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.