Transformer/amperage question

[Rysdad]

I understand step up/step down transformers and voltage/amperage changes, but I was wondering about the following scenario:

Lets' say you have a step up transformer with 50 windings on the primary and 100 windings on the secondary. The voltage would double and the amperage would decrease by half, right?

Ok, now let's say you wrap the secondary with another set of windings (on a completely separate circuit)--say 25 windings.

What happens? Would the interference cancel itself out? What would you end up with on the two secondary circuits?

Confused.

[Q.E.D.]

Originally Posted by Rysdad

Lets' say you have a step up transformer with 50 windings on the primary and 100 windings on the secondary. The voltage would double and the amperage would decrease by half, right?

Approximately. Voltage ratio is generally the same as turns ratio, but the exact relationship depends on a number of factors, such as core type and winding geometry. The current ratio may or may not be equal to the inverse of the turns ration; this is heavily dependent upon the various iron and copper losses present. In an ideal transformer, yes, you'd expect the current through the secondary circuit to be about half the current through the primary given the turns ratio above.

Originally Posted by Rysdad

Ok, now let's say you wrap the secondary with another set of windings (on a completely separate circuit)--say 25 windings.

Ok.

Originally Posted by Rysdad

What happens? Would the interference cancel itself out? What would you end up with on the two secondary circuits?

What interference?

If the 25-turn winding was in the same direction as the 100-turn secondary, you'd wind up with 1/2 the primary voltage across it, with the same polarity as the other secondary winding. The current through the primary would depend on the currents through each secondary. Assuming an ideal transformer, if the current through the 100-turn winding was 1 amp and the current through the 25-turn winding was 1/2 amp, the primary current would be 1/2 * 1 A + 4 * 1/2 A = 2.5 A

[Q.E.D.]

Originally Posted by Q.E.D.

...the current through the 25-turn winding was 1/2 amp, the primary current would be 1/2 * 1 A + 4 * 1/2 A = 2.5 A

Grrr. Got the primary and secondaries mixed up. That should read: "the primary current would be 1/2 * 1 A + 2 * 1/2 A = 1.5 A."

[Rysdad]

I mentioned interference because, if the current through the primary was going, say, south, then the current in the 100 winding secondary would head north, right?

If the 25-winding secondary was wound in the opposite direction as the 100-winding, then it would be 'being induced' by both the primary and the 100-wrap but in opposite directions.

I can't wrap my mind around it.

[Q.E.D.]

Originally Posted by Rysdad

If the 25-winding secondary was wound in the opposite direction as the 100-winding, then it would be 'being induced' by both the primary and the 100-wrap but in opposite directions.

Oh, I see what you're thinking. Don't do that, it'll give you a headache.

It's not necessary to think about the effects of the secondary windings on each other, because they effectively cancel out, regardless of winding orientation. You need only concern yourself with the relationship between the primary and each secondary. If the 25-turn secondary was wound opposite in direction to the 100-turn secondary, the only difference would be polarity. The voltage and current relationships would hold.

[Rysdad]

Ok...but just one more question...

How does each secondary winding know which one of the others is the primary?

Argh!

[Fridgemagnet]

Originally Posted by Rysdad

How does each secondary winding know which one of the others is the primary?

It doesn't - turn the transformer around and it will step up the voltage in the same ratio that it's stepped down the proper way round. The primary is just whatever winding has been designated as the input, though it should be noted that most transformers aren't designed to be bidirectional. Don't try connecting the secondary of a transformer across the mains, it will go bang.

[Q.E.D.]

Originally Posted by Fridgemagnet

Don't try connecting the secondary of a transformer across the mains, it will go bang.

Indeed. Unless you use a suitable current-limiting resistor in series with the driven winding (which now becomes your primary). The value of this resistor should be chosen so as to limit the current through the winding to whatever the rating plate calls for. This keeps the current density with design limits and also prevents core saturation. Of course, then you might have issues with corona discharge or insulation breakdown, because you're now exceeding the design voltage, but that's not something you can fix; proceed at your own risk.

[Rysdad]

Sigh.

OK--this may not be fair, but I need to change the scenario in order to get an answer I can understand. Here goes...

Primary: 100-wrap
Secondary A: 100-wrap
Secondary B: 100-wrap.

They're positioned like this:

.....O....
...O..O...

Primary is on the top. A and B are on the bottom. Imagine they make sort of an isosceles triangle.

Turn on the AC and........what happens?

The primary is inducing A. It's also inducing B. But isn't A inducing B, and vice versa? If not, why not?

If I input X amps into the Primary, what ends up running through A and B?

[Q.E.D.]

Originally Posted by Rysdad

The primary is inducing A. It's also inducing B. But isn't A inducing B, and vice versa? If not, why not?

Well, yes. A is inducing a voltage across B and vice versa. But, they effectively cancel each other out of the equation. So, don't worry about it. The only important considerations here are the turns ratios between the primary and each secondary, and the resulting voltage ratio.

Originally Posted by Rysdad

If I input X amps into the Primary, what ends up running through A and B?

Uh uh. It doesn't work like that. The current through the primary is dependent upon the currents through the secondaries, which in turn is determined by the secondary voltages and the connected loads--Ohm's Law and all that fun stuff. There is also a loss current through the primary called the magnetizing current; it's an iron loss due to the energy required to magnetized the core and is present regardless of loading.

[Rysdad]

Originally Posted by Q.E.D.

Well, yes. A is inducing a voltage across B and vice versa. But, they effectively cancel each other out of the equation. So, don't worry about it. The only important considerations here are the turns ratios between the primary and each secondary, and the resulting voltage ratio.

Let's say that, for whatever reason, you wanted the turns ratio equal on all three just because you wanted to separate the circuits for whatever reason.

How does B know to cancel A and not the primary? I mean, could I just put a primary in the middle of 173 secondaries and have equal voltage pushing through all of them?

Uh uh. It doesn't work like that. The current through the primary is dependent upon the currents through the secondaries, which in turn is determined by the secondary voltages and the connected loads--Ohm's Law and all that fun stuff. There is also a loss current through the primary called the magnetizing current; it's an iron loss due to the energy required to magnetized the core and is present regardless of loading.

I understand V = I x R, but if the load on all three circuits is equal, then is the amperage then equal? It seems like I'm violating TAANSTAFL.

[Q.E.D.]

Originally Posted by Rysdad

I mean, could I just put a primary in the middle of 173 secondaries and have equal voltage pushing through all of them?

Yep. If all 173 secondaries have the same number of turns, their no-load terminal voltage will be the same.

I understand V = I x R, but if the load on all three circuits is equal, then is the amperage then equal? It seems like I'm violating TAANSTAFL.

No, if all three windings have equal turns, then the primary current will be the simple sum of the secondary currents. Power in must equal power out. So, the primary current needs to satisfy the equation, V_in * I_in = V_out * I_out. So, I_in = (V_out * I_out) / V_in. You can apply the realtionship between turns ratio and voltage in this equation in an ideal transformer and work out the current based on secondary currents and turns ratio, but I leave that as an excercise for the reader (aka, Q.E.D. is feeling lazy ).

[Sage Rat]

(Am I the only one who was thinking they might find out how many amps Optimus Prime uses in this thread?)

[Q.E.D.]

Originally Posted by Sage Rat

(Am I the only one who was thinking they might find out how many amps Optimus Prime uses in this thread?)

Two. They're both Fenders.

*rimshot*

[Sage Rat]

*laughs*

[Rysdad]

1. I spelled TANSTAAFL wrong.

2. After sleeping on it (and reading QED's explanation ) I think I get it.

Thanks.