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Assume an ideal transformer with the same number of windings, the primary windings generate 50VAC, 10Amps (5Ohms), the secondary side has a 10 Ohms resistor, if the voltage induced on the secondary side is 50VAC, does this mean 5Amps is flowing through the secondary circuit, but why is the power different? Is it lost somewhere?

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  • $\begingroup$ The voltage on the secondary remains 50V like the primary due to equal number of turns. The power in the primary will be the same as the secondary. So find power in secondary P, then current in primary is I=P/50 Amperes. The current in primary changes to match the power taken at the secondary till the iron saturates or one of the coils is burnt. $\endgroup$ – Riad Dec 30 '18 at 9:06
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An ideal transformer has an impedance that's purely inductive and dissipates no power. If you apply 50V at the primary winding, 50V will also show up at the secondary winding. The current is determined solely through the load at the secondary winding. In your case of a 10 Ohm resistor, that would be 5A. The current is the same in both windings.

An ideal transformer with a 1:1 ratio is basically a pass through. They are often used as isolation transformers to decouple ground references or different potentials, but the AC current and voltage is the same on both sides.

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