You not at all reading what I am saying. I am saying MORE turns in the secondary coil with LESS resistance. = More turns with more current. Obviously you can't get free energy, so where would the loss come from? I am assuming from the flux being unbalanced in favor of the secondary coil overpowering the primary and eliminating your gains.
Hi,
Let me run this by you. You make a transformer with the following characteristics using ordinary copper wire:
Primary wire diameter 0.02 inches
Secondary wire diameter 0.06 inches
Primary wire turns:100
Secondary wire turns: 200
So we have a step up transformer that someone who did not know how to design transformers created. The secondary turns are greater than the primary turns so it's a step up, yet the secondary wire diameter means the secondary wire can handle more current. So that doesnt make sense either because the secondary will have 1/2 the current of the primary current, so we dont need a wire diameter that is 3 times higher, we can actually get away with a wire diameter that is less than the primary because of that current fact.
The bottom line here is that the secondary current Is will be Ip/N where N is the turns ratio from primary to secondary, so for a step up as above we'd have N=2 so the secondary current would be Is=Ip/2 no matter how good the wire was that we used.
If we did design a transformer like that with primary wire 0.02 inches and 100 turns and we had wire that was 3 times less resistive (the unobtainium here) on the secondary then we would use a diameter that is LESS than the primary because it only has to handle 1/2 the primary current.
Another example, we design a transformer with 100 turns primary and 100 turns secondary. We use copper on the primary, we use silver on the secondary. We've wasted the silver wire because the current in the secondary will be the same as the primary.
Let's do a few calculations.
Here, r is resistivity and RT is total resisance and V1 is primary wire voltage drop with load and V2 is secondary voltage drop with load.
r1=6.8e-7 primary resistivity in Ohm inches
r2=3.4e-7 secondary resistivity in Ohm inches
D1=0.02 primary wire diameter in inches
D2=0.02 secondary wire diameter in inches
L1=120 length of primary winding wire in inches
L2=120*2 length of secondary winding wire in inches
RT1=4*r1*L1/(D1^2*pi) total resistance of primary wire in Ohms
RT2=4*r2*L2/(D2^2*pi) total resistance of secondary wire in Ohms
Ip=1 primary current in Amperes
Is=1/2 secondary current in Amperes
V1=Ip*RT1 primary wire voltage drop with load in volts
V2=Is*RT2 secondary wire voltage drop with load in volts
P1=V1*Ip primary wire power loss in watts
P2=V2*Is secondary wire power loss in watts
Doing these calculations we get:
P1=0.25974 watts
P2=0.064935 watts
So we see that the secondary has much less power being lost in it, so this would not be as good of a design as it could be because we are using a wire diameter that is too large for the secondary, and thus wasting the unobtainium wire material.
Knowing this maybe we use trial and error to get to the right value. To start with, we decrease the wire diameter of the secondary by a factor of 1/2, making it now 0.01 inches in diameter. Now we do the calculations over again and we get:
P1=0.25974
P2=0.25974
so now we loose equal power in both windings. In real life we might increase the primary wire diameter slightly because we now have more room in the core window area, so we make the transformer more efficient and also get a higher power rating for it that way.
The main point however is still the same, we calculate the secondary current by using:
Is=Ip/N
and that has nothing to do with the wire itself, for the most part.
Maybe you have some specific reason for asking this question, as to why you think the secondary should somehow violate nature. It's just fine to have curiosity like that but you should explain why you think this in more specific terms so we can address those issues directly. Perhaps you've discovered something new, and in that case we'd want to know what it is.