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If I have 6 speakers, each at 4ohms. If I wire three in series (x2) and do parallel, what is my end ohm to the amp? Im not sure if the calculation goes something like this



4ohm + 4ohm + 4ohm = 12ohm



4ohm + 4ohm + 4ohm = 12ohm


Thats 24ohms, and when in parallel i divide 24/6 = 4ohm


Or do I divide just the top number (12) by 2 = 6ohm


Or something else Im not doing correctly


thanks guys!
 

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If I understood your question correctly, you would have two 12 ohm loads and could run 12 ohm per channel on a stereo amp, or wire the groups in parallel for a single 6 ohm load on a mono amp.
 

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Discussion Starter #3
ok thanks. so when running things in parallel is the formula to divide by 2? so like the example above, i had 12ohm on 3 speakers in series, 12ohm on the other 3, then running in parallel i divide 12 by 2 = 6ohm? or is this not always the case? thanks
 

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Quote:
Originally Posted by AV Doogie  /t/1418273/quick-question-on-ohms-for-speakers#post_22183046


1/4 + 1/4 + 1/4 = 1/x = 4/3 Ohms
Apparently you are the recipient of a US public school education. I know you understand what you are doing, but the mathematical notation is just plain wrong.
 

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Quote:
Originally Posted by Roachforlife  /t/1418273/quick-question-on-ohms-for-speakers#post_22182611


...when running things in parallel is the formula to divide by 2?
No, you add the reciprocals of the impedances of the speakers, then take the reciprocal of that, as AV Doogie attempted to show.


Assume three 4 ohm speakers. The reciprocal of 4 is ¼. So, add ¼ + ¼ + ¼ and you get ¾. The reciprocal of ¾ is 1⅓. So, running three 4 ohm speakers in parallel is like having a 1⅓ ohm speaker.
 

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Oh, I understand what you were getting at. But there is a problem with your mathematical notation.


Your equation says:


1/4 + 1/4 + 1/4 = 1/x = 4/3 Ohms


By the transitive property of equality then:


1/4 + 1/4 + 1/4 = 4/3 Ohms


Do the arithmetic on the left side and you get:


3/4 = 4/3 Ohms


Which is obviously not true.


It is a common kind of mistake made by a lot of students in the USA because of the attrocious state of math education in the primary grades in most places.


I think what you wanted to say was:


1/4 + 1/4 + 1/4 = 1/x


x = 4/3 Ohms


You could also write it:


x = 1/(1/4 + 1/4 + 1/4) = 4/3
 

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Lest we confuse the open, he never mentioned running all the speakers parallel or even in two groups of parallel. The 4/3 ohms configuration isn't what he asked about, although it is another option he could consider.
 

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In any case, the formulas are the same.


Assume we want to parallel two sets of three 4 ohm speakers in series. Three 4 ohm speakers in series gives us 12 ohms. The impedance when two sets are paralleled is:


1/(1/12 + 1/12) = 6


That happens to be 1/2 the impedance of one set in this case (divide by two as OP noted), but that is not always the case. In fact, you can count on it not being the case most of the time, unless you are paralleling two sets with the same impedance.
 

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Sure. Of course I never tried to describe the math, I just gave him the correct answer.



For most cases, the math can be shortened to the following, which may be useful to the op:


For series, add the resistances together.


For parallel, dived the resistance of one by the number being wired in parallel.


That holds true for cases where all speakers are the same impedance rating, or you are working with groups of speakers where all groups have the same effective impedance and you treat the group as one effective speaker. Which covers the majority of cases, including this one.
 

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1 / ( 1 / (4 +4 +4) + 1 / (4 + 4 + 4) )


1 / (1/12 + 1/12)


1 / (2/12)


1 / (1/6)


6




Series -> add the impedances

Parallel -> take the reciprocal of the sum of the reciprocals.


If you have two identical loads wired in parallel, yes you can just divide by 2.
 

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Yeah...



It's just:


Zseries = Z1 + Z2 + Z3...


Zparallel = 1/(1/z1 + 1/z2 + 1/z3...)


So in the poster's example...4+4+4 = 12 (as he says)

and 1/(1/12 + 1/12) = 1/(2/12) = 1/(1/6) = 6


As has been stated.


With this formula you can calculate speakers that are different in resistance as well.



B.
 
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