditions. Ohms law works just fine with zero voltage.
I'm sure that you will agree with my explanations above that I have NOT discovered NOR admitted that Ohm's law has its limitations. Where you say that Ohm's law works "just fine with zero voltage," I must again caution against confusing "voltage drop" in a circuit with "potential source" in a circuit. In an "active" circuit, Ohm's law does work fine with a zero "voltage drop." That is, across any two points in a circuit, if we know that the resistance is zero (such as in an ideal conductor,) then we know that the voltage drop will be zero volts (if V=IR, then V will be zero - there will be zero voltage drop) regardless of the current flowing in the circuit (I*0=0.) If by zero voltage, you mean an inactive circuit with no potential source, then again, why do we need to apply Ohm's law.

Ohm's law fails under zero resistance conditions. And that applies to any circuit or portion of a circuit with zero impedance.
Wow, that is a stretch, and not true. I'm not quite sure what your reasoning is here. So, let me explain (although I unknowingly just explained this concept in the last paragraph.) In an active circuit, if you know that there is zero resistance across two points in that circuit, then you will have zero voltage drop for any level of current flow. I*R=V will equate to I*0=0 for all current levels. So Ohm's Law does NOT fail in zero resistance conditions. Let me provide an example of how Ohm's Law will correctly apply to a portion of a circuit with zero impedance.

Imagine a 12V battery. This will be my potential source in my "active circuit. I will label the positive terminal as A. From A, I will route a one foot length of zero resistance conductor (this, together with the return conductor will be the zero impedance portions of the circuit.) The node at the load end of this conductor I will label as B. At node B, I will also connect one end of a 12 Ohm resistor, and the other end of the resistor, I will label as terminal C. And from C, back to the battery, I will route another one foot length of zero resistance conductor, and label the negative terminal of the battery as D.

This circuit just described has an "active" element, the battery source, and three "inactive" elements, the two conductors and the resistor. The total load resistance (from A, through the resistor, to D) is 12 Ohms. There will be 1 Amp flowing in the circuit. So let apply Ohms Law between each of the nodes in the circuit in order to determine the voltage drops between each of the nodes. Between A and B, we will have 1 Amp of current flow, and we know that the resistance between A and B is 0 ohms. So applying Ohms law, we will see a voltage drop between A and B of V=IR=1 Amp * 0 ohms, which equals a zero volt drop between A and B. Between B and C, we have 1 Amp flow and 12 ohm resistance, so the voltage drop between B and C will be V=IR=1 Amp * 12 ohms = a 12 volt drop between B and C. And similar to A-B, there is a zero volt drop between C-D.

So you can see, Ohm's Law applies just fine to portions of circuits with zero impedance. I hope these discussions and the examples provided have been clear enough to understand.

#### crossman gary

##### Senior Member
You seem to be expecting that if you put two potential sources in series, such as the two 12v batteries that you drew in your example, that during a short circuit that these two sources must still have 12V drops across them. This is not so.
To the contrary. The very definition of an ideal source requires the voltage across the terminals to be maintained no matter whether zero current or infinite current flows. It maintains its potential regardless of load conditions, including that of a short circuit. The physics texts will point this out, for example: "An ideal battery with an emf of one volt would maintain a one volt potential difference between its terminals regardless of the current passing through it." > p. 519, Fundamentals of Physics, 2nd ed, Halliday and Resnick

If that were so, then in the example that I gave of the 75kVA transformer, the voltage in the secondary should still be 208V during a short circuit. However, we can measure (in the real world) that the voltage drops during a short circuit.
In the real world xfmr example, the secondary windings have impedance, i.e. the source has an internal impedance. If the load conditions on the xfmr result in 190 volts at the secondary terminals, then 18 volts is being dropped inside of the secondary windings.

190v + 18v = 208v

The 208 volts still exists, none of it magically dissappearred. It must be accounted for in KVL and Ohm's Law. You being an engineer, I am sure that you have seen this internal resistance of a source modeled in a circuit diagram as a resistor in series with the source. This internal resistance must be added as a series component to the rest of the circuit ohms, then ohms law is applied. Again, this is all in the physics texts, for example, p. 521 of the above mentioned text.

This is the flaw in your reasoning. Your expectation that the voltage drop across the potential source will remain at 12V is wrong.
The 12 volts in each battery must be accounted for in KVL and Ohm's Law. The voltage must be dropped somewhere. Either the 24 volts of the 2 batteries in series is dropped entirely in the sources, entirely in the external circuit, or in both (in the case of both, obviously the two voltage drops must equal the original source voltage).

I did not arbitrarily insert 0V across the potential sources, I arrived at 0V across the sources because that is what Ohm's law tells me they should be.
You are forgetting that the batteries are a voltage source. The 12 volts in each battery is due to the chemical action contained within. It is not due to the voltage drop across the source. Voltage drop across a source indicates how much of the source voltage is LOST. Your calculation of 0 x infinity = 0 volts dropped is exactly what it says. None of the emf of 12 volts is being dissipated or dropped or lost in the battery. All 12 volts is being applied to the external circuit.

David, the above is seriously correct. Please think long and hard, maybe even do some drawings of various real life and theoretical circuits and do the calcs.

For example:

Take a 208 volt source with 2 ohms internal impedance. Connect 50 ohms impedance across the source.

Total impedance of the circuit/source = 50 + 2 = 52 ohms.

Compute current by ohm's law: I = 208/52 = 4 amps

Use 4 amps to compute voltage drops:

External circuit voltage drop: 4 amps x 50 ohms = 200 volts dropped across external impedance

Internal voltage drop: 4 amps x 2 ohms = 8 volts

Notice how the voltage drop calculation for the internal resistance of the source results in how much voltage is LOST inside the source, and the remainder is applied to the external circuit.

Your calculation of the voltage drop across the 12 volt ideal source was zero. If you apply that information correctly, as we applied it in the xfmr example above, that means none of the 12 volts is lost inside the source. All of the 12 volts due to the chemical action in the battery is applied to the external circuit. And this is exactly as expected for an infinite ideal source.

I do not see any contradiction in this.
I seriously hope you do now. If you don't, please grab your physics texts and electrical texts and rethink your position. Search the net for "internal resistance" or "internal impedance" and see what is happening. Consult with your mentors and see what their view is. Please do not mistake this as rudeness. Please consider this as an opportunity to see the circuit in its reality (theoreticalness?).

Thanks for listening, David. I am thoroughly enjoying the conversation and your side of the story. :smile:

Edit: David, I grabbed your original message as a quote, I think before you edited it. I do not know if you deleted any of the quoted material above.

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#### david luchini

##### Moderator
Staff member
David, in all honesty, we have been doing this very thing all along.
Crossman gary, this was not directed at you. I have enjoyed your example circuits, and the curiosity with which you approach the subjects at hand. I think I share that same curiosity. One should never be done with trying to learn new things.

But there has been a poster who has insisted that I have been contradicting myself, or that I have mistated Ohm's law, but offered no evidence or examples to back up those claims. Or for another example, I have said that I believe that a voltage drop does not HAVE to exist between two points in a circuit for current to flow between those two same points in that circuit (like the ideal conductors connected between the battery and resistor.) Many posters have disagreed with my statement, but I have not seen any of them try to give an explanation as to why that statement might be wrong, or provide an example to prove mathematically that the statement is wrong. Like I said, I believe that this forum should be an educational tool, but if I make the statement that, oh, I don't know, PI=3.141592..., but I don't provide an explanation of WHY PI=3.141592..., or an example demonstrating that PI=3.141592..., then I have not created a condition for learning, but have just made a statement and expect everyone to accept it as fact. (And no, I will not provide an example demonstrating the value of PI.)

Okay. Please allow me some rope to hang myself by using the logical device of reductio ad absurdum to disprove one of your statements. Please endulge me for a moment...

Thought experiment...

I have two externally identical "black boxes", Box A and Box B.

Each box has two ideal, zero impedance conductive terminals protruding from the top. There is a voltmeter mounted on top of each box, with the leads attached to the terminals in a manner to measure the potential difference between the two terminals. The meters are perfect, ideal, infinite impedance, and infinitely precise.

Both voltmeters read exactly 0 volts.

You have two perfect, ideal, zero impedance conductors, each with a built-in, perfect, ideal, zero impedance ammeter. You take one of these conductor/ammeter combinations and attach across the terminals of Box A. You take the other and attach across the terminals of Box B.

What predictions might you conclude about the ammeter readings in each of the conductors?
I'm not sure which of my statements you are trying to disprove, but I don't think that there is enough information in your thought experiment to provide an accurate answer (though I may not be reading the problem correctly) but I will give it a shot.

You are giving me two black boxes, so I don't know what is happening inside, but you have told me that the voltmeters across the terminals of each box read zero volts. I will take that to mean that the black boxes are "inactive" electrical networks, that there are not potential sources within the black boxes (or there may be sources within the boxes, but with "open circuits" so that there is not potential measuring on the terminals.)

If I connect these two boxes together via the conductor/ammeters, I would expect that no amps flow through the conductor/ammeters, because there is not potential source in the complete circuit. So the ammeters would read zero.

Alternately, if there were potential sources within the boxes, but within the black boxes, there is an "open circuit" such that the potential on the black box terminals is measured as zero volts, then connecting the two boxes does not create a completed circuit, so no current will flow, and the ammeters would read zero.

Or more alternately, I am completely missing some detail in your thought experiment, and I have it completely wrong.

#### mivey

##### Senior Member
I think this forum is an important learning tool, and some people in this discussion seem to have the incorrect idea that the "voltage drop" along a conductor is the "force" that causes the current to flow through that conductor. I have attempted to show through various discussions, including providing examples, why this belief is untrue...then provide a counter argument with examples to back up your argument. By doing this, this forum will be an educational opportunity for all of us.
You have requested examples to help clarify points but there already have been many posted. You seem to be fixated on identifying exactly where the voltage source is located in the circuit. I can't understand why you can't get beyond that. As far as the conductor is concerned, the voltage source is at its terminals. The conductor is just another circuit element, but with a relatively small resistance.

The potential force as measured at the conductor's terminals is what is moving the current through the wire. THAT is fundamental electical theory. You seem to have become confused because you started substituting a wire with zero resistance.

You don't seem to like the fact that the Radio Shack power supply is attached at a terminal somewhere else in the circuit. Fine, take a conductor of appreciable length and connect it directly to the power supply. Now tell me the voltage applied to the conductor terminals is not moving the current through the wire.

But wait, there's more (RIP Billy). The real surprise is that the terminals of the power supply are not really the voltage source you are looking for either. What about the output resistor of the power supply? Now you are going to have to move inside the Radio Shack box to your potential source. But you are not going to find it there either.

If you did not like the conductor across the power supply, just use a resistor. Hook a resistor across a power supply. The potential at the resistor terminals is what makes the current flow through the resistor. Now hook three resistors in series. The potential at their terminals is what makes the current flow through them. The conductor is just a very small valued resistor. In fact, some conductors are used as resistors.

#### mivey

##### Senior Member
Mivey, again you are missing the point...In his circuit, there is no potential source, and therefore, no current flowing in the circuit...
And once again, you have ignored a point. I don't recall CG having stated there was no current. 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. You can only say that current will flow unimpeded. If there is a current, it will continue.

There does not have to be an active potential source in the superconducting loop for current to be flowing.
...Even though it sure looks like you just discovered and admitted that ohm's law has limitations, and it feels sort of good to me, I am going to say that the logic is wrong..... Ohm's law does not fail under zero voltage conditions. Ohms law works just fine with zero voltage.

Ohm's law fails under zero resistance conditions. And that applies to any circuit or portion of a circuit with zero impedance.
I'm not sure that anyone has been saying Ohm's Law never fails. It doesn't apply to everything. It is just a model. But it has been expanded beyond its original use with linear resistors.

We often expand our models beyond their original use. We can use them, extrapolate, modify, and even replace them. If they fail, so what? We just make a new model.

It is not such a big deal that we need to determine whether some deity supports it or not. If we can define parameters such that the model works: great. If it doesn't: move on.

#### crossman gary

##### Senior Member
I know that you have been saying since post 130 that Ohm's law will fail under certain situations, but I still do not agree with that.
Your original implication in post 130 was that Ohm's Law is universally true. You have since narrowed that down to "Ohm's Law applies only in "active" circuits as defined by you. Isn't this an admission that ohm's law is not universally applicable as I have stated?

I have not given up on showing you that ohm's law does not apply universally. In some situations, ohm's law and strict adherence to mathamatical principles leads to absurd results. principally, the absurd results come about when we deal with zero resistances.

Active circuit elements can deliver power into a circuit, while passive circuit element can absorb power in the circuit.
So a zero resistance conductor is neither a passive nor active circuit element?

Concerning the term "potential difference". To me, that term does not necessarily mean voltage drop. It means that a potential exists from one point to another that will cause electrons to move from one point to the other.

Your discussion with other members of voltage drop versus potential source is interesting. However, I have not been involved in that. So suffice it to say, if I say potential difference, I am meaning that, as I said, a potential exists between two points which, under proper conditions, can cause electrons to move from one point to the other.

Now, the fact remains that if you have a voltage drop, then a resistor can be connected across that voltage drop, and current will flow through it, certainly implies that voltage drop has the potential to move electrons. I understand that you may disagree or not see it exactly like that. And, that is not my argument. My argument is that zero resistance conductors are only useful for simplifying real world calculations, and when a person starts applying ohm's law to the zero resistance conductors to prove something, then the argument becomes arbitrary. For example, a statement similar to "2 amps is flowing in the conductor, even though the potential difference from one end of the conductor to the other is zero" is simply not in line with reality. It is confusing a mathamatical tool for the real phenomenon.

But if you think about Ohm's law logically, it is used to predict either voltage drop, current flow, or resistance across any two points in a circuit when two of these components are known values.

You have described a circuit (your loop of zero resistance conductor) where we already know that the voltage drop across any two points in the circuit will be 0 Volts, the resistance between any two points in the circuit will be 0 Ohms, and the current between any two points in the circuit will be 0 Amps. If you know that V=0, R=0, and I=0 across any two points in the circuit, why do you need to try to apply Ohm's Law? That is why I say that Ohm's law should only be applied to active circuit. Have I explained this concept in a clear manner? I hope that I have.
All we know about the superconducting loop is that R = 0 ohms and E = 0 volts. You are assuming that the current is zero. We need to be mathamatically rigorous, we cannot depend on emotion here. This results in E = I x R ---> 0 = I x 0 and the solution set is the entire set of real numbers and I think the mathamaticians say that the solution is arbitrary, undefined. We cannot be certain exactly what the current is. Perhaps there is several hundred amps circulating in the loop. (and this is mentioned in the referenced physics text on page 508. Ohm's law fails for super conductors because it cannot predict the precise amount of current flowing in them.

If ohm's law does not work for the above, regardless of the personal feelings of "why would you want to", then ohm's law does not apply universally.

I'm sure that you will agree with my explanations above that I have NOT discovered NOR admitted that Ohm's law has its limitations.
What about the "passive" circuit limitation? Passive circuits have a voltage source which is equal to zero. So we can't use a source with zero voltage? Is this a limitation of ohm's law? Certainly zero voltage exists? Is ohm's law so limited that it doesn't work in a circuit with zero voltage? No? So is ohm's law limited or not?

If by zero voltage, you mean an inactive circuit with no potential source, then again, why do we need to apply Ohm's law.
Why shouldn't we? If ohm's law has no limitations, then we can apply it to any closed loop and it will accurately predict the answers, from 0 volts to infinite volts. If we can't, then it is not universally applicable.

#### crossman gary

##### Senior Member
Crossman gary, this was not directed at you. I have enjoyed your example circuits, and the curiosity with which you approach the subjects at hand. I think I share that same curiosity. One should never be done with trying to learn new things.[/QOUTE]

Thank you for those thoughts.:smile:

Or more alternately, I am completely missing some detail in your thought experiment, and I have it completely wrong.
Sorry I wasn't desriptive enough. What I am meaning is that we are dealing with the boxes individually, not connecting wires between them. So, Box A has two terminals. We take a zero-resistance conductor/ammeter combo and connect between the two terminals on that individual box. Then we do the same for Box B. There is no interconnection between the boxes. They are seperate experiments.

And, while we are at it, let's add a third box C and do the same thing with it.

I am trying to show that zero resistance leads to absurd results.

Again, David, thank you for being a good sport in this.

Epiphany: I was reading some about the superconductors. Now, there wasn't a mention that the resistance was exactly equal to zero, but it was extremely extremely close. If a 0 resistance superconductor were to exist, and we used it as the connecting elements from a battery to a resistor as in your previous examples, then you know what, I am going to agree with you that current can flow in those superconductors without a voltage drop on them or a potential difference on them. So, based on the premise of the zero ohm superconductor, I retract my objections to that.

But, I am sticking with my "zero resistance leads to the failure of ohm's law" statement.

So.... 3 black boxes, each one is its own experiment. We are shorting the individual boxes' terminals with the zero resistance conductop/ammeter combos.

#### crossman gary

##### Senior Member
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.
errrr.... that was MY point!!:grin:

I'm not sure that anyone has been saying Ohm's Law never fails. It doesn't apply to everything.
It sure seemed like some people were.

If we can define parameters such that the model works: great. If it doesn't: move on.
Exactly. But we need to discover the parameters before we move on. I am looking for the parameters.

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#### mivey

##### Senior Member
errrr.... that was MY point!!:grin:
Well, if you agree with me, you must be right. :grin:

#### crossman gary

##### Senior Member
The potential force as measured at the conductor's terminals is what is moving the current through the wire. THAT is fundamental electical theory. You seem to have become confused because you started substituting a wire with zero resistance.

If you did not like the conductor across the power supply, just use a resistor. Hook a resistor across a power supply. The potential at the resistor terminals is what makes the current flow through the resistor. Now hook three resistors in series. The potential at their terminals is what makes the current flow through them. .
Mivey: All this makes perfect sense to me.

#### crossman gary

##### Senior Member
Mivey, I read in a physics text of a vague report of a circular superconductor. Once the electrons were set into motion, they continued with no external force. Two years later, they were still moving around the circle.

My question is, how did they measure the current flow to know the electrons were still moving? Any real world device which can measure current flow is going to absorb energy from the circuit and introduce impedance into that circuit, probably by inductance. And once you introduce the impedance, bang, the electrons stop. Hall effect sensors? Wouldn't those still interact with the magnetic field of the moving electrons? And this would absorb energy from the electron motion?

I'm not so sure that the two year thing was an actual occurance, but was an extrapolated result. Hmmm.... guess I should search the net.

#### david luchini

##### Moderator
Staff member
Your original implication in post 130 was that Ohm's Law is universally true. You have since narrowed that down to "Ohm's Law applies only in "active" circuits as defined by you. Isn't this an admission that ohm's law is not universally applicable as I have stated?
Crossman gary, what an excellent post. I like the way you have you have provided counter points to my earlier post, and explanations to go with your counter points. I have reread post 130, and I do not see where I have implied that Ohm's law is universally true. I believe the point that I was trying to make in post 130 was that voltage drop across a conductor is not required to have current flow in that conductor. Therefore, voltage drop is not creating the current flow.

Since I'm not sure that I have implied Ohm's law is universal, I'm not sure that I have "admitted" anything by suggested it should only apply to "active" circuits. If you want to apply it to "passive" circuits, go right ahead (and I'll get to that later.) But I don't think you have shown be an example yet where it does not work.

I have not given up on showing you that ohm's law does not apply universally. In some situations, ohm's law and strict adherence to mathamatical principles leads to absurd results. principally, the absurd results come about when we deal with zero resistances.
Please do not give up, because this has been thouroughly educational. Yes, we may be getting absurd results in some of the more absurd examples, but it seems to me that the results have thus far been consistent with Ohm's Law.

So a zero resistance conductor is neither a passive nor active circuit element?
No, I said a passive element CAN absorb power from a circuit, not that it HAS to. And conversely, an active element CAN provide power into a circuit, but not that it has to. But a passive element CANNOT provide power into a circuit (yes, I know that certain reactive "passive" elements can store power from a circuit and then deliver it back to the circuit, but it cannot provide power into a circuit without first absorbing power, so it is passive.) And I guess I would say that "active" elements can also absorb power from another active element in a circuit, (such as a car battery providing power to the headlights, etal, but absorbing energy from the alternator to recharge the battery.)

Concerning the term "potential difference". To me, that term does not necessarily mean voltage drop. It means that a potential exists from one point to another that will cause electrons to move from one point to the other.

Your discussion with other members of voltage drop versus potential source is interesting. However, I have not been involved in that. So suffice it to say, if I say potential difference, I am meaning that, as I said, a potential exists between two points which, under proper conditions, can cause electrons to move from one point to the other.

Now, the fact remains that if you have a voltage drop, then a resistor can be connected across that voltage drop, and current will flow through it, certainly implies that voltage drop has the potential to move electrons. I understand that you may disagree or not see it exactly like that. And, that is not my argument. My argument is that zero resistance conductors are only useful for simplifying real world calculations, and when a person starts applying ohm's law to the zero resistance conductors to prove something, then the argument becomes arbitrary. For example, a statement similar to "2 amps is flowing in the conductor, even though the potential difference from one end of the conductor to the other is zero" is simply not in line with reality. It is confusing a mathamatical tool for the real phenomenon.
"Now, the fact remains that if you have a voltage drop, then a resistor can be connected across that voltage drop, and current will flow through it..." Didn't you just fall into a trap there? You had just said "potential difference" and "voltage drop" do not mean the same thing in your mind, and that "potential difference" means that potential exists between two points which, under proper conditions, can cause electrons to move from one point to the other. Yet in the very next paragraph you describe a "voltage drop" between two points which, if you insert a resistor, current will flow. Under your definitions, you mean "potential differnece" in that statement, not "voltage drop." When you insert the resistor between the two points, the "potential difference" between the two points causes current to flow through the resistor, creating a "voltage drop" across the resistor.

All we know about the superconducting loop is that R = 0 ohms and E = 0 volts. You are assuming that the current is zero. We need to be mathamatically rigorous, we cannot depend on emotion here. This results in E = I x R ---> 0 = I x 0 and the solution set is the entire set of real numbers and I think the mathamaticians say that the solution is arbitrary, undefined. We cannot be certain exactly what the current is. Perhaps there is several hundred amps circulating in the loop. (and this is mentioned in the referenced physics text on page 508. Ohm's law fails for super conductors because it cannot predict the precise amount of current flowing in them.
Well, I am not a mathematician, I am an engineer. I have already said that a potential source is REQUIRED in a circuit in order for current to flow. Therefore, in the circuit that you have created, the loop of zero resistance super conductor, we already know that the resistance across any two points in the loop will be zero ohm's, the voltage drop across any two points will be zero volts, and the current flowing between any two points in the circuit will be zero amps. So in you circuit, the result of Ohm's Law is not V=I*R---> 0=I*0. Instead, the result will be V=I*R---> 0=0*0.

If ohm's law does not work for the above, regardless of the personal feelings of "why would you want to", then ohm's law does not apply universally.
My point about "why would you want to" apply Ohm's Law to the circuit you described, is because we already know that V=0, I=0 and R=0.

What about the "passive" circuit limitation? Passive circuits have a voltage source which is equal to zero. So we can't use a source with zero voltage? Is this a limitation of ohm's law? Certainly zero voltage exists? Is ohm's law so limited that it doesn't work in a circuit with zero voltage? No? So is ohm's law limited or not?
Why would Ohm's Law be limited in the "passive" circuit. I have said that I didn't see much sense in applying it to the "passive" circuit that you described, but if you want to apply it go ahead. Ohm's law in the circuit you described say that 0=0*0. (Again, you are trying to say that I have argued that Ohm's Law is universal. I don't think I have, but we haven't found a case in any of our examples where it does not apply.)

Why shouldn't we? If ohm's law has no limitations, then we can apply it to any closed loop and it will accurately predict the answers, from 0 volts to infinite volts. If we can't, then it is not universally applicable.
"Why shouldn't we?" Because there isn't much point if we know V=0, I=0 and R=0. We CAN apply it to the passive circuit you described: V=I*R--->0=0*0, but we have we gained by applying Ohm's Law in this case?

#### david luchini

##### Moderator
Staff member
Originally Posted by mivey 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.
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. My point is that that is an exercise in futility. There is no reason to try to use Ohm's Law in that way since the very definition of a "passive circuit" tells us that the current flow will be zero. Why use the Ohm's law equation to determine the value of the current flow when we already know that the current flow is zero?

Originally Posted by mivey I'm not sure that anyone has been saying Ohm's Law never fails. It doesn't apply to everything.
It sure seemed like some people were.
Again, I don't think I have argued that Ohm's Law applies to everything, but I don't think it has failed in any of the examples we have been discussing.

#### winnie

##### Senior Member
Hi, just popping my head up when I really should not be spending time on the intarwebs 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.

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. In order to understand these devices, you need to think AC, not DC.

The superconducting magnets are wound as solenoids with a superconducting shorting switch at the terminals. The solenoid is topologically equivalent to a single loop of wire, and in theory a single loop could be used.

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.

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.

-Jon

#### crossman gary

##### Senior Member
"Now, the fact remains that if you have a voltage drop, then a resistor can be connected across that voltage drop, and current will flow through it..." Didn't you just fall into a trap there? You had just said "potential difference" and "voltage drop" do not mean the same thing in your mind, and that "potential difference" means that potential exists between two points which, under proper conditions, can cause electrons to move from one point to the other. Yet in the very next paragraph you describe a "voltage drop" between two points which, if you insert a resistor, current will flow. Under your definitions, you mean "potential differnece" in that statement, not "voltage drop." When you insert the resistor between the two points, the "potential difference" between the two points causes current to flow through the resistor, creating a "voltage drop" across the resistor.
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.

Well, I am not a mathematician, I am an engineer. I have already said that a potential source is REQUIRED in a circuit in order for current to flow. Therefore, in the circuit that you have created, the loop of zero resistance super conductor, we already know that the resistance across any two points in the loop will be zero ohm's, the voltage drop across any two points will be zero volts, and the current flowing between any two points in the circuit will be zero amps. So in you circuit, the result of Ohm's Law is not V=I*R---> 0=I*0. Instead, the result will be V=I*R---> 0=0*0.
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.

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.

My point about "why would you want to" apply Ohm's Law to the circuit you described, is because we already know that V=0, I=0 and R=0.
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 in the circuit you described say that 0=0*0. (Again, you are trying to say that I have argued that Ohm's Law is universal. I don't think I have, but we haven't found a case in any of our examples where it does not apply.)
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.

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.

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