I was wondering if there was a way to figure out the current loss for various gage wires and lengths. I looked online bulk of what i see is voltage loss not amperage. Maybe ohms low or something on the source and load end measuring the drop in voltage you could find the lost amperage. Not exactly how would do that, or if there is a simpler way.

I redid a cable for our backup generator from a 4 FT 10AWG wire to a 12 ft 6 AWG wire. Generator puts out 30 A 120,120, 240 and gnd so 4 wires.

I think i'm fine with the 6 AWG cable, but was still curious what difference it could make. Maybe i'm totally wrong, but i know with length current also drops, i've seen it my dad ran 100 ft of 6AWG 4C cable to the garage 100 FT away while it works even with 60A breaker at house he couldn't hardly use his 50A stick welder, and i guess generator putting out 220 @30A couldn't run it either. So i think that's evidence amperage also drops with length.

(I know this isn't really an AC wiring job forum, but i'm more interested in the theory, and how to perform it. And if somebody wants to explain they can use those figures.)

 

Well, whiz, current doesn't work that way. With smaller gauge wire there is less electron flow because there is more resistance, but no current is "lost" in the wire. Power is lost in the wire, and this can be calculated from the resistance of the wire (usually given in ohms per foot), the length of the wire, and Ohm's law. Your new generator cables are a little bit longer than the originals but have significantly less resistance, so you will see less voltage drop across them at high loads, and they will not get as warm.

ak

 

Hello there,

Here is a little formula for calculating the current with the wire in place:
I2=(pi*92^((72-2*N)/39)*V)/(pi*92^((72-2*N)/39)*RL+(816*L)/625)
where
N is the wire gauge (AWG),
V is the source voltage in volts,
RL is the load resistance in Ohms,
L is the length of the wire in feet.

This formula gives you the current with the wire in place, so to calculate the current that was 'lost' due to the wire simply subtract this current from the current you would get with no wire:
I=V/RL-I2

So for example if you calculate 9.9 amps with V=100 and RL=10, this would give you 10-9.9=0.1 amps.

This is useful when you want to estimate the power lost in the LOAD, but remember this assumes a resistive load. This is also for an ambient temperature of 20 degrees C and assumes NO temperature rise in the wire to keep it simpler.

For a more widely applicable formula we would include ambient temperature, and also allow reactive loads. It's up to you if you want something more advanced like that. I thought i would show you this formula first and see if you think you would like to use something like this.

Note it's a little more complicated because we have to calculate various things about the wire. It can be simplified quite a bit if you dont mind using the wire diameter instead of the wire number gauge (AWG).

 

Well, whiz, current doesn't work that way. With smaller gauge wire there is less electron flow because there is more resistance, but no current is "lost" in the wire. .............

Don't understand what you mean about "less electron flow". For a given current the electron flow is the same, regardless of the wire resistance or size since current is defined as the number of electrons/second. The electrons just have more collisions going through a smaller wire which causes an increase in the voltage drop (more resistance).