KlayBuRn

OK, heres my deal, i have 2 15' Rockford Fosgate Power DVC's, 500 RMS a voice coil

i have a bd1000a1, as i was told that would be the best...how would i hook this up? what Ohm would the amp be running? both subs are 4 ohms...what would be the total wattage(RMS) going to each coil?

and with the bd1000a1, what is the lowest OHM it is stable at? and those mono amps, do they just have the terminals on just mono or what?

cbpldc

Wow, that should sound pretty nice. Wire them in parallel (both pos go to pos in amp) and they will draw a 2 ohm load from the amp. I think that amp should be 2 ohm stable.

Chris Payne
'92 RS
'84 S/E

KlayBuRn

yea, so its 2 ohms when you have Single Voice Coil subs and you wire it with boths pos's to mono pos and both neg's to mono neg? thats 2 ohms right?

or...was you talking about running both the positive's from the DVC to the mono on the amp?

i guess i just really need an OHM lesson...
heh

FormulaJoe

An ohm is a mearsure of resistance in current. With two 4 ohm subs in this case, you can get a 2ohm load by wiring them in parallel, or you can get an 8 ohm load by wiring them in a series configuration. The equation for parallel is R1R2/(R1+R2) or (4x4)/4+4= 16/8=2ohms. This goes for any 2 resistances in parallel. If the two resistance values in ohms are equal (4ohm and 4ohm) then you simply cut it in half (4/2=2)

FOr a series configuration, you simply just add the two resitances to achieve the total resistance of the circuit.



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MelloYello

I was wondering if anyone knew the formula to determine the increace of watts due to reduced resistance (ie 300w @ 4Ohms = Xw @ 2 Ohms = Yw @ 1 Ohm). If there is a formula, then thats what I would like, simply because I dislike guessing or just assuming that halfing the ohm load would double the output (it doesn't, right?). Anyway, thanks for all your help

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[This message has been edited by MelloYello (edited March 23, 2001).]

Justins86bird

P(watts)=V^2/Z(impedance)

for example:
V=10 volts
Z=4 ohms
therfore:
10^2/4=25watts

if Z is half of 4=2
then:
10^2/2=50watts

This is ideal. There are other factors to take into account, but saying half the ohms will double the power output is somewhat true.

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