Wiring options for three dvc 2 ohm subs

[lnsomnia]

Hey guys! So this is a noobish question, but I would like to get an answer anyways.

I'm trying to figure out what my wiring options would be for three DVC 2 ohm subwoofers. Right now I have them at 1.3 ohms. What is the option below that? .33 ohms? I plan on wiring them this low on a SAZ-2500d for sound offs.

Also, if someone could link me to an explanation on how this works, that would be great. Maybe I just need to study up more on Ohms Law...

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[bbeljefe]

Finding the load on parallel resistances is pretty easy, actually. You simply divide 1 by each resistance, add the products together and then divide one by that product.

For example: three 4Ω impedances would be: 1÷4=.25 + 1÷4=.25 + 1÷4=.25 = .75 and then, 1÷.75=1.33.

For three 1Ω impedances: 1÷1=1... X3 = 3 and then 1÷3=.33.

Not that you would see this with loudspeakers but, 1Ω, 5Ω and 9Ω in parallel would be: 1÷1=1 + 1÷5=.2 + 1÷9=.111 = 1.311, then 1÷1.311=.763.

Hope that helps... I'm no math whiz so I'm sure it's not expressed properly.

 

[lnsomnia]

Finding the load on parallel resistances is pretty easy, actually. You simply divide 1 by each resistance, add the products together and then divide one by that product.
For example: three 4Ω impedances would be: 1÷4=.25 + 1÷4=.25 + 1÷4=.25 = .75 and then, 1÷.75=1.33.

For three 1Ω impedances: 1÷1=1... X3 = 3 and then 1÷3=.33.

Not that you would see this with loudspeakers but, 1Ω, 5Ω and 9Ω in parallel would be: 1÷1=1 + 1÷5=.2 + 1÷9=.111 = 1.311, then 1÷1.311=.763.

Hope that helps... I'm no math whiz so I'm sure it's not expressed properly.

That is actually the best response I could have ever asked for lol. Thanks. That was perfect!

 

[Chrispy 10+ year member]

Finding the load on parallel resistances is pretty easy, actually. You simply divide 1 by each resistance, add the products together and then divide one by that product.
For example: three 4Ω impedances would be: 1÷4=.25 + 1÷4=.25 + 1÷4=.25 = .75 and then, 1÷.75=1.33.

For three 1Ω impedances: 1÷1=1... X3 = 3 and then 1÷3=.33.

Not that you would see this with loudspeakers but, 1Ω, 5Ω and 9Ω in parallel would be: 1÷1=1 + 1÷5=.2 + 1÷9=.111 = 1.311, then 1÷1.311=.763.

Hope that helps... I'm no math whiz so I'm sure it's not expressed properly.

Isn't dividing the ohm by the amount of speakers/VCs a lot less complicated... 4ohm/3=1.33 or 1ohm/3=0.33

Now that wouldn't work with mismatched ohm loads but who's mismatching anyway.

 

[BoomTaco64]

Hey guys! So this is a noobish question, but I would like to get an answer anyways.
I'm trying to figure out what my wiring options would be for three DVC 2 ohm subwoofers. Right now I have them at 1.3 ohms. What is the option below that? .33 ohms? I plan on wiring them this low on a SAZ-2500d for sound offs.

Also, if someone could link me to an explanation on how this works, that would be great. Maybe I just need to study up more on Ohms Law...

use this page for your diagram.

Subwoofer Wiring Diagrams MTX Audio

input your number of subs and then the impedance of the subs