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  #1  
Old 03-06-2011, 09:01 PM
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Quick and easy question

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When combining two cabinets I know the impedence is divided (two 8 ohm cabs go down to four) but what happens to the wattage.

For example, two 8 ohm cabs both rated at 200 watts. Are the wattages added, subtracted, multiplied, divided or any combination of those?

Also for the sake of this example I currently own a head that pumps out 450 watts at 4 ohms, so it should easily power two 8 ohm cabs rated at 200 watts (each, correct?

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  #2  
Old 03-06-2011, 09:08 PM
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Endorsing Artist: Knucklehead Strings
 
Join Date: Jul 2009
Location: concord, nc
Quote:
Originally Posted by justAlex13 View Post
When combining two cabinets I know the impedence is divided (two 8 ohm cabs go down to four) but what happens to the wattage.

For example, two 8 ohm cabs both rated at 200 watts. Are the wattages added, subtracted, multiplied, divided or any combination of those?

Also for the sake of this example I currently own a head that pumps out 450 watts at 4 ohms, so it should easily power two 8 ohm cabs rated at 200 watts (each, correct?

Thanks for putting up with me
Yes. My amp is 350 at 4ohms and 250 watts at 8ohms.
What I would worry about is what is the impedance of the cab and what at watts.

Do can use higher wattage cabs with a lower wattage amp but you might blow your speakers if the wattage of the amp is higher than your cab.

Also, are you running your cabs in serial or parallel configuration?
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  #3  
Old 03-06-2011, 09:15 PM
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sorry, i had to read your post again. the wattage stays the same on the cabs.

If you two 200watt cabs you still have two 200watt cabs. only the amount of impedance changes not the wattage.
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  #4  
Old 03-06-2011, 11:03 PM
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1. Impedance (measured in Ohms) -- for a number of cabinets wired in parallel (the usual situation), the impedance decreases as the number of cabinet increases. Total impedance = 1/[(1/R1) + (1/R2) + ...
This is easy when the impedances of the cabinets are the same, e.g. 2 x 8 Ohm cabinets = 1/[(1/8) + (1/8)] = 4 Ohms.

2. Power handling -- the "wattage" figure quoted on the cabinet's spec sheet and/or back panel is the maximum power it can handle before the voice coil suffers heat damage (i.e. "thermal rating") and bears little real relationship to the power the speaker will handle before it distorts/farts, etc, although this is typically about half what is cited. The power output from the amp will be divided between the cabinets according to the impedance they present to the amp's output; i.e. if both cabs have the same impedance (e.g. 8 Ohms), then each will receive half the amp's power output (i.e. in your case ~225 Watts to each cabinet). In practice, you can pretty much ignore the power rating stamped on the cabinet, and listen for the nasty sounds a speaker makes before it craps out, then back off the volume, and/or the bass end of your EQ. Bass frequencies require much more of a speaker in terms of coil/cone motion and therefore heat generation. Much of your "tone" comes from the middle of the frequency spectrum anyway, so not too much of a big deal to back off the low end a bit, and it will allow you to get much more volume from your speakers.

More information than you asked for, perhaps, but I hope helpful nonetheless.
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  #5  
Old 03-07-2011, 01:32 AM
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Join Date: Apr 2006
Location: Northern CA.
Quote:
Originally Posted by justAlex13 View Post
Also for the sake of this example I currently own a head that pumps out 450 watts at 4 ohms, so it should easily power two 8 ohm cabs rated at 200 watts (each, correct?

Thanks for putting up with me
Yes, your amp should push both cabs just fine.
A simple rule for for figuring ohm load is; Parallel divides, Two 4 ohm cabs = 2 ohm load; Two 8 ohm cabs = 4 ohm load; Two 16 ohm cabs = 8 ohm load
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