|> | Well, the answer goes back to the theory. Remember that when an |> isolated |> | load is connected to any secondary of an energized transformer, the |> | magneto-motive force created by the secondary current is opposing the |> MMF of |> | the primary current. |>
|> This I know. But there is some ambiguity to this. You said force and |> then |> current. The problem I see is that I don't know which it really is that |> determines the orientation. Given 2 identical windings, if the _voltage_ |> ends up being the same on the same ends, then the 2 _currents_ are flowing |> in opposite directions. This is explained by the fact that power is going |> in on the primary and power is going out on the secondary. Understanding |> the "opposing force" would help. But due to the loose usage of terms when |> many people speak about theory, especially "force" vs. "current", I just |> can't be sure what is going on there. | | The 'force' I spoke of is Magneto-Motive Force, not voltage. MMF is | measured in ampere-turns. It is directly proportional to the *current* (not | voltage) through a winding.
Is this a flux density of the magnetic field?
| Rule # 1: The MMF generated by any current flowing in the secondary is | *always* opposite the MMF generated by current flowing in the primary. From | this simple rule, you can figure out all your answers.
So how can that be? How do the electrons ... and whatever it is the magenetic field is made of ... know which is primary and which is secondary?
|> | two). Point the thumb of that hand (for me, it would be the 'left |> hand') in |> | the same direction along the core. Now your fingers of that hand wrap |> | around the core in the direction of current flow in the secondaries. |> | Regardless how many secondaries there are, and which way they are |> wrapped, |> | the current in*all* secondaries will go around in the direction of your |> | second hand's fingers. You're done. |>
|> I'm sure all the secondaries are like any other secondary. I'm sure all |> the |> primaries are like any other primary. It's the relationship between the |> primary and the secondary that I haven't pinned down. |>
|> Looking at the core from one end, I believe the direction of wire wrap, |> e.g. |> clockwise vs. counter-clockwise, is the issue. Whether the wire starts at |> the bottom and ends up at the top, or starts at the top and ends up at the |> bottom, is not. Or a wire could wind CCW going from top to bottom and |> then |> continue winding CCW in a new layer going back to the top, repeating until |> the needed number of windings are done. It would still be CCW from the |> referenced view. | | This is true. Did I say something to make you think otherwise??
I'm paraphrasing it to look at it in a different perspective to be sure my understanding of it is consistent.
|> For consistency, I would bundle all the windings together as a "cable" and |> wind this cable as described. While that may be a lousy way to construct |> a real-life transformer, I think it clearly shows the idea of everything |> in |> the same orientation. |>
|> When I energize the transformer, the primary current is opposed by the |> field. |> Whether that is an actually a current opposition or a voltage opposition |> would not matter (yet) since it affects the same winding. But when we |> look |> at the 2nd winding as a secondary, there would be a voltage potential |> there, |> but not being connected to a load, no current. Now if a load is |> connected, |> is the _current_ on the 2 windings going to be in the same direction? | | No. See Rule # 1 | | The varying magnetic field (created by varying MMF of primary) induces a | voltage in primary that opposes the applied voltage. Since the secondary is | wound right along side the primary (in your 'cable' example), the induced | voltage in the secondary would have the same polarity at any instant as the | applied voltage on the primary. If a load is connected to the secondary,
It would seem to me that the induced voltage on the secondary would have the opposite polarity as the applied voltage. The reason I say this is the magnetic field is going to induce a voltage on one wire, then it will induce a voltage on another wire running along with it in exactly the same polarity. Since that polarity is the opposite of the applied voltage, it would be opposite in the secondary as well.
It gets back to the same ambiguity. How can a magnetic field that is inducing voltage in two wire do it differently on one vs. the other just on the basis that one is a primary and one is a secondary.
| the induced voltage in the secondary creates a current flow to the load. | The MMF *must* of the secondary *must* be opposite the MMF of the primary | (see Rule # 1). Since you have them wound in the same direction, the only | way to get opposite MMF is to have opposite direction of current flow. | | So for some instant in the cycle when the electrons are traveling CW around | the core in the primary winding, the electrons in the secondary winding are | traveling CCW.
Maybe I'm misunderstanding your description of induced voltage. This makes sense about the current because it is consistent with wiring practice. That is, we run 2 wires along side each other to power a load, and the magnetic fields between them cancel out nearly to zero.
| No. The currents would be traveling in opposite directions (see Rule # 1).
I can see that. I just can't see it from Rule #1 yet.
|> If the currents are the same then the |> voltage on the secondary will be opposite because power is being taken out |> instead of being put in. | | The currents are not 'traveling in the same direction', so the voltage on | the secondary is the *same* polarity as the primary. (in your winding | example)
Then I must be misunderstanding because I got a different meaning than this from a previous statement.
|> But is that so, that both primary and secondary currents go in parallel |> when |> power is drawn from the secondary? It seems that can't work because it |> would |> increase the field strength, and something I read suggest the secondary |> has |> to be tapping into the field strength, effectively lowering it, for power |> to |> be taken out. | | Congratulations, you've talked yourself out of your mistake. The currents | do *not* travel in the same direction around the core, so the field does not | increase.
What mistake?
A lot of this posting is paraphrasing what others post. If what I say seems wrong, then one of:
- What they posted was wrong.
- What they posted was poorly written.
- I just didn't understand what they posted.
- I understood what they posted by paraphrased it poorly.
- My paraphrasing was misunderstood.
I've always believed, and your post here is confirming, that the currents will be in the opposite direction. Thus if I wire 2 windings together the same way, such as I could do by using a 2-wire cable, with the winding only going from top to bottom (for clarity, not for practical construction) then when at some instant the current is going from top to bottom on the energized winding (primary), it is going from bottom to top on the secondary.
That also tells me if the voltage is - at the top on one where the power is coming in, it will be - on the top of the other where the power is heading out.
| How much clearer do you want me to word this? The *current* in the | secondary travels in the opposite direction around the core as the primary | *current*.
OK, that's clear. But what I read from you about the voltage seems to be the opposite of what I expect. I would expect the voltage polarity to be the same at the same instant, when the current is reversed. But you have it stated the other way.
At least if we agree about the current, I'm half way to understanding this.
| No, you have to mentally 'cut the core' and unwrap it. Once you open up a | toridal core and straighten it out, or 'slide' the secondary winding over to | the same side as the primary, you have a simple solution.
I sort of did that. I just did it the other way around, starting with a straight core and bending it around.
|> Here is a more complex scenario which is one of those things that has |> given |> me the want to find out precisely about this. Suppose I have an "E" core |> transformer, which has 3 vertical bars crossing between a top bar and a |> bottom bar. This is the typical core used for a 3-phase transformer. |> Label |> the three vertical core segments A, B, and C. If I wind a primary around |> core A and energize it, the field will loop around through cores B and C. |> If I wind a secondary around core C, now what happens? | | Well first of all, you don't have a three-phase transformer with just one | primary winding and one secondary winding. Your example is a single-phase | transformer on a three-phase core. Bad, bad, very bad.
Certaing it is not a three phase transformer; it's just the core from one and other windings added for single phase.
| But the answer is still there. When you have current flowing through the | primary around the 'A' leg, suppose that on some particular half-cycle, the | MMF of the primary is such that your 'thumb' points upward. So 'lines of | flux' will rise through the iron from the top of the coil, turn with the | iron and 'flow' downward through the iron in legs B and C. When a load is | connected to your secondary winding, its MMF *must* oppose this 'flow of | lines of flux'. | | Looking down from the top, you can think of the 'lines of flux' rising up | toward you in the 'A' leg, turning and going down away from you in the 'C' | leg. So if the current in the primary is flowing CW at a particular instant | (as viewed from above), then the current in the secondary will also be | flowing CW at the same instant (as viewed from above). The reason it | *seems* to be the same direction this time is because the magnetic field did | a u-turn in the iron. So we still have an opposition (rule #1 still | applies).
I can see that from the reversal of the orientation because the field is turned around in C with respect to its origin in A.
| BTW, with no load on the secondary, the voltage output of the secondary is | *not* going to be what you'd expect from the turns ratio. If the 'split' of | the magnetic flux between the B and C legs were perfect, you would get half | the voltage that the turns ratio alone would predict. Can you see why?
No, I don't see why. In this scenario I see ambiguity. That means I just don't know what would happen. Or why.
|> Will the secondary |> around core C be able to get full power out, despite B being present? Or |> will the B core reduce the available power in some way? | | The 'B' core will greatly reduce the amount of power available on the 'C' | winding. This is because when the secondary current begins to flow through | a connected load, the secondary's MMF (which remember opposes the primary's) | will simply 'divert' much of the magnetic flux of the 'A' leg that was | originally 'flowing' into the C leg of iron, over to the B leg of iron. | This results in much less magnetic flux in the C leg and correspondingly | less induced voltages. (i.e. the voltage output of the C winding will drop | drastically as it is loaded) | |>
|> Suppose I wind a tertiary winding around core B, and monitor the voltage |> by drawing a trivial few microamps. Will this voltage change as more and |> more power is drawn from the winding around core C? |>
| Yes. The exact 'split' of magnetic field from the primary A winding will | vary drastically with load. So a heavy load on the 'C' winding will shift | the 'split' such that the field strength in C is reduced and the field | strength in B is increased. This is why you almost *never* make a | transformer like this. Three-phase transformers are *not* wound as you're | describing.
I would not expect them to be wound that way.
|> What happens if I put a big load on B, or even solidy short it? How will |> that affect the power I can get from C? What if I put a resonant circuit |> on the B core, peaked at a high impedance at the power frequency? | | Now, you can have all sorts of fun. A shorted turn on B would effectively | reduce the amount of primary flux that goes through the B leg, forcing more | through the C leg. Or, you can 'notch' the iron in the B leg so small | amounts of flux pass as before, but 'large' amounts saturate the iron in the | area of the notch. So limiting the amount of power that can be drawn | through that leg. | | A ferro-resonant transformer with a tuned tertiary winding is just such a | 'critter'. With a high-reluctance path (an air-gap) to the winding with a | tuned resonant circuit (a series circuit tuned for *low* impedance at power | frequency) draws little power in the tertiary because of this air-gap, | leaving most of the power for the secondary. But a slight increase in | primary MMF (as caused by a rise in supply voltage) will increase the power | drawn by the tertiary and thus the power supplied to the secondary remains | fairly constant. Such things are also known as constant-voltage | transformers, or by the brand name Solo-Transformers.
OK, this does sound like fun, though I was more interested in playing with harmonics.
|> If I had 3 windings around a single core, I can better visualize what |> might |> be happening, despite a few lingering doubts or ambiguities about how it |> all works. But with the E-core, things are "stranger" here. | | Because E-core are not used for simple, single phase transformers. When an | E core is used for a three-phase transformer, *two* windings are wound | around each leg. A primary, and the same phase's secondary. The three | primaries (one on each leg) are connected similar to how three single-phase | transformers might be connected (either wye or delta). And the three | secondaries are also connected in much the same way (either wye or delta). | | Putting a single phase primary on one leg, and two different secondaries on | the other legs is *not* something I've run across in my many years (except | the ferro-resonant transformer I described earlier).
Well, let me take the normal three phase transformer and vary it just a bit. Instead of varying it by construction (let's leave it at normal), let's vary it by placing a solid short on ONE of the secondary windings. Would this then cause an _increase_ in the MMF of the other two phase cores, thus _raising_ their secondary voltages?
This would be consistent with your statement from above, repeating:
| Yes. The exact 'split' of magnetic field from the primary A winding will | vary drastically with load. So a heavy load on the 'C' winding will shift | the 'split' such that the field strength in C is reduced and the field | strength in B is increased. This is why you almost *never* make a | transformer like this. Three-phase transformers are *not* wound as you're | describing.
I have seen a scenario where, during a storm, I saw the voltage rise quite a bit for a couple seconds, from around 120 to an estimated 140 to 150. A 60 watt light bulb got brighter to about the level somewhere between a
75 watt and a 100 watt bulb. A few more seconds after this I heard from a distance a loud boom like some primary distribution had shorted. My speculation is that a different phase shorted, and the transformer that was supplying power on all these phase raised the voltage on the others due to the high current of the one that was shorted. Plausible?
I'm going to save this whole post, because it must be me who is simply misunderstading an earlier induced voltage reversal description. This I'll have to re-read a few more times to be sure.