Originally Posted by
Tom Cornish
A long run of wire actually reduces the current drawn - it does not increase the likelihood of a tripped breaker.
Let's say the resistive equivalent of our tool is 10 ohms (around 1.3 HP or so). If we plug this tool into a power source that has an infinite current supplying ability without any voltage drop, we can calculate that this load will dissipate 1440 watts (P = V^2/R). The current flowing through the tool is 12A (I = V/R).
Let's say we now take this same tool and plug it into 100' of orange BORG 16ga extension cord. The resistance per foot of 16ga wire is .00473 ohms/ft (which may be optimistic considering how many splice points there are if you use 4 25' cords instead of 1 100' cord for example). Multiplying this by 200' (need to count the round trip distance of the wire) gives us a total resistance of .946 ohms.
This gets added to our tool load, so now our effective resistance of the tool plus wiring is 10.946 ohms. Using Ohm's law again, we find that our current has actually dropped from 12A without counting the wiring losses to 10.9A. Obviously a 10.9A load is less likely to trip a breaker than a 12A load.
The reason for using large wire and short runs is so that we aren't wasting power in the wiring and starving our loads. Again taking the 100' extension cord example, we can find that our tool which really wants 120V of input power for maximum performance will only be operating on 109V, consuming 1188 watts, due to the losses of the small wiring. We've lost 250 watts of potential work output due to the higher load impedance of the circuit and the resistive losses of the wiring.