# MWBC= more heat or Less heat?

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#### LarryFine

##### Master Electrician Electric Contractor Richmond VA
Gary, I mean no disrespect, but this is way off topic from this thread.
You're right. The answer is: less.

Fewer conductors overall, and fewer CCC's.

#### crossman gary

##### Senior Member
Rick,

Thank you for being cordial. I was just adressing post 130:

Rick said: "there must be a voltage drop across the wire for current to flow through the wire"

David said: "Rick, you truly surprise me. If this statement was true then Ohm's law would not work in an ideal case. Consider a circuit with a 12V battery with terminals + and -. From + we run an ideal conductor (with no impedance) to a 12 ohm resistor (terminal R1.) On the other end of the resistor (terminal R2) we route an ideal conductor to the battery terminal -. According to Ohm's law I=V/R, there will be a 1 Amp current flowing through the circuit. However, if we isolate the conductor from terminals + to R1, it is plain to see that there is no voltage drop from + to R1, yet there is clearly 1 Amp flowing along this conductor. From your statement above, this could not be happening, so Ohm's law must be wrong. Let me correct your statement - it should say "there must be a current flow through a wire for there to be a voltage drop across it."

That is what I am addressing, and it seems pertinent to what the discussion has evolved into. If this line of thought needs to be moved, perhaps a Moderator would be kind enough to split the thread.

I firmly believe the statement you made. David said if it were true, then ohm's law must fail under extreme conditions. I am attempting to show that ohm's law does indeed fail under extreme conditions.

#### crossman gary

##### Senior Member
Man, we seem to be arguing all over the place.
And the intent of the above was directed to David who in some posts has said that current can travel through a conductor even though no potential difference exists across it, and in other posts he has said that current cannot flow in a conductor unless a potential exists across it.

#### hurk27

##### Senior Member
I am so confuse with why is it necessary to have the resistance of the wire in a hypothetical circuit to show the out come of an equation? in ohm's law you would still add the wire resistance together with the loads resistance to form one resistance figure to use in the equation as in any series circuit, so what is the difference if the wire has 1 ohm and the load is 50 ohms or you just use 51 ohms in your equation as you would do anyway?:-?

what would be the difference of saying your wire is 0 ohms, and your load is 51 ohm's or the wire is 1 ohm and the load is 50 ohms, it makes no difference of the out come?

and arguing of the current of a circuit of a source of infinity with a resistance of 0 would equal a current of infinity. that is the only answer obtainable no argument it is a fact:-?

#### Rick Christopherson

##### Senior Member
And the intent of the above was directed to David who in some posts has said that current can travel through a conductor even though no potential difference exists across it, and in other posts he has said that current cannot flow in a conductor unless a potential exists across it.
I agree. David is all over the place and is contradicting himself in ways that he is not even realizing, which is the basis for several of my postings. The divergence into super-conductivity was in fact the result of one of his contradictions.

By the way, I don't believe that Ohm's Law fails under super-conductivity, it just becomes "undefined", as does many mathematical equations when they encounter either zero or infiinity as one of their parameters.

The infinite flow of current flowing through a super-conductive loop is not the result of an electric field, but due to inertia in the absence of resistance to stop them. Once a body is in motion, it will stay in motion unless acted apon, and even electrons with their small mass still need to follow that law. An external force got the electrons moving, such as a magnetic field, and once that force was removed, there was no opposing force present to make them stop.

#### mivey

##### Senior Member
And finally, if we take our circuit in its current state, 75Volts and 0 ohms impedance, with infinite current flowing, and turn the voltage down to zero, the current in the circuit will be zero. Both parts of Ohm's law must be applied to any circuit - the proportionality to the voltage and the inverse proportionality to the impedance.
No. You have allowed the resistance and voltage to reach zero at different rates so you have residual current. Rick stole my reply before I got back:
...The infinite flow of current flowing through a super-conductive loop is not the result of an electric field, but due to inertia in the absence of resistance to stop them. Once a body is in motion, it will stay in motion unless acted apon, and even electrons with their small mass still need to follow that law. An external force got the electrons moving, such as a magnetic field, and once that force was removed, there was no opposing force present to make them stop.

#### Rick Christopherson

##### Senior Member
I am so confuse with why is it necessary to have the resistance of the wire in a hypothetical circuit to show the out come of an equation?
You are absolutely correct, that is, until you suddenly forget that you made this assumption and begin looking at the conductors themselves. That's the mistake that David made that launched this whole discussion. He assumed ideal conductors, which is perfectly acceptable, but then started looking at the behavior of the individual conductors while still using this same assumption that they were ideal.

You use this assumption when examining the load is your primary goal, but when you switch over to examining how/why current flows in the actual conductors, you can no longer treat them as being ideal conductors.

#### Rick Christopherson

##### Senior Member
Rick stole my reply before I got back:
Send me \$20 and I will give it back to you.

#### mivey

##### Senior Member
I am so confuse with why is it necessary to have the resistance of the wire in a hypothetical circuit to show the out come of an equation? in ohm's law you would still add the wire resistance together with the loads resistance to form one resistance figure to use in the equation as in any series circuit, so what is the difference if the wire has 1 ohm and the load is 50 ohms or you just use 51 ohms in your equation as you would do anyway?:-?

what would be the difference of saying your wire is 0 ohms, and your load is 51 ohm's or the wire is 1 ohm and the load is 50 ohms, it makes no difference of the out come?

and arguing of the current of a circuit of a source of infinity with a resistance of 0 would equal a current of infinity. that is the only answer obtainable no argument it is a fact:-?
I think it evolved from the fact that adding or removing a neutral in a circuit that is constrained to be balanced has no impact on the circuit.

As for the ohms: The difference is in whether you want the delivery to the load to be the same with and without the wire's resistance. If you want stable delivery to the load, you have to vary the voltage. A critical point for some loads...so critical in fact that we add regulators to compensate for the loss.

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#### mivey

##### Senior Member
Send me \$20 and I will give it back to you.
That's some mark-up. My thoughts are only worth \$0.02 :grin:

#### hurk27

##### Senior Member
I agree, but the conductors only come into play when one needs to know the outcome at the loads, it needs no place in the discussion of a MWBC in a single phase circuit, the other problem I saw was the mis-understanding of the polarity of each single winding of the transformer, they are in fact in-phase with each other and must be treated like such. we know that no current flows across a balance neutral point because there is no voltage across this point no matter what the resistance of the neutral wire is, 0 to infinity ohms, there will be no voltage, un balance the circuit and you will have a different story:grin:

#### gar

##### Senior Member
090710-1631 EST

Two more things to chew on.

1. Make a closed ring of copper. At any temperature above absolute zero there is a current flow. This is a result of the thermal activity in the resistance and produces random noise voltage. The background hiss you hear in a high gain amplifier. The noise level increases with temperature.

2. Heat a filament. Edison was probably the first to observe this. Electrons boil off of this heated element.

.

#### hurk27

##### Senior Member
I have to say that over the years, we tend to forget the many things we learned in school, between the electronics and studies of lightning and all of the math we had to learn, it boils down to what we use in our every day life to what we retain, as an electrician most of the equations I have learned have long gone out the window to the point I have to look them up when I need them. but as long as we can keep the basics, such as ohms law etc. we can get by with looking up what we need.

#### chris kennedy

##### Senior Member
I want to know if there would actual be less heat produced with a MWBC?
Yes........

:grin:

#### mivey

##### Senior Member
I agree, but the conductors only come into play when one needs to know the outcome at the loads, it needs no place in the discussion of a MWBC in a single phase circuit, the other problem I saw was the mis-understanding of the polarity of each single winding of the transformer, they are in fact in-phase with each other and must be treated like such. we know that no current flows across a balance neutral point because there is no voltage across this point no matter what the resistance of the neutral wire is, 0 to infinity ohms, there will be no voltage, un balance the circuit and you will have a different story:grin:
The discussion started with more heat vs less heat. It has been shown that a MWBC will have less heat. You have moved closer to tapping into the equilibrium sweet spot at the load.

If you go further and balance the MWBC, you have reached an equilibrium sweet spot that uses the least power, and have reduced the neutral current to zero.

#### hurk27

##### Senior Member
090710-1631 EST

Two more things to chew on.

1. Make a closed ring of copper. At any temperature above absolute zero there is a current flow. This is a result of the thermal activity in the resistance and produces random noise voltage. The background hiss you hear in a high gain amplifier. The noise level increases with temperature.

2. Heat a filament. Edison was probably the first to observe this. Electrons boil off of this heated element.

.
And the fact that every thermocouple in a furnace uses this to generate enough power to open a gas valve without any external power supplied

#### mivey

##### Senior Member
090710-1631 EST

Two more things to chew on.

1. Make a closed ring of copper. At any temperature above absolute zero there is a current flow. This is a result of the thermal activity in the resistance and produces random noise voltage. The background hiss you hear in a high gain amplifier. The noise level increases with temperature.

2. Heat a filament. Edison was probably the first to observe this. Electrons boil off of this heated element.

.

Did not know about the ring and amp hiss.

#### Rick Christopherson

##### Senior Member
That's some mark-up. My thoughts are only worth \$0.02 :grin:
What, are you nuts?!?! Your words are not worth \$0.02, I had a much higher markup on them than that.

#### mivey

##### Senior Member
What, are you nuts?!?! Your words are not worth \$0.02, I had a much higher markup on them than that.

A penny for my thoughts? That seems to be the going rate.:grin:

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