Pop quiz: If Brian has a bass cabinet with two 16 ohm 300 watt drivers wired in parallel, he'd have an 8 ohm cabinet rated at how many watts? a) 600 watts b) 150 watts c) 300 watts d) I'm too stoned to think What's the answer? Please show all work. Thank you, Brian

Okay, The square root of 45 * 6 / 13 = um....2! Yeah, 2. 2 divided by the average length of a guitar solo, 124 bars, then you add the wattage bridged twice by the reverse parallel of the second square unit, all over 3. Next, you have to bridge the wattage of the RMS and the PMS. PMS being greater than the sum of the 3rd entry to the fourth string. Assuming the bass is tuned E-A-D-G# then you must add in the frequency of the converted left ventricle, -45 htz, plus 1. Add pi to the first number and then subtract by the weight of the cabinet, without casters. What was the question again?

a. If you use speakers together, their powers add. Regardless of impedance (even different impedances), being in the same cab, being in different cabs, being connected to different amps, the same amp, using long wires, short wires, A fender bass, a Conklin GT7 bass, playing with a pick, or fingers, or teeth for that matter. When you got two speakers, just add their powers. Some say I'm sarcastic !!!

Joris, But wouldn't two different impedances change that? Say I have one 10" speaker rated at 100 watts that is 8 ohms and one 10" rated at 100 watts but is 4 ohms. If they are wired in parallel and you run them w/ an amp that is 200 watts at the 2.6667 ohms the cabinet is, the 4 ohm speaker would get twice the power of the 8 ohm speaker so it would get 133 watts and the 8 ohm would get about 67 watts so the 4 ohm would be over powered w/ 200 watts so it isn't a 200 watt cabinet. Is that correct? Now, w/ speakers of the same impedance, no matter how they are wired, you can just add them.

The answer is not 'D'. Here is a copy of the very elegant work posted by Mr. Oscar T. Grouch on R.M.M.B. I checked his work and I think he's right on. > Simplifying this to a time-invariant problem (DC), and assuming the > amplifier provides maximum voltage and any demanded current yields: > > P = EI > > E = IR (Ohm's law) > > therefore: > > P = I**2/R or I = sqrt( P/R) > > Let us further assume that the limiting factor in terms of speaker handling > is it's ability to handle current, which is a wild assumption, but we've > already considered the speaker to be a resistor so bear with me... > > Given P=300 and R=16, yields I = 4.33A > > Remembering Ohm's Law, and treating V as constant: > > V = IR > > I1 * R1 = I2 * R2 > 4.33 * 16 = I2 * 8 > I2 = 4.33 *16 /8 > > I2 = 8.66 > > Plugging this back into the equation P = I**2R > > R=8, I = 8.66 > > P= (8.66)**2 * 8 = 600 W > > In other words, each branch of the parallel circuit can handle 300W. There > are 2 circuits so the total power handling capacity is 2 * 300 = 600W. > > QED

Now the only question I have that still bugs me is this: Does it not matter whether the circuit is wired in paralell or series in this case? If I wire two 4 ohm 300 watt drivers in series, will I still end up with a circuit that can handle 600 watts?

Its a lot simpler to just think "add the wattages" than to do all of the calculations Brian Gordon did. But Brian basically did the proof so now we all know why we can just think "add the wattages".

All true. But mixing different impedance in one setup (I mean 1 1-channel amp) is always a very bad idea (but I reckon a lot of people do this anyway) because of the power distribution issue. Unless you really know what you're doing. If you hook those 2 speakers up to two different amp (which is what you should be doing anyway) you can just add wattages. Brian, another remark: a given amp will deliver different maximum power to different impedances. If you hook up an 8 ohms cab to an amp it will give you, say, 100 watts. With a 4 ohms cab you get 200 watts. All in theory, of course. In real life the 4 ohms rating will be more like 150-160 watts.