A Quick question ??

[Mark_ii]

If I have an 8 ohm cab and a 4 ohm cab can I run them together through a dual channel amp e.g. the Genz Benz Shuttle Max.

I know you can run 2 x 4 ohm loads on this amp but I'm not sure if it is ok to run 2 dissimilar loads on the 2 channels ? Any input and comments would be greatly appreciated.

Many thanks,

[Lozz196]

I`m sure I`ve read on here that if you do this, the load will be about 5.6 ohms, so in theory that should be ok.

However I`m not sure if I`d do it myself, as I don`t really understand the full techie stuff and would be worried about messing things up - but am sure someone will be along soon to provide accurate info.

[xgsjx]

Yes. According to the website, you can run a 4Ω cab on amp A & an 8Ω cab on amp B. Bear in mind that the 8Ω cab is gonna get 375 watts & the 4Ω 600w.

Edited to check: It is the S Max 12 you have?

Edited April 27, 2012 by xgsjx

[flyfisher]

[quote name='Lozz196' timestamp='1335517844' post='1632391']
I`m sure I`ve read on here that if you do this, the load will be about 5.6 ohms, so in theory that should be ok.
[/quote]

I'm not familiar with that amp but if it's two separate power amps (channels) then I can't see a problem as it's just like having two separate amps only they're in one box.

I don't understand the 5.6 ohms thing though because the two speakers would not interact with each other. One channel would be driving into 8 ohms and the other into 4 ohms, with the power implications that [b]xgsjx[/b] described above.

[xgsjx]

I think a 4Ω & 8Ω load combined is 3.2Ω.

It wouldn't apply to that amp though, unless you put both cabs on the one channel (such as all on amp A) & then you would risk damaging your amp.

[icastle]

[quote name='xgsjx' timestamp='1335527067' post='1632606']
I think a 4Ω & 8Ω load combined is 3.2Ω.
[/quote]

Closer

It's actually 2.67Ω

[leftybassman392]

[quote name='icastle' timestamp='1335547011' post='1633053']
Closer

It's actually 2.67Ω
[/quote]

Yup. For parallel connections you get 1/8+1/4 = 3/8 , then turn the result upside down to get 8/3 = ....

For future reference, the formula for 2 parallel connections is [i]1/R[sub]1 [/sub]+ 1/R[sub]2[/sub] = 1/R[sub]n[/sub][/i] where R[sub]1[/sub] and R[sub]2[/sub] are the the individual values and R[sub]n[/sub] is the resulting net impedance.

[Lozz196]

So it was 2.6 I was remembering - I knew there was a third digit but that one completely escaped me.

[leftybassman392]

[quote name='Lozz196' timestamp='1335579440' post='1633439']
So it was 2.6 I was remembering - I knew there was a third digit but that one completely escaped me.
[/quote]

To be totally and unnecessarily pedantic, it's actually 2.6 recurring. That means you write down as many 6's as you can be bothered with, then top it off with a 7.

It means (quite literally) almost nothing in the real world, but I thought I'd say it anyway.