Audio question - running 8 speakers off a 2 channel amp
#11
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three 8 ohm loads would be
(8 x 8) / ( 8 + 8) = 64/16 = 4
then take that load and parallel it to the other 8 ohm
(4 x 8) / (4 + 8) =32/12 = 2.66666667 is the load for 3 8 ohm loads
(8 x 8) / ( 8 + 8) = 64/16 = 4
then take that load and parallel it to the other 8 ohm
(4 x 8) / (4 + 8) =32/12 = 2.66666667 is the load for 3 8 ohm loads
#12
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Originally Posted by Bigyellowcat
the formula is for 2 loads not 3. you have to do 2 loads at a time then you can combine them.
#13
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Originally Posted by Bigyellowcat
three 8 ohm loads would be
(8 x 8) / ( 8 + 8) = 64/16 = 4
then take that load and parallel it to the other 8 ohm
(4 x 8) / (4 + 8) =32/12 = 2.66666667 is the load for 3 8 ohm loads
(8 x 8) / ( 8 + 8) = 64/16 = 4
then take that load and parallel it to the other 8 ohm
(4 x 8) / (4 + 8) =32/12 = 2.66666667 is the load for 3 8 ohm loads
Whoops you beat me to it! Thank you. It drives me crazy when I can't figure this kind of stuff out!
Last edited by onesickpantera; 01-09-2006 at 09:43 AM.
#14
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Originally Posted by noboat
Now, let's say hypethetically I add two more speakers to each channel and wired those two in series as well for a total of 3 "8 ohm speakers". Now I wire all three in parallel. My brain is telling me 8 ohm load/3 = 2.67 ohm total load. !
just because that worked this time I would always use the formula because, that worked out right that time and some times it will work out easy but usuallly it is wrong if you just divide.
#15
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Originally Posted by Bigyellowcat
just because that worked this time I would always use the formula because, that worked out right that time and some times it will work out easy but usuallly it is wrong if you just divide.
So in my example 3 4 ohm speakers on a mono amp would be:
(4x4)/(4+4) = 16/8 = 2
(2x4)/(2+4) = 8/6 = 1.33 ohms total?
#16
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I guess going to school for degrees in mechanical engineering, electrical engineering and acoustical engineering finally paid off.
#17
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Originally Posted by Bigyellowcat
I guess going to school for degrees in mechanical engineering, electrical engineering and acoustical engineering finally paid off.
#18
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Also, is there a way to figure out power(wattage) that an amp poduces between the 2 and 4 ohms? In other words, if an amp is rated at 720 watts RMS at 4 ohms and 960 watts RMS at 2 ohms how do you figure out what power it would have at 2.67 ohms?
#19
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Originally Posted by Bigyellowcat
I guess going to school for degrees in mechanical engineering, electrical engineering and acoustical engineering finally paid off.
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#20
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Originally Posted by Audiofn
Ya but what is the formula for shear
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