I don't think you are looking at this the right way.
You can ignore the resistance side of it. Any electrical device is designed to work at a set power rating (watts, or Kilo-watts for example).
This figure is derived from the resultant of the supplied voltage, and the required amperage. (power = volts X Amps)
So if a laptop uses 240 watts to run, it will be supplied with 240 Volts, and will draw 1 amp. If you then drop the supply voltage to 120 V, the laptop will then have to draw 2A to run correctly. (this is why US mains power devices draw twice the current as the equivalent UK item)
Ausatco, There is a simple flaw in your logic. you stated that
"Say your lappy's designed to run on 15 volts, and say that when given 15 volts it draws 5 amps"
In this case, the total power would be 90 W.
"If you now apply 6 volts, then using V=IR, 6=I x 3. That is I, the current with a lower voltage of 6, will be 2 amps"
The total power used would now only be 12 W. ie. The laptop would now only be getting 1/6 of the power required to run it.
Most laptop power supplies are regulated, so the output current is limited at the top end.
voltage drops, current escalates. It's generally the cause of many a component burnout in my humble experience.......
That sums it up perfectly.
Hope this helps.
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