Me bear of ickle brain and long ideas much bother me, but...
The heat gained by the water must surely depend upon the total energy gained, itself reflected by the power consumed. The current, as it were, will in any case vary inversely as the resistance, and directly with the voltage, no?
In effect, for the same resistance (the heating element in the kettle), twice the voltage (UKian), would pump energy into the water at twice the rate, therefore (in theory at least) bringing it to the boil in half the time.
As far as energy efficiency is concerned, the issue would be that of an enclosed element (the Brit kettle), versus the open flame/hob of a kettle placed on a heater. I'd suggest that the enclosed element, ceteris paribus, would result in less loss of heat into the nearby environment.
But it's been an age since I did electrical circuits in school, so I could be deadwrong.
cheer
the sunshine warrior
