MWBC= more heat or Less heat?

[mivey]

Your first statement said that if you increase the impedance in a circuit (or replace a superconductor with a real conductor) that you will need to increase the voltage to get current to flow through the circuit.

That is not what I said. I said to get "the current from the rest of the circuit" to flow through the conductor you would need to add voltage. I also gave a supporting statement for clarification. You have, and should, agree with my clarification. Why ignore the clarification and completely redefine what I said? Stop extracting part of my post and adding your own clarification to that extracted piece and you will hear what I am saying.

I have heard what you are saying and you agree that for the same current to flow to the same load, but across an added resistance, you will need to step up the voltage. You have already agreed with my point. No need to keep doing so.

 

[LarryFine]

You are missing the point of that example. It was put forth by someone that there has to be a potential difference from source-n to neutral-n in order for current to flow in the neutral. In that example, there is clearly NOT a potential difference from source-n to neutral-n, but there IS current flowing in the neutral.

Of course, with an intact neutral, there will be no potential difference between source-N and load-N (in theory; we're not discussing conductor voltage drop here.)

What I've been talking about is, if there's no potential difference between source-N and load-N without an intact neutral, adding one has no effect on current flow.

To me, the 2-line node and the 3-line node w/no neutral current function the same. The neutral carries current to keep load-N at the same potential as source-N.

My original point is that the neutral has not been bypassed, it is still an active part of the circuit. Nor is there no current flowing in the neutral because there is "no potential difference along the neutral in a balanced circuit" as has been suggested. It just happens that there is zero current flowing on it. That is not the same as an open circuit at the neutral.

No argument.

I guess the only place we disagree is in your post #119 where you say that if there is zero voltage difference between two points, a conductor between them will carry no current.

Already addressed above.

 

[mivey]

...because "infinity" does not exist in the real world...

On the contrary: What is at the end of the universe? What is at the beginning and ending of time?

 

[david luchini]

OK, I'm lumping a voltage across something and a potential across something as the same thing. Maybe there is a point you are making that I'm missing.

Yes! The potential source in circuit and the voltage drops along the circuit are different things. The point I have been trying to make, because some in this forum have argued that it is the voltage drop in the circuit (ie, the potential difference from N-source to N-load along the neutral conductor) that causes current to flow in the neutral, is that it is the potential source that causes current to flow in the circuit. Whether you have a voltage drop along the neutral conductor, or whether you have no voltage drop along the neutral (in the ideal circuit) you WILL have current flow in the neutral when you create a closed loop circuit with a load impedance.

That is to say, the voltage drop along the conductor is NOT causing the current to move along the conductor (the potential source is causing the current to move in the conductor,) but instead the voltage drops along the circuit conductors and circuit load are CREATED by the current flowing through them. I think it is a fairly simple point to understand, but I seem to be having a hard time of getting my point across.

Are you saying there is no current flow if you have a series capacitor or are you saying there is no potential across a capacitor unless current is flowing?

Aw, man, you're killing me.:smile: I haven't thought about capacitive circuits in years. I'm not saying either of those two statements, what I am saying is that the voltage drop along a conductor is not REQUIRED for current to flow along the conductor, nor does the voltage drop along a conductor CAUSE a current to flow along the conductor.

 

[david luchini]

Of course, with an intact neutral, there will be no potential difference between source-N and load-N (in theory; we're not discussing conductor voltage drop here.)

What I've been talking about is, if there's no potential difference between source-N and load-N without an intact neutral, adding one has no effect on current flow.

To me, the 2-line node and the 3-line node w/no neutral current function the same. The neutral carries current to keep load-N at the same potential as source-N.

No argument.

Already addressed above.

Larry, Amen, brother.

 

[mivey]

Crossman gary, I'm afraid you are wrong on this matter. You are looking at the Ohm's Law equation of V=I*R and saying - aha, if V = 0 and R = 0, then I = anything I want it to.

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

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.

 

[david luchini]

On the contrary: What is at the end of the universe? What is at the beginning and ending of time?

Ah yes, but can you tell me where to find the end of the universe? Or when did time begin and when will it end? These, too, are theoretical concepts. If you apply infinity to any of these concepts, you are still in the theoretical world.

 

[david luchini]

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, I must disagree. Ohm's law: "The current through a conductor between two points is directly proportional to the potential difference or voltage across the two points."

If we have not applied a potential source to the circuit, then the potential difference across any two points on a conductor in the circuit will be zero amps.

If the current through the conductor between two points is "directly proportional" to the potential difference between the same two points, then the current must be zero.

 

[crossman gary]

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

Exactly. And that is why the mathamatics in this instance fails to describe reality. Even with calculus, we never let delta x actually equal zero. It may be very very close to zero, but it is never actually zero.

 

[mivey]

Yes! The potential source in circuit and the voltage drops along the circuit are different things.

If you are saying the voltage source is different than the measured potential, then I agree. One is a physical object, there other is a measure of something produced by a physical object.

I'm saying the quantity of voltage required for the x current to flow through a conductor in a circuit is above and beyond what is needed for x current to flow through a superconductor in the same place in the circuit.

The voltage required is measured at the conductor terminals. I don't think anyone is saying the conductor terminals have turned into a voltage source. The original source, or an additonal source can supply the extra voltage. But without it, something must give.

 

[crossman gary]

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

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

David, still waiting on an answer. The above leads to logical and mathamatical absurdities with your theoretical view.

 

[crossman gary]

Ohm's law: "The current through a conductor between two points is directly proportional to the potential difference or voltage across the two points."

So, in other words, for current to flow through a conductor, there must be a potential difference from one end of the conductor as compared to the other end?

 

[david luchini]

That is not what I said. I said to get "the current from the rest of the circuit" to flow through the conductor you would need to add voltage. I also gave a supporting statement for clarification. You have, and should, agree with my clarification. Why ignore the clarification and completely redefine what I said? Stop extracting part of my post and adding your own clarification to that extracted piece and you will hear what I am saying.

I have heard what you are saying and you agree that for the same current to flow to the same load, but across an added resistance, you will need to step up the voltage. You have already agreed with my point. No need to keep doing so.

I have not ignored your clarification, nor have I redefined what you said. If you want to maintain the same level of current in the circuit then you must increase the voltage in the circuit. We both agree on this.

But, if a take a circuit with a 12v battery, a 12 ohm resistor and two 0.5 ohm impedance conductors, with 0.923 amps flowing through the circuit, and I decide to take out one of the 0.5 ohm conductors and replace it with a 1 ohm conductor, leaving the rest of the circuit the same, then I will get 0.888 Amps flowing through the new conductor from "the rest of the circuit" without increasing the voltage in the circuit.

 

[david luchini]

David, still waiting on an answer. The above leads to logical and mathamatical absurdities with your theoretical view.

Crossman gary, didn't I respond to this in post 178? Or is this a different situation. These theoretical short circuits (with even no impedance in the potential source) would theoretically provide infinite current in the circuit.

 

[mivey]

...the voltage drop along a conductor is not REQUIRED for current to flow along the conductor, nor does the voltage drop along a conductor CAUSE a current to flow along the conductor.

How about: The force that must be applied for current to flow through a conductor is measured by the voltage drop along the conductor.

 

[crossman gary]

Crossman gary, didn't I respond to this in post 178? Or is this a different situation. These theoretical short circuits (with even no impedance in the potential source) would theoretically provide infinite current in the circuit.

Perhaps I missed that. Thanks for the "re-response." Let me make some drawings to post and I will present my case of the absurdities.

Now, post 192 says alot.

 

[david luchini]

So, in other words, for current to flow through a conductor, there must be a potential difference from one end of the conductor as compared to the other end?

Post 192 does say alot. Here is a reply.

No, there must be a potential difference (provided by a potential source) across the entire completed circuit. The conductor will only be one component of the system. Say two conductors connecting a resistor to a battery. The potential (from the source) is provided along one conductor, across the resistor and along the second conductor. The potential is provided across the ENTIRE circuit.

When current is flowing in the circuit, there will be a measured potential drop from one end of a conductor to the other. This is voltage drop caused by current flowing through the impedance of the conductor. This potential difference, or voltage drop, is not a requirement for current to flow along the conductor.

Don't confuse the "potential source" in the circuit with the "voltage drop" along the circuit. According to KVL, the sum of the voltages in the circuit loop must be zero. If the potential source is a 12V battery, and the load is a 12 ohm resistor, and the connecting conductors have 0.5 ohm impedance each, then 0.923 Amps will flow in the circuit (by ohm's law I=V/R = 12 Volt/ 13 ohm.) The voltage drop along each of the load impedances (the conductors and resistor) would be 0.462 Volt drop for each of the conductors, and 11.076 Volt drop across the resistor (again by ohms law.) By KVL, the voltage drops along each of the load components must equal the voltage of the potential source, which we can see,
0.462V+11.076V+0.462V = 12V, is true.

 

[crossman gary]

What was quoted in 192 and what I wrote in 192 are exactly the same thing.

 

[crossman gary]

Regarding David's answers to my questions:

Example 1: 12 volt source with 0 ohm circuit, current = infinity

Example 2: 480 volt source with 0 ohm circuit, current = infinity

Looks like ohm's law has gone out the window. Raising the voltage saw no corresponding increase in current.

Ohm's law at the extreme edge of theory fails.

 

[mivey]

Mivey, I must disagree. Ohm's law: "The current through a conductor between two points is directly proportional to the potential difference or voltage across the two points."

In the classic equation, Ohm's Law holds true for ohmic media.

Another way to state it is to say the added force required (voltage) to overcome the resistance to the flow of moving charges (resistance) is proportional to the product of the current and resistance. If the resistance is zero, there is no added force required and the charge will flow unimpeded. If R = 0 then delta V = zero as well. This can happen in a superconductor where the voltage drop is zero.

Also, I have used the term ideal conductor for a zero ohm conductor but and ideal conductor is really one who's resistance doesn't change. I guess sticking with superconductor or maybe perfect conductor is better.

I have seen claims of superconductor loops that have had a current running through them for years with no appreciable loss. Zero voltage and zero resistance means the system remained unchanged.