Of course not. 10000W is only 500V on a standard 20A constant current circuit. 2.5mm^2 conductors are sufficient for a 20A circuit.
UEC requirements state that the overvoltage protection shunt be set at no less than 600V, and that all wiring devices be rated to properly short 600V when loads are removed.
I don't understand your question about 'overcurrent protection'. In a constant current system, you don't get over-current. Instead open circuit faults cause excessive voltage; that is why you use overvoltage shunting devices; they safely short out the circuit in the event of an open, preventing dangerous overvoltages.
Seriously, Text, you have made the mistake of thinking that everyone on the net is from the same country and using the same electrical code as you. In addition, even when the code version is known, you would also need to describe the context in which you are applying it. What voltages are you talking about, what sort of load is being supplied, what sort of operational duty cycles, etc.
I doubt that you are working with constant current DC circuits, but such things are _possible_, fun to BS about, and sometimes similar to systems that actually do show up (runway lighting, eg). Without knowing the context of your question, it is almost meaningless.
Where are you located?
What code do you use?
What voltage will be used to supply the load?
What are the characteristics of the load?
[This message has been edited by winnie (edited 05-24-2004).]