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>OK, I'll try once more and then I give up ;-) . > >E = I/R > >As you decrease the voltage (resistance constant) the current will also >decrease a proportionate amount. If it worked the way you suggest >(decreasing the supply voltage increases current) disconnecting a >battery (dropping the voltage to zero) would cause infinite current. >This is silly. Impediance and reactance only enter into it if the >voltage applied to power the card is ac. yes, i understand your problem. i don't mean to change this list into a physics lecture, so unless i have this wrong, let this be the end. given your E=energy, similar to V=energy dissipated per second, (which is the way the locals do it here), your ohms law must have served you great, my one works great too, it is E = I * R or V = I * R Voltage, is actually the amount of energy converted (ie, work done) to other forms per unit charge flowing through a conductor. conveniently, Current is a measure of charge flowing through a conductor per second. Lets say the amount of work needed to be done in a device is X. that means it needs a fixed number of charge to pass through it to do the amount of work it needs to do. If current is higher than usual, then more charge passes thru per unit time, and hence, less of this charge per unit time must be converted (used to do work) which is a direct implication that there is less potential drop (voltage drawn). Likewise, lets say this SO-DIMM requires 5volts to be supplied. that means that 5 Joules of work will be done for each charge the flows through the memory device, for it's intentional operating voltage. If the device is supplied 3.3volts, but was designed to run at 5volts, it will still need to dissipate 5 Joules of work in the same time as it would with a 5 volt supply. how can it do this? by drawing more current, since only 3.3 Joules of work would be done for each unit of charge passing through, the charge must be supplied at a higher rate. Current (I) = Charge per Second with no voltage applied, this means that no work is done for each unit of charge passing through the device. zero of course, always made special cases :) did i get it right this time? sam