Brain Fart: Series vs Parallel

[Longhair]

I'm having a brain fart today about the differences between series and parallel.

Let's say I have a Mesa Carbine M9 head, Mesa 2x10 & 1x15 cabs - both 8 Ohms and rated for 600 watts.

Layout 1: Head to 2x10 to 1x15.
Layout 2: Head to 2x10 and Head to 1x15.

Is layout 1 series at 8 Ohms and layout 2 parallel at 4 Ohms?

[wcriley]

Both are parallel.

[Longhair]

My brain fart must be worse than I thought

*shakes fist at Windows SBS 2001

Back to my example layouts, the Ohms were correct but they are both parallel - not series and parallel?

I guess series and parallel only really matter when it comes to wiring multiple (2 or more) speakers up in the box.

[edbass]

It matters a bunch regardless of where it's used. If your example was paralled the load would be 4 ohms, but in series the load would be 16 ohms, not 8.

[Bass_Pounder]

Unless the cab or the head were specially wired for a series connection, you would need a special set of cables to run your cabs in series.

Both layouts you have are wired parallel.

I think you are confusing "daisey chaining" (layout #1) with series.

[JTE]

PHYSICAL connection arrangement means nothing. It all depends on how the speaker jacks are wired. And, almost every speaker jack on amps and cabinets is wired in parallel. In fact, the only exception I know if is the original Music Man amps (the HD-130 and the 65 series, as well as the short-lived successors the HD-150 and 75). Those AMPS had the speaker jacks wired in series, but the speaker cabinet jacks were all wired in parallel.

Therefore, because the jacks are wired in parallel, regardless of whether you daisy-chain from the amp to one cabinet and then to the second cabinet, or if you connect both cabinets to the head, the ELECTRICAL wiring is all parallel, and your amp will see a 4Ω load from the two cabinets.

BTW, if the two were really in series (as you thought with layout 1), it would be a total load of 16Ω, not 8Ω. Formula for series impedance is the sum of the impedances. Parallel impedance (assuming all loads are the same!) is the individual impedance divided by the number of loads. So, two 8Ω loads in series is 16Ω because 8 + 8 = 16. Two 8Ω loads in parallel is 4Ω because 8/2 = 4.

The formula for mixed loads in parallel is a bit more complicated, but not hard at all. It's
(R¹ * R²)/(R¹ + R²).

So, for two 8Ω loads the math becomes (8*8)/(8+8) = 64/16 = 4. For an 8Ω and a 4Ω mismatched load, the math becomes (8*4)/(8+4) = 32/12 = 2.67. Not really that hard to figure.

John

[Longhair]

Long day at the office...

Thanks for the explanations everybody

[BassmanPaul]

Originally Posted by Longhair

*shakes fist at Windows SBS 2001

As a Mac user I sympathize!

[Rick Auricchio]

Originally Posted by BassmanPaul

As a Mac user I sympathize!

But we don't gloat...