fusible link questions

[Raymond C Drouillard]

>requires closer to 2.5 HP. Pushing the same current at higher voltage
>quickly oveheats the alternator due to the increase in these I^2 R
>losses, which increase at the rate of a power of two with voltage
>increase - double the voltage, 4 times the heat with the same current.

I beg to differ, but the I^2 R losses are independant of the output
voltage of the device

P = I^2 R (hense the name)

That is why power is transmitted using the "high tension" (High Voltage)
lines.

If you are stuffing current through a resister, the power dissapated by
that resister is dependant only on the current through the resister and
the resistance.

P = I^2 R


Let me explain...

Power is defined as voltage * current

P = IV

That is, the current through the resister * the voltage across the
resister.  The voltage across the resister can be calculated using Ohm's
law.

V = IR


Taking the origional P = IV equation and substituting IR for the V yields

P = I * (IV) = I^2 R



What I am proposing uses exactly the same princable that has always been
used by the electric suppliers to get power from the source to the
substations, and finally to your home.


One thing that neither of us mentioned is that some of the heat is
generated by hysteresis losses in the iron core.  That will somewhat
reduce the ultimate power that can be supplied by the system, but it'll
still be more than the unmodified system will provide.



Ray Drouillard, BSEE

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