Check out this wikipedia entry...it's got good diagrams and a good explanation. In brief, multiplying Volts time Amps gives you "apparent power" in kVA.
http://en.wikipedia.org/wiki/Apparent_power I get this type of issue all the time as an engineer when I get handed a catalog sheets with value for watts for a load like a big chiller and some %#&^ decides that is enough information for me to design a feeder.
Watts is the power that most folks know about; kVA is the only viable source of calculating the amperage actually needed to operate a load.
Dividing watts by volts works fine for heaters, toasters and other purely resistive loads, but most electrical loads have motors or other non-linear components that make things a bit trickier.
[This message has been edited by ghost307 (edited 09-15-2006).]
