i am struggling to understand something, and reaching out for help. if i am running an 8 ohm cabinet and a 16 ohm cabinet using the same amplifier, what kind of ohm load am i presenting the amp with?
i have been told that running an 8 ohm and a 4 ohm cab from the same head results in a 2.67 ohm load, or something very close to that, but i really dont want to run anything lower than 4 and i am considering rewiring a 4 ohm cabinet in series to make things a bit safer...just not sure how much power i'm going to lose by doing this
First of all, stop worrying about power. Seriously. Even if you lost half the maximum power of your amp, you'd barely notice it. You'd be hard pressed to hear the difference in volume between 100 watts and 150 watts even though we're talking about a 50 watt difference. Power is related to volume exponentially, it is not a linear relationship.
Ok, since you put impedance in the title, Ed won't be answering this. Darn.
Calculating parallel impedances of two cabs or speakers is very easy as there is a shortcut. It'll be the product divided by the sum. Here is an example:
One 8 ohm and one 16 ohm.
Product = 8 * 16 = 128.
Sum = 8 + 16 = 24.
Product / Sum = 128 / 24 = 5.33
So our total impedance would be 5.33 ohms.
Not enough difference and certainly not enough volume loss compared with 4 ohms to notice.
