fusible link questions

John Hess johnhess at cris.com
Sat Feb 6 22:04:48 GMT 1999


Actually, you haven't really considered that:

while P=IE (the basic formula)
E=IR, therefore substituting IR for E:
P=I*IR or I^2R
I=E/R, therefore substituting  E/R for I
P=E*E/R or E^2/R, etc.

To say that one is not related to the other is to refute Omm's Law. 
Note
that this is not electronics theory, it is _LAW_!


Raymond C Drouillard wrote:

> >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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