Re: Speaker Attenuation


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Posted by Paul Tyzack on November 10, 2000 at 01:21:12:

In Reply to: Re: Speaker Attenuation posted by Mickster on November 07, 2000 at 16:51:41:

Deman.
The resultant resistance depends on whether the speakers are wired in parallel, or in series. Series meaning one resistance connected after another, just like a string of sausages, parallel is a more difficult picture to conjure up but it may be thought of as a row of toll gates next to each other having originated from a single road!

So Series calculation is Rt = R1+R2......+Rn, where Rn is the last resistance. so for your two speakers of 8 Ohms in series Rt=8+8, which =16Ohms.

For speakers in parrallel (1/Rt= 1/R1+1/R2....+1/Rn) where rn is the last resistance, so again for our two speakers of 8 ohms (1/Rt=1/8+1/8), 1/8 as decimal is 0.125, So 0.125+0.125 = 0.25.

So we get 1/Rt = 0.25, to find Rt you need to find the reciprocal of 0.25, which is not as horrific as it sounds, it is simply Rt = 1/0.25, which happens to equal 4, try it on your calculator.

Or for the purists out there:
Rt =(r1*R2)/(R1+R2) (8*8)/(8+8) which = 64/16 = 4!

There is one thought though.....all this works with DC circuits.....AC, now thats a whole new kettle of fish!!!!!


Hope that helps

Paul.



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