Just a thought, I think you will find as resistence increases, current decreases, Mr Ohm might say E=I x R, and P = V x I.
Yes, but the problem is that power dissipated is dependent on resistance and current.
It is true that P=V*I
However, V = I*R
so, P= I(squared)* R
Therefore, in order to get the same power dissipation you have to reduce the current by a factor of 4 if you increase the resistance by a factor of 2.
The only way to do that is to reduce the voltage.
Since the voltage for any serious electrical supply can be assumed to remain constant then current will also remain purely a factor of resistance from Ohms' law.
You will find that a higher resistance will increase the power dissipated through the fault given a constant voltage.
