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Originally Posted by Atomic
I think someone asked earlier but never got an answer...
In the calculations, if your using a Dual voice coil sub and its at 4ohms, but you have it bridged to 2 ohms, do you use the 2 or 4 in the calculations?
For instance my amp puts out 400wx2@2ohms bridged and im using a 400w rms sub (jl w6v2) which is dual 4 ohm.
So would my equation be V = sqrt (400x2) or V = sqrt(400x4)?
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I answered your PM.
You would want to use the actual ohm load that the sub will be wired to. so sqrt(400*2)
Just remember what i said in PM...if your amp isnt 2ohm stable I HIGHLY recomend that you do NOT run it at that low of an impedance.