So guys i have a basic (and probably a stupid) Question about trasformators.
I understand that the power on both sides of the transformators should be (ideally) the same, which makes sense.

So if we have on the left side low voltage and high current and want to get a high voltage on the right side, we will get a low current.
What i dont understand is how can you have a high voltage and a low current, if we know that current and voltage are propotional V=RI?



for example in my simple example, should I2 be very hight, since the voltage is also very hight? This is confusing me.. please help

 

Your example is too extreme!
If this is an "ideal" transformer, on the secondary side, ten thousand amps would flow through R1 (I = V/R). The voltage ratio of the transformer primary to secondary is 10,000:1 so the primary current would be 10,000 x 10,000 = ONE HUNDRED MILLION AMPS!
That would be one almighty transformer!!!!
Try an example that is a little more realistic then it will be much easier to understand.

 

Power = Watts = Volts X Amps.
You can't get out more Power than You put in, it will always be somewhat less.

Power doesn't care whether it's in the form of more Voltage and fewer Amps, or vice-versa.

The amount of Power is limited by the capacity of the Transformer-Core to
transfer "X" amount of Power from
the Primary-Winding to the Secondary-Winding with acceptable efficiency.

The Power-capacity of the Transformer-Core generally increases with
increasing Frequency, and increasing weight/mass,
but, as with anything, there is no "Free-Lunch",
and other compromises start coming into the picture and
changing what works, and what doesn't work.

Most, if not all, inefficiencies in a Transformer show-up as increased HEAT.
.
.
.

 

Your output circuit (on the transformer secondary) obeys Ohm’s law, so current=voltage/resistance.
power=current * voltage.

Ohm’s law has nothing to do with the operation of a perfect transformer. It is governed by Faraday’s law, so the flux is proportional to the voltage* and the number of turns, so it steps the voltage up or down in proportion to the ratio Of numbers of turns on each winding.
It‘s a perfect transformer, so there are no losses, and power can‘t be gained from nowhere, so power in equals power out.

current in= power /input voltage
current out = power/ output voltage


(* rate of change of voltage to be perfectly accurate)

 

1) Ideal transformer. Take turn count=50*V/S(cm2) for standard grid frequency of 50 Hz. For any other frequency apply the multiplicator =50/f(Hz). For example, 50 kHz will have 1000x less turn count. If meander signal (bridge) apply the surplus factor 1.1x. If core is not 0.8 Tesla steel, but 1...1.2....1.6 Tesla steel, then apply the according surplus factor =0.8/B(max). For example, many steel cores may stand the factor not 50V/S but 48. 45, 42 and some even 38. Vice versa, if use the ferrite core having no more might than 0.36...0.3...0.2 Teslas, then use more turn count, accordingly. Worth to add, tah using formula dU=di/dt for calc a trafo gives giant mistake what may have a value between 10x and 100x. Never use that formula for any transformer, because it inductance is heavily dependant on current.
2) Real transformer: Voltage loss in primary=R(wiring)*I(nominal), thus the effective grid voltage will be Grid Voltage minus voltage loss. Therefore secondary will get just slightly less voltage. For example: 220-->22 Volt transformer have primary voltage drop 22 V. Then Secondary (transformed) voltage drop will be 2.2 V or output 19.8 V. But there is also the secondary voltage drop into secondary wires. Calc it resistance and at real secondary current calc the straight secondary droop. Then add both droops together and have real output voltage. However, practically always that droop is less than few % thus may be ignored.
3) Immaginary currents in trafo: normal transformers have linkage factor between sides about 99%, thus the 1% leaking inductance may be ignored. But some cases, as an example AC welding transformers with veeeery loooong coils movable on the core, have significant leakage inductance. Then it can be calced in as well, making an imaginary part into trafo equation, but that is longer story.

 

To restate what has already been said in a completely barebones way:

The absolute value of the voltage or current is immaterial to the concept of trading one for the other. It will always be a matter of \( \mathsf{W=V \times A} \) no matter what \( \mathsf{V} \)AND \( \mathsf{A} \) are within the other limits placed on the system.

So, there will always be losses due to resistance, capacitance, and inductance which will limit the the numbers from the simple \( \mathsf{W=V \times A} \) calculation. Aside from that, in the median cases, yes: 1V@10A will get you 10V@1, etc. This doesn’t change (concerning the magnitude of loss, outside of actual electrical breakdown) if we have 1V@1MA and 1MV@1A.

I think you probably need to avoid imagining how “big” some numbers are in the physical world. Yes, a MV is going ot have all sorts of practical problems but if those are accounted for, and the losses considered, it’s the same idea as the smaller cases.

 

In an ideal transformer the ratio of the primary voltage to secondary will be proportional to the number of turns on these respective windings.

In an ideal transformer the power to the load and the power from whatever is driving the transformer will be the same. Since the power is the same on both windings and the voltages are proportional to the number of turns the ratio of current drawn by the load supplied by the power source to the primary will be proportional to the reciprocal of that turns ratio.

Be aware with real world transformers there are factors that cause significant departures from the above.

 

Generally, the larger the transformer the closer it approaches the ideal.

 

On second thought; the example given is so ridiculously extreme and no response to our answers was received, so was this just posted to get a reaction?