Why is the total length of wire in a circuit used to find the voltage drop between a battery and a load? Suppose the load is a distance l away from the battery. Wouldn't the voltage drop between the battery and the load be due to a length l of wire and not 2l? I understand that the current flows through a total distance of 2l of wire, but if the load is only a distance l away, why would it "see" a voltage drop due to 2l of wire? Is it possible to explain this with ohm's law and a simple circuit diagram?

  • 2
    \$\begingroup\$ What about the voltage drop in the return wire (one end is ground, the other end is almost a ground voltage). \$\endgroup\$
    – MarkU
    May 8, 2015 at 6:52

1 Answer 1


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.

  • \$\begingroup\$ Yes, that makes sense. So the problem is that there would be less than 12V to power the load due to the dissipative resistance in the wires. In the example above, how much less voltage is available to the load? Or, what supply voltage does the load "see"? \$\endgroup\$
    – Val
    May 8, 2015 at 8:51
  • \$\begingroup\$ Methinks that final 1.7 ohms should read 1.7 volts \$\endgroup\$ May 8, 2015 at 9:31
  • \$\begingroup\$ @JImDearden: Heh, heh, methinks you're right. :) \$\endgroup\$
    – EM Fields
    May 8, 2015 at 10:11
  • \$\begingroup\$ @Val: Since we started with 12 volts out of the supply and lost 1.7 volts in the wires, we'd have 10.3 volts left for the load. \$\endgroup\$
    – EM Fields
    May 8, 2015 at 10:15
  • \$\begingroup\$ Thanks EM, that makes sense. Would it also be accurate to say that the resistance in the wire reduces the current in the circuit and therefore reduces the potential difference across the 12 ohm load? (Since the potential difference across the load is proportional to the current in the circuit) And it is this reduced potential difference across the load that would cause a problem if a device required the full 12 volts to operate properly? \$\endgroup\$
    – Val
    May 9, 2015 at 8:16

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