Originally Posted by Shinbone
I have an Ampeg PF-115HE. This is an 8 ohm cab. It has a 15" speaker + a tweeter. Each of these is labelled as "8 ohms". Given that resistances in series add and resistances in parallel divide, how do they get the cab to come out as 8 ohms? I suspect the crossover has something to do with that but I'm not sure. Can anyone enlighten me?
Also, if I were to remove the tweeter, how would that affect the resistance? Would it be the same as just turning the tweeter off?
A crossover is basically applying two filters to the input signal - a low pass filter and a high pass filter. The load that the amp sees is the crossover, not the individual drivers connected to it.
At frequencies lower than the crossover point the tweeter side of the crossover presents a very high impedance, so it doesn't really affect the sytems impedance if connected in parallel with an 8 ohm woofer. At frequencies higher than the crossover point, the woofer side presents a higher impedance so doesn't really affect the systems impedance and at the crossover frequency the impedance of both drivers is usually higher than 8 ohms anyway...impedance isn't simple resistance, it is frequency dependant and is all over the place, but a nominal 8 ohm impedance means that the system will never be lower than that.
And yes, removing the tweeter will be like switching it off, but don't do that. Remove the cable to the tweeter if you must, but removing it will leave a hole in the baffle and screw up the cabs tuning...at worst making a blown driver more likely