speaker ohms

  • Thread starter Thread starter rgroh
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rgroh

New member
I just got a marshall micro stack with 2 10inch bottoms,
both cabs are 16 ohms each, if I hooked up a 12 inch 8 ohm bottom with one of the 10 inch 16 ohm speaker what would the total ohms be???
I can't go any lower than 8 ohms total.
Any help would be appreciated.
Russ
 
Resistance in series:
R = R1 + R2 + R3


Resistance in parallel:
1 / R = (1 / R1) + (1 / R2) + (1 / R3)


Use these formulae to determine a speaker connection combination that yields the total load you require.


Bruce Valeriani
Blue Bear Sound
 
I'm no math wiz, but I've always found the parallel formula easier viewed as

Rt= 1 / 1/R1+1/R2+1/R3 etc

Same answer of course. Just a simpler process for my lacking math skills.

(Even better viewed with the first 1 on top and the other values below but the formatting gets all messed up)
 
Hey Emeric... how's things in the 'burbs?

Yup... you're right... but when I typed it that way originally I thought it looked even more complicated, so I wrote it "full-hand"!

Bruce
 
So to answer your question, the two 16 ohm cabinets together give you a total load impedance of 8 ohms. An 8 ohm cabinet in parallel with a 16 ohm cabinet yeilds a total load impedance of about 5 ohms. So you need to know if the speaker jacks on your amp are wired in parallel. For quick reference, as you can see, the total impedance of two equal resistances in parallel is equal to 1/2 of either of the single resistances.

If they are not equal, the total impedance will always be less than the smaller of the two.


Twist
 
Thanks for the replies. Is it possible to convert a 8 ohm cab to 16 ohms?
Russ
 
Nice going Twist...

... just go ahead and give out ALL the answers! That's the trouble with these kids today.... they don't think for themselves anymore!!!!!!!

:D :D :D

Bruce
 
rgroh,
Whats in the cabinet for speakers?

Bruce,
This is one of the only questions I've ever been able to answer, and my post count is pathetic, so........


Twist
 
rgroh,

Yes you can convert an 8 ohm cab to 16 ohms but it is not efficient. If you place an 8 ohm resistor in series with the 8 ohm speaker (or cab) you will end up with 16 ohms total. Here's the problem: that resistor must be capable of handling half the total RMS wattage of your amp (at least if you plan on cranking it near max volume). Otherwise, poof! you toast the resistor, open the path, and push your final amplifier stage through zero load, which is not healthy. Actually, the burn-out won't be too fast. You will smoke the resistor kind of slowly depending on how underrated it is. You can find some very high power resistors, but not at Radio Shack.

This configuration is inefficient because half the power will be dissipated by the resistive load.
 
and beware of that recast equation above which does not include the needed parentheses. If you follow that simpified equation using the standard order of mathematical operations you will get hosed. It should be:

Rt= 1 / (1/R1+1/R2+1/R3)

just being a prick!
 
I've never seen this formula written with brackets before. I could see the use of brackets if it were something like X=1 /(1/a+b)+(1/a/b)+(1/b) etc.. otherwise it's unnecessary clutter.
 
Not really. I understand this stuff, but I never did it out with equations before (never really cared to). The parentheses do serve a purpose.

Rt= 1 / 1/R1+1/R2+1/R3 etc

The 1 in the beginning would only be affecting the first part of the equation in this sense. Not the whole thing, which is what you want.

Rt= 1 / (1/R1+1/R2+1/R3 etc)

Now thats better! The 1 is now affecting everything within the brackets (like you want it to)


Jake
 
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