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Speaker wire consists of twisted copper strands, around AWG 22 thru 10. As good consumers know, one must consider resistance when wiring speakers. How do the wire terminations contribute to the circuit's resistance? Two questions:

1) Quantitatively, is any non-negligible contact resistance is added by the wire connections? Loose (low surface area) connections vs. tight (high surface area) connections?

2) If one were to cut the wire and leave only one tiny copper thread of length 1cm, and screw it down at the speaker post.... would this add significant resistance?

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1, An ideal screw connection with a soft metal like copper gives an almost perfect metal-metal bond. There is a potential problem of corrosion and pollution forming and some people cut the ends of the wires and redo them periodically. Or use gold plated connectors and tinned wires.

2, No the resistance of a thin wire in series with the speakers will add to the overall resistance. But Copper is about 18nOhm-m so a thin single strand of wire with a diameter of around 0.5mm has a cross section area around 0.2mm^2 . If the piece is 10mm long it will have a resistance of 0.8milli-Ohm, much less than the 4-8 ohm typical speakers

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  • $\begingroup$ why don't u use latex? $\endgroup$
    – ABC
    Commented May 19, 2013 at 2:04
  • $\begingroup$ @007 feel free to edit answers to improve them $\endgroup$
    – Martin Beckett
    Commented May 21, 2013 at 4:10

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