Voltage and current for travel kettle I used a 500 watt travel kettle in France last year and it tripped the electrics in my hotel room so I stopped using it. I assumed the current was too high, even though its wattage rating wasn't that high.
However I realised the voltage switch on the kettle was set to 120V and not 240V (the voltage in France).
Would the travel kettle have used more current as it was set to 120V? If it didn't have a voltage switch, it'd surely use less current at a higher voltage, but I'm wondering if the voltage switch being on the wrong voltage caused an issue here?
 A: I believe this is a good question and I think the key is understanding what the voltage switch on the kettle does. Realize that current in the kettle is not bound to the power of the kettle by $P=VI$ rather the actual voltage and current is related by (approximate) Ohm's Law: $$V=IR$$
Notice that I will show show the voltage switch changes the resistance of the kettle circuits. Think in your situation when you put the voltage knob at $120V$ then normally for the kettle to work at $500W$ power it should be able to have a current of $500/120=4.17A$. This would mean that the resistance of the kettle should be $$R=V/I=120/4.17=28.7\Omega$$.
Hence now in France the voltage is actually now $240V$ and at the $120V$ setting your resistance is still $28.7\Omega$. But the kettle is only built to withstand $4.17 A$ at this setting. But notice due to this resistance at $240V$ you get a current of $$I=V/R=240/28.7=8.36 A$$.
Thus you see that it is this reason why the circuit malfunctioned and produced high current as the current is not bound to the power rating but the resistance of the kettle circuit which is changed by the voltage knob.
Notice that your kettle would indeed work in France if the voltage knob was at $240V$ as then the resistance rating would be to ensure that $P=VI$ or $500=240*I$ or $I=2.083A$ is maintained. Then the resistance needed would effectively increase. $$R=V/I=240/2.083=115\Omega$$. Hence at $240V$ knob setting(which really means a resistance of $115\Omega$) would cause the current to be low enough for the kettle to work and not malfunction. 
A: Suppose that at a voltage of $2V$ the kettle element consists of two parts in series each of resistance $R$ ie a total resistance $2R$.  
The power dissipated in the element is $\dfrac{(2V)^2}{2R}= \dfrac{2V^2}{R}$ 

If the element is to dissipate the same power when the voltage is halved to $V$ then a switching arrangement can used to now have the two parts of the element in parallel and thus having an effective resistance of $\dfrac R2$  
The power dissipated in the element at this lower voltage is $\dfrac{V^2}{\left (\frac R2 \right)} = \dfrac{2V^2}{R}$ , the same as that at the higher voltage setting.  

If whilst in the lower voltage $V$ setting the kettle is connected to a higher voltage $2V$ then the power dissipated in the element will be $\dfrac{(2V)^2}{\left (\frac R2 \right)}= \dfrac{8V^2}{R}$ which is four times the power rating.  
If this happened then the current would be four times higher than it should be and this would trip the circuit breaker. 
