Generaly...

If you hava a battery it has an internal resistance. I think you want
to pull out the maximum power from this power source. In this case
your external resistor sould be equal with the internal resistance. If
you increase the value of the external resistance, the current will
drop on it ( altough the voltage will be higher ). If you decrease the
value of the external resistance, the voltage will drop on it (
altough the current will be higher ). So if you multiply the voltage
with current the result will be the actual power on the external
component.

For example:
U battery =3D 600V
R internal =3D 50 Ohm
R external =3D 50 Ohm

I =3D U / ( Ri + Re ) =3D 6 Amper
U ext =3D 300V
P ext =3D 300V * 6 Amp =3D 1800 Watt

------

U battery =3D 600V
R internal =3D 50 Ohm
R external =3D 30 Ohm

I =3D U / ( Ri + Re ) =3D 7.5 Amper
U ext =3D 600V * 30 / ( 50 + 30 ) =3D 225V
P ext =3D 225V * 7.5 Amp =3D 1687.5 Watt

------

U battery =3D 600V
R internal =3D 50 Ohm
R external =3D 100 Ohm

I =3D U / ( Ri + Re ) =3D 4 Amper
U ext =3D 600V * 100 / ( 50 + 100 ) =3D 400V
P ext =3D 400V * 4 Amp =3D 1600 Watt

If you have a solar panel, you can't know the exact value of the
internal resistance, because sunlight doesn't come countinuesly and
the clouds disturb the sunlight also. So you have to found the maximum
point where you can pull the maximum energy from the solar cell or
other power source because it depends on the illumination.

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