240v debate....

[david luchini]

Show the simple math on the 240V system Mivey. You can't solve the equations without using double negatives for your choice of reference.

Vload+Vbn-Van=0. Where is the double negative?

 

[weressl]

I was on top of an "A" frame ladder about 16 feet above the concrete floor when I fell.

That, of course,:lol: explains a lot.....

 

[jim dungar]

Vload+Vbn-Van=0. Where is the double negative?

Have you solved all of my loop questions?

Take two unequal size resistors, connect them in series, power them from a 240V 2-Wire source.
Does the loop current have the same phase angle as the voltages across each resistor?
Do the voltages across the resistors have the same phase angle as the source voltage?
Is the loop current in phase with the source voltage?

Take the exact same resistors in series, except power them from a 240V 2-Wire source created from (2) 120V supplies connected in series.
Does the loop current have the same phase angle as the voltages across each resistor?
Do the voltages across the resistors have the same phase angle as the (2) supply voltages?
Is the loop current in phase with each of the supply voltages?

If we can't agree on the answers to this simple connection of loads and sources, it will be impossible to agree on the answers to anything.

 

[david luchini]

Have you solved all of my loop questions?

Take two unequal size resistors, connect them in series, power them from a 240V 2-Wire source.
Does the loop current have the same phase angle as the voltages across each resistor?
Do the voltages across the resistors have the same phase angle as the source voltage?
Is the loop current in phase with the source voltage?

Vl1+Vl2+Vba=0

Take the exact same resistors in series, except power them from a 240V 2-Wire source created from (2) 120V supplies connected in series.
Does the loop current have the same phase angle as the voltages across each resistor?
Do the voltages across the resistors have the same phase angle as the (2) supply voltages?
Is the loop current in phase with each of the supply voltages?

If we can't agree on the answers to this simple connection of loads and sources, it will be impossible to agree on the answers to anything.

Vl1+Vl2+Vbn-Van=0. What's the issue?

 

[pfalcon]

Vload+Vbn-Van=0. Where is the double negative?

First determine your load direction. Presumably from your equation Vload = Vab

Vab = Vnb + Van = - Vbn + Van :: Vab + Vbn - Van = 0
240<0 = 120<0 + 120< 0 = - ( -120<0 ) + 120<0 :: 240<0 + ( -120<0 ) - ( 120<0 ) = 0

So since you asked nicely, there ya go. A double negative.

 

[jim dungar]

Vl1+Vl2+Vba=0
Vl1+Vl2+Vbn-Van=0. What's the issue?

I hate to be putting words in your mouth but:
Note: I did apply mathematical techniques to reword your answer as Vl1+Vl2 = Van-Vbn

Do the voltages across the resistors have the same phase angle as the (2) supply voltages? - I believe your answer is NO, based on your choice of supply reference point.

Does the loop current have the same phase angle as the voltages across each resistor? - I believe your answer is YES, based on your formula Vl1+Vl2.

Is the loop current in phase with each of the supply voltages? I believe your answer is NO, based on your formula Van-Vbn.

 

[david luchini]

First determine your load direction. Presumably from your equation Vload = Vab

Vab = Vnb + Van = - Vbn + Van :: Vab + Vbn - Van = 0
240<0 = 120<0 + 120< 0 = - ( -120<0 ) + 120<0 :: 240<0 + ( -120<0 ) - ( 120<0 ) = 0

So since you asked nicely, there ya go. A double negative.

From my equation, Vab=Van-Vbn

240<0= 120<0 - 120<180... There you go, no double negative anywhere to be seen.

Of course you could force a double negative into the discussion of two 120V sources viewed from the perspective of having the same phase angle: Van=120<0 and Vnb=120<0.
Vba+Van+Vnb=0 :: Vba + 120<0 + 120<0 = 0
120<0 + 120<0 = -Vba :: 120<0 + 120<0 = -(-240<0) :: Vba=-240<0

 

[jwelectric]

I just got done talking to Eli down at the Ice factory. He said that I came to the right place if I wanted to see a phase shift between the voltage and the current. He said that only ELI the ICE man could do such a feat.

 

[david luchini]

Do the voltages across the resistors have the same phase angle as the (2) supply voltages? - I believe your answer is NO, based on your choice of supply reference point.

Yes, the voltages across the resistors have the same phase angle as the supply voltages. You could reference Vl2 as going in the opposite direction, if you want, such that Vl1-Vl2+Vbn-Van=0, but that doesn't change anything. If for example your two resistors were 10ohm and 20ohm, and Van=120<0, Vbn=120<180 and your loop current is considered as "leaving" A to the first resistor, then Iloop=8<0A.

.....The voltage from A-B across the loads is Vl1 + Vl2 = 10(8<0) + 20(8<0) = 80<0 + 160<0 = 240<0.

.....The voltage from A-B across the source is Van - Vbn= 120<0 - 120<180 = 240<0.

Does the loop current have the same phase angle as the voltages across each resistor? - I believe your answer is YES, based on your formula Vl1+Vl2.

Again, the answer is yes. The loop current is 8<0. The voltage across each resistor is 80<0 and 160<0.

Is the loop current in phase with each of the supply voltages? I believe your answer is NO, based on your formula Van-Vbn.

Again, yes the loop current is in phase with each of the supply voltages. The loop current defined as "leaving" from terminal A is 8<0. That same loop current would be seen as "entering" terminal B. When referenced as "leaving" terminal B, the loop current is -8<0, or 8<180. So the loop current referenced from Supply Voltage A (120<0V) is 8<0A. The loop current referenced from Supply Voltage B (120<180V) is 8<180A.

 

[jim dungar]

Again, yes the loop current is in phase with each of the supply voltages. The loop current defined as "leaving" from terminal A is 8<0. That same loop current would be seen as "entering" terminal B. When referenced as "leaving" terminal B, the loop current is -8<0, or 8<180. So the loop current referenced from Supply Voltage A (120<0V) is 8<0A. The loop current referenced from Supply Voltage B (120<180V) is 8<180A.

You chose the loop current as "8<0" so it appears you have chosen Supply Voltage A as your reference for solving this loop. So now you have a 100% resistive current at 0? 'driven by' a source at 180?.

 

[david luchini]

You chose the loop current as "8<0" so it appears you have chosen Supply Voltage A as your reference for solving this loop. So now you have a 100% resistive current at 0? 'driven by' a source at 180?.

No, the loop current at 0? is driven by a source at 0? (Van=120<0, Vab=240<0.)

From earlier:

So the loop current referenced from Supply Voltage A (120<0V) is 8<0A.

 

[cowboyjwc]

I didn't do this much math when I built my time machine, but that was tomorrow so maybe I did.

 

[mivey]

Are you saying that a center tapped transformer is not two sources in series?

Back to the basics: To be a series circuit, the currents will be the same in all parts of the circuit. Without the neutral, this is the case. With the neutral in the circuit, the currents in the sources can be, and almost always are, different.

 

[mivey]

When have I said you must do something?
I have stated that the presence or absence of a load neutral should not require a change in the way the sources are viewed.

It is a simple fact that a two wire source has only one voltage but the three wire source has two voltages (three if we include the larger voltage). They are simply not the same. Anyone who has lost a neutral can tell you that.

I have asked you to solve a simple loop that would be typical of a 120/240V center tapped transformer that looses the load neutral connection.

And how do you suppose that losing a neutral is the same circuit that you had before? We have gone from a source with two smaller single-phase voltages and one larger single-phase voltage to a source with just one single-phase voltage. Hardly the same at all.

Oh yeah, I have suggested that it should be explicitly mentioned that the neutral point is your reference when you say 120<0 and 120<180 combine to be a 120/240V source.

I think it should be explicitly mentioned what you use as a reference whether or not that reference is the neutral point.

 

[jim dungar]

No, the loop current at 0? is driven by a source at 0?

But you said that one of the source voltages(Vbn) is at 180?, so for the purposes of solving this loop you have a resistive current out of phase with a source voltage.

I have not said the math prevents it, but it definitely is not intuitive, so why would you do it?
Would this happen in a center tapped transformer winding, or any other transformer with a single core?

My choice has been to say a center tapped transformer is equal to a series connection of two in-phase sources, which prevents the I<0 and V<180 issue, and also models the construction of a single core transformer with windings that can be paralleled or series connected.

 

[mivey]

No, the potential was converted to kinetic and then to deformation. The potential is gone.

Since energy can neither be created or destroyed, it was definitely converted to some other form of energy.

Since due to inertia and friction the Earth did not move at all, his testimony is correct. To move the Earth he wouldn't be he.

Who knows? The Earth might have started spinning just a tad faster.

Yes there is a defined direction for r. It's an absolute distance and therefore never negative.

If two balls of equal mass were positioned in deep, empty outer space within appreciable ranges of their space distortion, the gravitational forces will pull them together. Now who is to say which one moved in which direction? You have to define a reference frame and there is no "right" or "wrong" reference frame.

Show the simple math on the 240V system Mivey. You can't solve the equations without using double negatives for your choice of reference.

Looking ahead, I see David already answered that.

 

[mivey]

So the loop current referenced from Supply Voltage A (120<0V) is 8<0A. The loop current referenced from Supply Voltage B (120<180V) is 8<180A.

Electricity 101. I still find it incredible that so many are still struggling with it. The only thing that comes to mind is that they are trying to change fundamental electrical theory to fit a model that they have become accustomed to over the years. I can understand someone getting in a rut, but that excuse will only stretch so far.

 

[__dan]

So, when I move the black lead to the top of the stack and the voltage display goes negative, do you want to say "it's 180 out of phase".

From the transformer looking out, the magnetic flux direction the the winding turn direction always match exactly, the same. From the load looking at it's source, you have two sources and a range of connection choices, series, parallel, and polarity reversal - having the windings in series either add or subtract voltages.

 

[mivey]

So, when I move the black lead to the top of the stack and the voltage display goes negative, do you want to say "it's 180 out of phase".

Since when does a constant, direct current have a phase?

From the transformer looking out, the magnetic flux direction the the winding turn direction always match exactly, the same.

Do you really understand flux or are you just throwing that out there because it "sounds" good? How do you propose that it is going to dictate the voltage direction? I'm open to studying any theory you might be proposing but I suspect you are mis-interpreting some of the known theory. Feel free to post the theory that shows the voltage directions must be the same as I do not recall the one you might be referencing.

The fact is, the flux does not determine the voltage direction because the choice of a voltage reference is just that: a choice.

From the load looking at it's source, you have two sources and a range of connection choices, series, parallel, and polarity reversal - having the windings in series either add or subtract voltages.

FYI: Polarity and voltage direction are not the same thing.

 

[mivey]

Do you really understand flux or are you just throwing that out there because it "sounds" good?.

I apologize for that. It sounds snippish and mean.