impedance question
#1
Instructor
Thread Starter
Join Date: Jul 2003
Location: San Gabriel Valley, CA
Age: 48
Posts: 193
Likes: 0
Received 0 Likes
on
0 Posts
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!
#3
Instructor
Thread Starter
Join Date: Jul 2003
Location: San Gabriel Valley, CA
Age: 48
Posts: 193
Likes: 0
Received 0 Likes
on
0 Posts
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
#4
Registered AssHat
Join Date: Feb 2003
Location: Portland, OR
Age: 46
Posts: 3,777
Likes: 0
Received 0 Likes
on
0 Posts
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
11-16-2015 08:30 PM
DerrickW
3G TL Performance Parts & Modifications
9
11-15-2015 05:52 PM