I have 12 volt and 135 ampere battery. Now I want to connect a device approximately 300 feet away from the battery.

When I connect it at this distance, my device does not work. When I connect the same device to the same battery at a shorter distance, the device does work.

What can I do?

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    \$\begingroup\$ What is the gage of the wires you are using? How much current does the load actually draw? and what is the minimum voltage needed at the load? \$\endgroup\$ – The Photon Aug 10 '13 at 15:58
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    \$\begingroup\$ Your question is vague. What device? How much power does it draw? How is it connected 30 feet away? Do you mean 135 Amp-hour capacity battery? \$\endgroup\$ – Bobbi Bennett Aug 10 '13 at 16:23
  • \$\begingroup\$ I don't know what is the gage of conductor but I am using conductor like 8 pair internet cable wire. \$\endgroup\$ – Zakiuddin Aug 10 '13 at 19:28
  • \$\begingroup\$ And the power used by the device? What is that? I down-voted because both bits of information are needed to describe what is going on. \$\endgroup\$ – Bobbi Bennett Aug 11 '13 at 0:00
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    \$\begingroup\$ @TheTerribleSwiftTomato as long as they stay away from any elephants, no complaints from me. \$\endgroup\$ – Passerby Aug 11 '13 at 16:15

Any real conductor has an effective resistance per unit length. The voltage drop over that length of conductor, and the power dissipated in that conductor, is dependent on the amount of current you draw through that conductor. For a current draw of I, and a Resistance per Meter R you will experience:

(I^2 * R) watts per meter of power lost as heat in the conductor


(I * R) voltage drop per meter of conductor

If your endpoints cannot tolerate the worst case voltage drop, or if your power budget does not account for the losses in the cable, you can expect them not to function as intended.

This PDF is a reference I have used for stranded wire.


The battery is too far away. The voltage drop seems to big. You should have a bigger voltage on the battery side (DC-DC boos converter) or thicker cables. Voltage drop is affected by the current flowing through the cables, length of the cables and cable thickens. You can look for more info here http://www.buildmyowncabin.com/electrical/copper-wire-resistance.html . It even has a calculator, so you can just put your values in and get the voltage drop.


As others have mentioned we need more info to be exact, but assuming your "internet cable" means something like CAT5, and assuming you are only using one pair, then you probably have a serious voltage drop at 900mA.

A rough calculation based on a typical CAT5 cable with 24AWG conductors, using the typical loop resistance of 16Ω for 100m. We will estimate the routers load as 12Ω for a max of 1A:

300ft = 91.44m, so loop resistance = 16Ω * (91.44/100) = 14.63Ω

Total circuit resistance = 14.63Ω + 12Ω = 26.63Ω

Circuit current = 12V / 26.63Ω = 451mA

Cable voltage drop = 0.451A * 14.63Ω = 6.59V

Voltage available at router = 0.451A / 12Ω = 5.4V

So, although this is just a rough estimate (ignoring temperature effects, inductance/capacitance. Also we don't know what kind of regulation circuitry the router uses or it's input range) we can see that there is nowhere near enough power getting through to the router.

I would use at least 18AWG or thicker cable, preferably with some shielding and e.g. a ferrite bead. You can try using multiple conductors to lower the resistance. Also placing a largish capacitor (e.g. > 100uF) across the wires at the router end may help a little.


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