impedance question

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Old Feb 20, 2004 | 12:30 AM
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TWong1200's Avatar
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impedance question

For you audio/electronics gurus, I have a question. I'm pretty sure of the answer but am not positive. Here's my info. I have four, four ohm speakers. If I run them all in series to the same channel on an amp, the amp will still be only at four ohms, correct? What if I ran two of them in series, and the other in series as well, but then ran each pair with one another in parallel to the same channel on the amp, would that equal two ohms? I'm pretty positive but just want to make sure.

Thanks in advance!
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Old Feb 20, 2004 | 12:36 AM
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Lung Fu Mo Shi's Avatar
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Series = R + R + R + R = 4R

Parallel = 1 / (1/R + 1/R + 1/R + 1/R) = R/4

The other method as you described = 1/ (1/ (R + R) + 1/ (R + R)) = R
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Old Feb 20, 2004 | 08:28 PM
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From: San Gabriel Valley, CA
Originally posted by Lung Fu Mo Shi
Series = R + R + R + R = 4R

Parallel = 1 / (1/R + 1/R + 1/R + 1/R) = R/4

The other method as you described = 1/ (1/ (R + R) + 1/ (R + R)) = R
Thanks for the reply but What does that mean?
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Old Feb 20, 2004 | 09:58 PM
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Basically, if you put four 4 Ohm speakers in series on one channel, you'll have 16 Ohms (4*R). If you put them all in parrallel, you'd have 1 Ohms (R/4). If you did it the way you mentioned, you'd still end up with 4 Ohms.
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