impedance question
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!
Thanks in advance!
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
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
What does that mean?
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.
Thread
Thread Starter
Forum
Replies
Last Post
navtool.com
5G TLX Audio, Bluetooth, Electronics & Navigation
31
Nov 16, 2015 08:30 PM
DerrickW
3G TL Performance Parts & Modifications
9
Nov 15, 2015 05:52 PM



