37518transformer question wrt variable cross-sectional areas
- Aug 16, 2013Consider a U-I transformer where the primary and secondary coils are on separate 'legs' of the transformer.
The core cross-sectional area (2A) at the primary coil is double the core cross-sectional area (1A) at the secondary coil.
The number of turns of the primary is equal to the number of turns of the secondary; i.e. turns ratio is 1:1.
The primary coil creates a changing flux through its 2A section of the core that creates a reverse EMF equal to the AC voltage applied to the primary.
This same changing flux is 'funneled' down to the 1A secondary section to create the same voltage on the secondary.
Voltage on secondary only depends upon how much changing flux is passing through turns on secondary. It does not depend upon the cross-sectional
area except in the sense that you cannot surpass the core's saturation limit. We are assuming that the transformer is designed such that saturation
does not occur at the applied voltages and currents.
So, since flux = BA, and B = unI, we can double the amount of flux through the primary section of the transformer core simply by doubling the cross-sectional area of the core while maintaining a constant current, I.
So, when a load is placed on the secondary, causing a secondary current, I(s), causing a reverse flux, the primary only has to use half as much current, I(s)/2, to cancel out the reverse flux and maintain its original back EMF.
Every Ampere that is applied to the primary generates twice as much flux as each Ampere through the secondary simply because the cross-sectional area of the primary is twice that of the secondary.
So, if the voltages are the same, and the primary current is half that of the secondary current, doesn't that mean the input power is half that of the output power?
For the sake of simplicity, I have ignored all of the losses associated with a real transformer (hysteresis, eddy currents, resistive, etc.)
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