|
|
|
|
|
by OJFord
2084 days ago
|
|
Why's that a problem? Surely a kettle (ahem) boils down to a simple resistor (the heating element) that gets good and hot, boiling the water. P=VI, so if you halve the voltage (wrt the UK) you need to sink twice the current. I=V/R, so if you need to double the current (and voltage is fixed) you need to halve the resistance. Is the problem that they're rare enough that they're all import kettles safe at 110 but designed for higher voltages? |
|