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another cabinet and ohms question. Genz Benz and Avatar related.


justinbass

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No, you want an 8 ohm 2x10 to go along with a 4 ohm 4x10.

 

(Math review: 1/3 of 6 = 2, 2/3 of 6 = 4)

 

You want to send twice the wattage to the 4x10 as you are to the 2x10, since it has twice the speakers. That means a 4 ohm 4x10 and an 8 ohm 2x10.

 

If you send both cabinets an equal amount (or 50% of the power), then the speakers in the 2x10 are getting twice the wattage as those in the 4x10.

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No, you want an 8 ohm 2x10 to go along with a 4 ohm 4x10.


(Math review: 1/3 of 6 = 2, 2/3 of 6 = 4)


You want to send twice the wattage to the 4x10 as you are to the 2x10, since it has twice the speakers. That means a 4 ohm 4x10 and an 8 ohm 2x10.


If you send both cabinets an equal amount (or 50% of the power), then the speakers in the 2x10 are getting twice the wattage as those in the 4x10.

 

 

I don't think that's really how it works, though...

C7

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No, you want an 8 ohm 2x10 to go along with a 4 ohm 4x10.


(Math review: 1/3 of 6 = 2, 2/3 of 6 = 4)


You want to send twice the wattage to the 4x10 as you are to the 2x10, since it has twice the speakers. That means a 4 ohm 4x10 and an 8 ohm 2x10.


If you send both cabinets an equal amount (or 50% of the power), then the speakers in the 2x10 are getting twice the wattage as those in the 4x10.

 

This is correct, and my suggestion as well.

 

(step to the head of the class and collect your gold star MR. illidian) ;):thu:

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No, you want an 8 ohm 2x10 to go along with a 4 ohm 4x10.


(Math review: 1/3 of 6 = 2, 2/3 of 6 = 4)


You want to send twice the wattage to the 4x10 as you are to the 2x10, since it has twice the speakers. That means a 4 ohm 4x10 and an 8 ohm 2x10.


If you send both cabinets an equal amount (or 50% of the power), then the speakers in the 2x10 are getting twice the wattage as those in the 4x10.

 

 

I'll go along with this as well. You'll end up with a 2.6 ohm load with approximately 2/3 of the power going to the 410 and 1/3 to the 210.

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This is correct, and my suggestion as well.


(step to the head of the class and collect your gold star MR. illidian)
;):thu:

 

Agedhorse to the rescue. Logically, it made sense, but I've learned that very few things in the pro audio world work out the way I think they are supposed to, so I came here knowing someone would be able to shed some light. Thanks! :thu:

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For an application like this, even power sharing between drivers is probably the most important parameter. Say each driver is good for 100 watts, and as suggested each receives 100 watts from the amp at 2.66 ohms, all is "good". ;) (that smiley is for those of you who follow the pro audio fourm)

 

Now say we take the same power amp and choose a 4 ohm 210 cabinet, the power delivered by the amp will be greater due to the 2 ohm load (say roughly 800 rather than 600 watts total) and now it's divided up with 1/2 going to the 210 and the other 1/2 going to the 410. (ohms law) So the 210 will be receiving 400 watts (200 watts per driver) and the 210 will be receiving 400 watts (100 watts per driver) and guess what... 210 is overpowered and is the likely to fail under hard use (or accident/lack of good judgement).

 

The drawback is that the 8 ohm 210 will be about 3dB quieter than the 410, but it also has 3dB less power handling so that's expected. It will contribute to the overall sound and will also improve the low frequency extension a little bit.

 

Hope this helps.

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