OUTPUT POWER: 300w @ 8 ohm, 500w @ 4 ohm

and i had a cab that's available in 4 or 8 ohms and has a power handling of 800w. Would that mean that it's best to get the cab at 4 ohms to get the full 500w? or have i missed the mark completely.
Last edited by seanms at Sep 22, 2010,
Yeah, that's pretty much it.

Or you can get two of the cabs at 8 ohms and run two speakers at the full power.
Perfect, thanks Zipholblat
Yes, use the appropriate equations. Given that things happen at constant current, this is not hard.

To get the inputs, look at the current sketch to see that the capacitor charges at 10 mA for 2 ms, reaching the supply voltage Vc. It then holds for 2 more ms. There is no current, hence no voltage or energy change during this period. Then the capacitor discharges at 20mA for 1 ms, starting at Vc, before returning to the initial state.
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urls removed by moderator.
1) Take highschool physics
2) ???
3) Profit

Yo TS, most high school physics classes (Any respectable one anyways) discusses circuits, and subsequently, ohms law. If you take the class you will have an intimate understanding of how to rate heads for certain combinations of cabs. It was probably the one thing I learned in school that I applied outside