Understanding Transformer Output Current: Ohm's Law vs. Transformer Losses

[spidey]

so in transformer if we step up voltage,current is reduced and if we step down, current is increased based on p=vi...but based on ohm's law v=ir..so we step up,current should be increasing and and if we step down, current is decreasing right ..which formula is correct?

 

[Averagesupernova]

They are both correct. You are simply applying one of them wrong.

 

[berkeman]

so in transformer if we step up voltage,current is reduced and if we step down, current is increased based on p=vi...but based on ohm's law v=ir..so we step up,current should be increasing and and if we step down, current is decreasing right ..which formula is correct?

The open-circuit output voltage of a transformer depends on the turns ratio and the source voltage amplitude. The output current is zero in the open-circuit case.

Once you add a load resistance to the secondary of the transformer, then things change. You need to account for the source impedance and the current in the load resistance, before you can say what the current is. If the transformer were lossless, and the source impedance negligible, then yes, the secondary current is Np/Ns multiplied by the primary current (since there is no power loss in the ideal transformer). But when there are transformer losses, and when the source impedance of the driving voltage source is not negligible, then the output current is less than the ideal value.

V=IR will always hold for the secondary side, but the voltage that you are getting will be pulled down from the source's open-circuit voltage level, due to the drop across the source impedance and due to transformer losses (like voltage drop across the winding resistances).

Does that help?

http://en.wikipedia.org/wiki/Transformer

.

 

[spidey]

The open-circuit output voltage of a transformer depends on the turns ratio and the source voltage amplitude. The output current is zero in the open-circuit case.

Once you add a load resistance to the secondary of the transformer, then things change. You need to account for the source impedance and the current in the load resistance, before you can say what the current is. If the transformer were lossless, and the source impedance negligible, then yes, the secondary current is Np/Ns multiplied by the primary current (since there is no power loss in the ideal transformer). But when there are transformer losses, and when the source impedance of the driving voltage source is not negligible, then the output current is less than the ideal value.

V=IR will always hold for the secondary side, but the voltage that you are getting will be pulled down from the source's open-circuit voltage level, due to the drop across the source impedance and due to transformer losses (like voltage drop across the winding resistances).

Does that help?

http://en.wikipedia.org/wiki/Transformer

.

Thanks for the info...