pompeiisneaks wrote: ↑Mon Apr 08, 2019 2:28 pm
I would like an explanation of that. My understanding of resistance and impedance in series is that the first speaker takes the full load, then the second would, in theory, take the full load as well, whatever wasn't dropped by the first. Maybe I'm confusing resistance and impedance though. ?
The short answer is conservation of energy - or conservation of power in this case, still valid. If an amp is putting 20W into two speakers, they can't both receive 20W of power. That would be 40W total. Where'd the extra 20W come from?
The long answer makes use of the equation P = V^2 / R. Figure out what the voltage across a single 8 ohm speaker must be @20W (12.65Vrms), a pair of speakers in series for 16 ohms (V=18Vrms), and a pair in parallel for 4 ohms (V=9.0Vrms).
Now, for the 16 ohm case, realize that the 18V signal is divided equally between the two speakers (9 volts each), then the power to each speaker is P = 9^2 / 8 = 10W.
For the parallel (4 ohm) case, the power delivered to each speaker is ... we already figured it out: P = 9^2 / 8 = 10W.
All of the above assumes a tube amp with 4, 8, and 16 ohm outputs that allow you to load the amp identically in each case. In the case of a SS amp with vanishing low output impedance the output voltage won't change (much.) So if you get 20W into an 8 ohm load, then you'll get close to 40W with a 4 ohm load (20W per speaker) but only 10W with a 16 ohm load (5W per speaker.)
But, I'm assuming a tube amp with 4, 8, and 16 ohm output taps where, to first order, there is no difference between series or parallel connections of the speakers provided you choose the right OT tap.
Having said that, there are some people that claim the 4 ohm tap uses only some of the secondary windings while the 16 ohm tap uses all of them, hence the 16 ohm tap works better. I've never A/B'ed, so I can't comment on that...