question about wiring multi-speaker arrays (like Phil Jones bass amps, etc.)

[mexicanyella]

Hi all, wanted to ask something that's been puzzling me, and which my reviewing of series/parallel wiring info has not answered.

If I were to wire up an array of nine 4-ohm speakers to present a 4-ohm load to an amp, I could

a) wire groups of three each in series, giving me three 12-ohm groups, and then connect those three groups in parallel, bringing me back to 4 ohms...

OR

b) wire groups of three each in parallel, giving me three 1.33-ohm groups, and then connect these three groups together in series, also bringing me back to 4 ohms total.

The first method seems more intuitive to me, but I don't know why. Is there an advantage or disadvantage about either of these two connection schemes? Is there a problem with one or the other I'm unwaware of? Any direction here much appreciated...

Mexicanyellla

[Enzo]

Please explain your watts conclusion.

His two examples are already series/parallel, but the main point is that they add up to 4 ohms either way.  All the amp ever sees is 4 ohms, it never knows how many drivers there are.  And the drivers all share power equally.  So 90 watts from an amp would result in 10 watts in each speaker.

[J M Fahey]

c=a

[mexicanyella]

Enzo, what were you referring to about watts? Did I miss a post before it got edited or something?

I don't really understand how back EMF works yet, and don't know if that's a factor to consider here, or even if it would be different in either of my two series/parallel scenarios. The only reason I can think of to choose one approach over the other is that (I think?) if one of the drivers' voice coils were to blow and go "open," in scenario A that would take out one entire series-wired 3-speaker group, leaving me with 6 functioning drivers and a total impedance of 6 ohms.

If I were to blow a speaker and go "open" in scenario B, I'd still have 8 functioning drivers, but one of the parallel-wired 3-speaker groups would become a two-speaker group. If I've figured this right, I'd have two 1.33-ohm parallel groups and 2-ohm group, all in series, for a total one-speaker-blown impedance of 3.54 ohms.

If I have this figured right, it seems like scenario B might be a little safer if one expected to be driving the speakers pretty hard as long as the amp driving the speakers could handle a 4-ohm load, because losing one driver wouldn't also take two other drivers out of the circuit...and the odds of smoking multiple drivers together in scenario B seem more remote...right?

[Crystallas]

You're good. I had a brain fart, then made one of those before-coffee responses in the wee hours and was corrected. So I deleted my post, because it would have done nothing to help. I'll stay out of it. Good luck, hope your cab turns out the way you want it to.