B-wire - four cables or jumpers?

When you use one set of speaker leads all the current passes thru the same wires. When you split the leads for high and low frequencies, each carries only the current it uses because the high and low frequency drivers have different appetites for current. It is believed that separating the flows preserves the purity of the signal.

AS an aside-- Von S. uses Analysis+ Oval9 internal wireing. Does that tell you what brand you should be using? AND it should not be jumpers !!

Yes Rockvergo, I agree all the current passes down the cable. but the high and low frequencies are split at the crossover, not at the amp or the speaker binding posts. I know some say a low frequency signal will follow the line of least resistance, and that this is an arguement for mixing stranded and solid core conductors in the same cable. Hence it would, if this is really true, make sense that the second pair of a bi-wired set (the woofer pair)should be of a larger cross-sectional area (and therefore have greater current carrying capacity?) than the pair going to the mid/treble section. However, many very high quality cables do not do this, eg Nordost for one. And in any case, this could also be applied to the jumper leads.

So my basic puzzlement remains: if the current arriving at the speaker binding post is the same as that leaving the amp - which it is - then apart from the length, aren't the jumper leads simply acting as a second pair of cables in a bi-wired system?

I will check out the sites suggested by Bomarc and Audio_Buff, to see if these articles throw any light on this.

Using jumpers is not Bi-wiring. The choice would be between four cables or a pair of single internally bi-wire cable. Between those last two comes down to cost. Four cables will cost more.

You are best off experimenting with this yourself. If you buy used cables, you can resell your left-overs after you decided what you like.

Ok, let's add this to the puzzle: electrical devices draw current when processing electrical energy. Assuming an inexhaustible supply and an amplitude of signal to process, the load dictates the energy flow. No load, no flow. This explains why your electric bill doesn't skyrocket when you have the lights off.

More to the point, do we believe that if the crossover channels all the current, that the hypothetically-unused current gets soaked up and converted to heat, or maybe sent to ground and otherwise wasted by the crossover? Sure the crossover is a load in its own right. However, it's is my understanding that although the crossover processes the whole signal, it only allows the energy/current to pass to the device connected to it. The "rest" of the current never gets a chance to exist because there's nothing to draw it.