# another wattage question

Discussion in 'Amps and Cabs [BG]' started by rusmannx, Oct 18, 2004.

1. ### rusmannx

Jul 16, 2001
lets assume i have a 400watt, 4ohm amp.
if i use one 410 cab, and each speaker is 4 ohm, i would make two series pair, and put them in parallel with each other.
[(4+4)^-1]+[(4+4)^-1]^-1 = 4ohms.

my actual question is if the watts are divided up between the speakers (lets say 500watts rms per speaker). am i only sending my amps 400watts/4speakers = 100watts per speaker?

2. ### alexclaberCommercial User

Jun 19, 2001
Brighton, UK
Director - Barefaced Ltd
Yes.

Alex

3. ### rusmannx

Jul 16, 2001
ok.
so then if i toss on another 4ohm cab (like a 115) in parallel, would i be spending 400watts/2-4ohm cabs?
or would it be 400watts/5-4 ohm speakers?

the first equation would allow 50 watts to each 10" in the 410 cab, and the other 200 watts would go to the 15".

the second equation would equally divide 400watts/5speakers = 80watts per speaker.

i know that by putting the two 4ohm cabs in parallel i loose the total 4 ohm impedence, but i assume i could wire in a resistor in parallel with the cabs to correct this.

4. ### IvanMikePlayer Characters fear me...Supporting Member

Nov 10, 2002
Middletown CT, USA
here's some useful info

ohms

in your example of two 4 ohm cabinets each cabinet (not speaker) will get 1/2 of whatever you amp can put out at 2 ohms
that's assuming that your amp is capable of safley driving a 2 ohm load (which most are not)
wiring a resistor in parallel is a bad idea, biggerst reason being that your cabinets will only get 1/2 of what the amp is putting out, kind of defeating the purpose of having multiple cabinets in the 1st place. If it was a good idea, everyone would be doing this with their 4 ohm cabinets.