KJung said:
Actually, if I understand things correctly... a 2ohm load will 'allow' the amp to put out a lot of power... not require it to do so. ...
Actually, there's a big thread about this, but this is oversimplified. Yes, actually it will be "required" to do so.
In the virtual equivalent of the circuit (either thevenin (sp) or norton), the circtuit can be viewed as a sine wave generater and a series resistor (the amp (or battery)), and the speakers are in series with that.
The key is the "internal resistance" of the amp. For "optimum power transfer" you want to MATCH the internal and external. Even batteries can be viewed as having an internal resistance, which is why they get hot when shorted thru a very small load.
What this means is, if you lower the speaker resistor too much, the amp has to put out MORE current to maintain the same voltages across the voice coil (load), which determines cone movement, and of course, a given SPL.
The downside is, when you raise the current thru a resistor, more voltage is dropped, and the more heat is generated.
SO... to get the same voltage dropped across a 2 ohm load, you'll have to get TWICE that voltage dropped on the (internal) 4 ohm load, which raises the heat generated inside the amp.
In the case of the single higher impedence cab config, in a simplified model, you're basically dropping twice the voltage over one speaker, because it's trying to do the job of two, so you're still generating a lot more current out of the amp, and still heating it up internally.
Heat does two bad things... yeah it shortens the life of components, but it also can potentially alter your tone as well as lose "headroom" since the amp is basically out of reserve current.
It's all about matching... not under or over driving... if you want the "most" in output and reliability from an amp.
IMHO, anyway...