Let's say I have LM386 or some other audio amplifier. What determines the length, resistance, capacitance and inductance (if L C parameters matter) of the conductor that carries the output of the audio amplifier to the speaker?
The primary limiting factor will be the series resistance of the wire.
Let us work with a speaker of 8 ohms.
If the connecting wire were 18 AWG it has a resistance of 0.021 ohms per meter. If you were to connect the speaker with 50 meters distance from the LM386 (for a total wire path of 100 meters) the wire resistance would be (0.021 * 100) = 2.1 ohms. In such instance over 20% of the signal energy from the amplifier would be spent in the cable as heat and the rest going to the 8 ohm speaker.
(2.1 / (2.1 + 8) = 20.7%
You can use this example to evaluate other wire sizes and understand the effects.
Note: For audio frequencies do not fall victim to the marketing efforts of companies like Monster Cables that would have you believe that you need 8 AWG copper cable with gold plated connectors to connect up an 8 ohm speaker. Conventional low cost 16AWG lamp cord is almost always more than adequate to connect speakers at a reasonable working / listening distance.
If you really want to be serious about the subject, EAW do a calculator and Crest have a chart. I haven´t been able to find the calculator on the website, but it´s here, top of the list.
And the Crest table:
And Litz cable actually does work, especially for reducing losses at the higher audio frequencies, but there are other (cheaper) alternatives as well. They´re called graphic equalisers! :)