Jazzdogg
Less barking, more wagging!
Here's what it would look like. You would have an 8-ohm load (Crazy 8) in parallel with a 16-ohm load (Crazy 88 with drivers wired in series). With a voltage applied to that circuit, half the current would flow through the 16-ohm Crazy 88 than would flow through the 8-ohm Crazy 8. Correspondingly, half the power would be developed across the Crazy 88 than would be developed across the Crazy 8.
So, assuming linearity and equal sensitivity of the two cabs, the acoustic output of the Crazy 88 would be 3-dB down compared to that of the Crazy 8.
Now, let's look at each driver within the Crazy 88. Because it would be wired in series and because each driver has the same impedance, then the current flowing through each driver would be the same with each driver contributing equal shares of the "half-power" through the box. That is, the power developed across each driver in the Crazy 88 would be 1/4 of that developed across the Crazy 8.
Here are the calculations. Let's assume a 1-volt input to the circuit. We'll also assume that Ohms law (V=IR) applies and we'll eschew the reactive AC components.
What is the current across the Crazy 8?
It is 1/8, or 0.125 amps (I=V/R).
What is the power developed across the Crazy 8?
It is 0.125 watts (I2R or V2/R).
What is the current across the Crazy 88?
It is 1/16 or 0.0625 amps.
What is the power developed across the Crazy 88?
It is 0.0625 watts.
What is the power developed across each driver in the Crazy 88?
It is 0.0625 2*8=0.03125 watts (I2R).
So, you see, the power developed across each driver in the Crazy 88 is 1/4 of that developed by the single driver in the Crazy 8.
Thank you for teaching me how to fish instead of handing me a tin of canned fish, Drurb; that's exactly what I was looking for!
