alanh

Lets say you have two circuits, one with a 110V source and another with a 12V source. Lets assume DC to keep things simple. In order to have 20A flowing, you have: V/I=R. For the 110V circuit, 110V/20A=5.5 ohms of resistance. For the 12V circuit, 12V/20A=.6 ohms. So we have to have a different total resistance in each case to get 20A. The resistance will bascially consist of the load resistance plus the resistance of the wire. (Ignoring things like resistance of connections, the power source, etc.) If we have enough wire to get .1 ohm total resistance from the wire, then the 110V circuit load must be 5.4 ohms and the load of the 12V circuit must be .5 ohms to get the 20A. Of course, you can vary the length of wire and the resistance of the load in either case as long as the resistance of the wire plus the load resistance add up to a value that gives 20A.

One thing you do have to keep in mind with a higher voltage is that the insulation definitely needs to be rated to withstand that voltage without breaking down and leading to a short circuit.

Bumpin' Yota

hehehe as the poster above me alluded to, 20A @ 110v is different to the load being powered thna 20A @ 12v. At 110v you have, in theory, 2200 watts of power to play with. In the 12v scenario you have 240watts of power to play with.

To the wire it is still seeing a 20A load in either case... Votage is defined as the electrical potential from one area to another. So even a 32ga wire CAN be capable of being stable to 300volts but only a very very minimal amount of current...

runethechamp

But the loss of voltage in the wire is only a function of the current and the resistance of the wire, so if that's what you're worried about it doesn't matter. Ever wonder why electricity carried over long distances is high-voltage? The voltage drop, U = R*I, gets high with high currents, but much lower at low currents, so for a given power transport, P = U*I, they transform the voltage up and the current down.