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Jul 17, 2003 | 01:09 PM

DaDude

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Could someone use a resister to change the ohms of a sub..like change and 8 ohm sub to a 4 ohm.

Jul 17, 2003 | 01:29 PM

Ettore Casagrande Jr.

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Yes, but it won't help.

IE: Your sub will always get the same power, but the resistor will be eating power too.

Think of it this way. At 8 ohms, your amp puts out (say) 100W

Well, you use a resistor and lower the resistance to 4 ohms ... the amp NOW puts out 200W. Well, 100W of that power goes to the sub and 100w goes to the resistor. So, now you're wasting power heating a resistor for no reason.

Jul 17, 2003 | 01:40 PM

Paul Niwranski

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^ AND your sub will actually be getting LESS power.

Who knows why?

Jul 17, 2003 | 04:26 PM

Big Tom

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Quote:

Originally posted by Dukk:
^ AND your sub will actually be getting LESS power.

Who knows why?

I'd wager a guess that the resistor is a constant load whereas the speaker changes impedances so the 8ohm sub when playing would have a higher impedance and more of the power would be discipated through the lower load of the resistor

Jul 20, 2003 | 07:06 AM

Chadxton

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I'm assuming that's the function of an Accumatch, so really it proves no point having one? Am I right or wrong?

Jul 20, 2003 | 08:40 AM

Shaughn Murley

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Quote:

Originally posted by Chadxton:
I'm assuming that's the function of an Accumatch, so really it proves no point having one? Am I right or wrong?

Accumatch is an impedence match device, it's uses inductance in a constant resistance to create a higher or lower impedence to your amp. another benefit of it is it creates a constant load to the amp, isolates the fluctuating impedence of the woofer, as the woofer is no longer physically attached to the amp.

Jul 20, 2003 | 09:41 AM

Chadxton

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Cool. I like learning stuff. So it's a constant electrical impedance load that the Accumatch adjusts to, right? Therefore it does benefit giving it extra power I assume.

[ July 20, 2003, 10:41 AM: Message edited by: Chadxton ]

Jul 20, 2003 | 11:00 AM

Mr. Marco

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actually Tom, wouldn't it be the other way around? since the sub's impedence tends to also be a functions of its frequency response ( i hope i have the terminology right) it may drop below 8 ohms at certain frequencies, which would normally draw more power, but with the the constant impedence of the resistor, we have a constant amount of power..
correct?