Cable resistance vs amp loss

[jimmyhackers]

title says it all.....well not all but. this is purely my battery link cables and what power they sap.

i have a fair idea of how to calculate this... just wanted to make sure its valid (the internet seems to be overcomplicating the matter)

each length of wire between the batteries are around 10cm each and measure 0.1 ohms, i have 5 of them.

Vdrop = IxR ...... so my 5 - 20amps max x 0.1ohms = 0.5 to 2volts!!! just to my wires??

this is quite alarming.... and a fair waste of power i assume..... would swapping out to a wider lower ohm cable such as speaker cable improve efficiency?

 

[Deleted member 4366]

I think your calculation is a bit off. You should be using wire of about 14g, so 2 x 24" wires would be about 0.01 Ohms, which would give a drop of 0.2v at 20 amps or a loss of 4w. With 12g wire it would be about 2.5w.

 

[mfj197]

How are you measuring the impedance of your wires? It's difficult to do this accurately at such low impedances. If you are using a multimeter the impedance of the wires to the probes is much more than of the wire you are trying to measure.

Best to use an online calculator. d8veh has pointed me to a good one in the past here.

Michael

 

[jimmyhackers]

yeah......my bad guys.... multimeter reads minimum 8 ohms regardless.

ended up doing an online calculator http://www.bulkwire.com/wireresistance.asp

pretty cool as it gives all amp wire guages. So yeah.... at my length/AWG (around 14AWG) im loosing about 0.176 %. im slightly less worried/botherd about that kind aloss

thanks anyways chaps.