extra electric power / power amp.

[Kysmett]

Here is the drawing of the wave form that I mentioned.

[betajim]

Whoa! This is way off - in high school our physics class undoubtedly used both E and V in electrical-based equations, but they were in no way ever given as the meaning the same thing.

Well, here is what Kysmett posted above:

P=IE and E=IR

Replace E with V and you have Ohm's familiar law. E and V represent the same thing: electromotive force.
Any way, they are just symbols and in this context they do mean the same thing.

The power losses per transformer still look to be additive if both are fed back into one -

if V is stepped up considerably in one circuit - taking into account the original power is diminished because the circuit is split,
and I is stepped up in the other circuit,
and both those stepped-up values can be measured as those values (the current transformer being sold as such a meter in many cases), where do these stepped-up values go, if re-joining them ( after they are stepped-up ) will not equal both their values added together?.

A power loss is just that, a loss. You don't get that energy back. Some of the electrical energy flowing in the transformer is
converted to heat through resistive and eddy current loses. The setup you discribe creates no additional power.

Take care, Matthew.

[Kysmett]

Here is the shorthand I was taught and use:
I=Current
E=Voltage
R=Resistance
Z=Impedence
X_l=inducltive reactance
X_c = capacitive reactance
P=Power

The diodes are to keep backflow from the other transformer. ?It was just an idea, to protect the transformers. ?The reason I think it might smoke is that there will be a short of different voltages at the convergence point. ?This causes heat....

Try running a couple small transformers...power adapters for example. ?Run one forward and one backwards--- so that they are step up and step down respectively.
Keep it in a fire safe environment, as you are shorting two seperate voltages at the convergence point, and stay behind some plexiglass or something. ?If it doesn't blow or smoke, take a scope shot and let me know. ?I just realized that Voltage will lead the current through the transformers, and I'm not sure what effect, if any that will have. ?The diodes should be able to stand up to the voltages and currents. ?

Kind of skepticle, but dying to be supprised

[Kysmett]

Try it. 

Use simple power adapters as your transformers.  See what happens.  (Note: Do not read any sarcasm here...I'm truly interested)   Try it even without the diodes.   Right now I cant get my head around what the phase shift caused by the transformers will do to the wave form (voltage leads current by 90 degrees) 

Here is the information I think you were looking for.  Voltage adds only in series, otherwise averages.  Current adds only in parallell, otherwise averages.  I think thats right.  If there is a way that you could get the voltage and the current to add differently, then you may be on to something.

Look forward to hearing about the results.  I may even rummage around to find my stash of spare adapters to try it.  If I do, I'll let you know.  The only thing is that I don't have a frequency generator, so any work I do will be limited to either DC or 60 Hz.

[Kysmett]

Here is the relevance of voltage and current adding/ averaging, depending on configuration.

At the convergance point, you are putting the two outputs in paralell.? So, the currents would add, but the voltage will (or should if the experiments I've done in the past with bateries hold true for transformers and AC) equal 1/( 1/Voltage₁ + 1/Voltage₂)

So if you had 2 ten-to-one transformers and you started out with a 100V 10A signal(and assuming no loos in the transformer--which there will be) on one output you would have 1000V 1A and on the other you would have 10V 100A.? Where they meet, you would add the current so there would be 101A, but the voltage(V_out) would follow the above equation:

V_out=1/(0.001 + 0.1) = 9.901

Now if you multipy the Current and the Voltage you will still have 1000W