## 37528Re: FW: [free_energy] transformer question wrt variable cross-sectional areas

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• Aug 22, 2013
The mistake came from considering separately some elements of the magnetic circuit instead of taking it as a whole, while the formulae apply only to a closed circuit (by definition, a circuit is closed). They don't apply to parts of a circuit.
Phi = n*I/R where R is the reluctance of the magnetic circuit which is one and alone.

Physicists say "magnetic fields never diverge or converge. they go in closed curves". This means that the flux is conserved. This can be viewed here as a tightening of the magnetic field lines in the section of lesser area, while the number of lines remain the same.
No free energy.

--- In free_energy@yahoogroups.com, Keith Krasnansky <keithkras@...> wrote:
>
> No EEs, power supply designers, or physicists want to comment on this?
> I'm stuck as to why it would not work.
> Keith

> To: free_energy@yahoogroups.com
> From: keithkras@...
> Date: Fri, 16 Aug 2013 15:43:51 -0400
> Subject: [free_energy] transformer question wrt variable cross-sectional areas
>
>
>
>

> Consider 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.)
>
> Thoughts?
>
> Thanks,
> Keith
>
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