Ideal transformer paradox
For any given AC input voltage, the current an ideal transformer draws from the power supply is related to the load on the secondary. It's zero for no load.
If the primary voltage is Vin, then the secondary voltage is Vin*(n2/n1)
...where n2 is the secondary turns and n1 is the primary turns.
So the power at the secondary is Vin*(n2/n1)^2/RLoad
And the power at the primary is the same.
We say that the load resistance (or impedance) is reflected through the transformer by the square of the turns ratio \$(n_2/n_1)^2\$.
So a load of 100 ohms on the secondary of a 1:2 ideal transformer acts the same as 25 ohms directly on the primary voltage. Apply 10V and the power is 100/25 = 4W in either case. There will be 20V on the secondary of the transformer since it's stepping up the voltage 2:1.



simulate this circuit &ndash; Schematic created using CircuitLab
There is no contradiction. Once you fix the voltage the circuit determines the current drawn.