Ideal transformer output voltage

Question

I know that the output voltage and current of an ideal transformer depends on the number of wounding of the coils. It means that no matter how many resistors I put in a series in my secondary loop, the current would not change and it depends only on N (Number of turns in coil).

I have been thought that resistors define the current that flow through them and it is hard to look at the current independent of the load.

Could someone explain the physics behind v1/v2=N1/N2 So is it safe to assume that the current in secondary coil acts as an ideal current source because if I know the source current I can get the secondary loop current no matter what the load of the secondary circuit is

Answer 1

Could someone explain the physics behind v1/v2=N1/N2

In a transformer electric energy is converted into magnetic energy and then back to electric energy.

The formula V1 / V2 = N1 / N2 applies to this process without any losses.

Losses can occur due to:

• Resistance of the transformer wires

• Saturation of the magnetic core of the transformer

• Inductance of the transformer (a transformer is also a coupled inductor)

• frequency of the AC signal fed to the transformer

So the formula V1 / V2 = N1 / N2 does apply but it only applies to an ideal transformer.

Also an ideal transformer does not lose any power nor will it add any. So that means:

P1 = P2 and also V1 * I1 = V2 * I2

If you would connect a current source to one side of the transformer then the relation will still hold however you must make sure that the current on the other side of the transformer can flow so you must apply a load or short it.

This is the opposite of connecting a voltage source at the primary side, you are expected to have a load at the secondary side or leave it open, but you're not allowed to short the V2 side !