Voltage Drop Between Battery and Load
Let's say we have a 12 volt source and a 12 ohm load located 100 feet away from each other.

Let's further state that the wire connecting them has a resistance of 100 milliohms per foot and - since each of the wires is 100 feet long - each of the wires will have a resistance of 1 ohm.

That's a total of 2 ohms, and being in series with the load's resistance of 12 ohms, that's a total 14 ohms.

Then, from Ohm's law we have:

$$ I = \frac{E}{R} = \frac{12V}{14\Omega} = 0.857\ amperes $$

and, since:
$$ E = I R$$

the voltage drop in \$ \boldsymbol{each}\$ wire will be:

$$ E = I R = 0.857A \times 1 \Omega = 0.857 volt $$

Then, since there are two wires, each 100 feet long, connecting the supply to the load, the total drop in the wires will be twice that, or about 1.7 volts.