# is the resistance in ten meters of copper wire too high to use in a circuit powered by AA batteries?

I have a wireless security system at home, and the wireless node is too far from the receiver. I was thinking that I could splice about 10 extra meters of wire into the existing line to move the node close enough to the receiver.

From my university days, I remember the I = V / R equation, meaning that with a fixed voltage (from the two AA batteries), if I increase the resistance, the current that gets through will be smaller - possibly not enough to power the wireless node.

How do I determine the resistance of the wire? Is there a standard calculation I can use?

I haven't picked up the wire yet, so I can use another type of wire and a small gauge if that would help.

-
University days? I learned this in my first year of secondary school. –  stevenvh Jun 19 '12 at 6:22

10 meters is 32.8 feet

so 10 meters of 24 AWG wire is 990 milliohms. But you actually have twice that, because the current goes from the battery and back.

So 2 ohms for wire resistance.

AA batteries have a series resistance of about 0.5 ohms new, and more as they age, so the circuit is probably happy with a little series resistance.

If the wireless transmitter draws a peak of 100 mA (a guess) the voltage will dip an extra 0.2 V because of the wiring resistance.

I think it would work better with 18 AWG wire, which has about 1/4th the resistance of 24 AWG.

-

Resistance (R) is equal to the resistivity of the material (ρ, Greek rho), divided by the area (A), times the length (). The resistivity of copper is 1.68 × 10−8 Ω·m

I'd use a table to just look up what the resistance is, here's one I use occasionally.

To know if this is too much, one would need to know how much current you intend to draw.

-
quick elaboration - the current draw of your circuit will dictate how much voltage drops over the 10 meters of wire (V = iR). You probably have some minimum operating voltage for your circuit... –  vicatcu Dec 27 '10 at 18:46
@vicatu - Good point, the minimum operating voltage is what we're really looking for. –  Kevin Vermeer Dec 27 '10 at 21:21

You have to make that it's not too high. Work backwards from the receiver's specifications and see what cable the system can afford. You need to know the minimum operating voltage of the receiver, as well as its maximum current.

The AA batteries are probably not a good idea. Rechargeables like NiMH have a low 1.2 V rating, so you don't want to get that even lower. Alkaline cell energy is around a 1000 times more expensive than electricity from the wall. Use a 3 V wall wart instead, and you won't have to worry about sagging voltage.

Say the receiver's minimum operating voltage is 2.7 V, and that it uses maximum 100 mA. Then this 100 mA may drop maximum 300 mV (provided the wall wart effectively outputs 3.0 V). Ohm's Law says a maximum wire resistance of 3 $\Omega$ is allowed.

Copper has a resistivity of 16.8 m$\Omega$ $\cdot$ mm$^2$ / m, and we have 20 m to and fro, that's 336 m$\Omega$ $\cdot$ mm$^2$, then a cross section of (336 m$\Omega$ / 3 $\Omega$) $\cdot$ mm$^2$ = 0.112 mm$^2$ is the minimum required. That's bit less than 0.4 mm diameter. That's all non-USAers need to know. USAmericans still have to lookup in a table what AWG value this is, or calculate it (involves a couple of logarithms).

Everything in one formula:

$d = 2 \sqrt{\dfrac{2 L \cdot I \cdot 16.8 m\Omega \text{ } mm^2 / m}{\pi \Delta V}}$

where

$d$ = minimum wire diameter
$L$ = cable length
$I$ = maximum current
$\Delta V$ = maximum allowed voltage drop.

-