Tony Bones wrote: ↑Mon Apr 08, 2019 5:17 pm
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...
I still don't think this helps me understand it. Those are all great equations that state what a single entity takes, but doesn't show the math or explain the interaction of two speakers
Your statement "If an amp is putting 20W into two speakers they can't both receive 20W of power" is not related to my question. my point is, what if speaker one drops 20W and speaker two then drops 0W? no? yes? I.e. I thought I understood that in series, the load is going across the first one, and it takes as much as it takes, and what's left goes to the second one. That's not correct? Why?
I suspect the brain trust here has like 110% chance of being right, I just don't understand why.
I swear I'd heard before that if you had say a 15W speaker and a 100W speaker and sent 40W to the speakers the 15 watter may likely blow up because when the signal hits it (if it's first) it will drop all of the power hitting it first or at least some mathematically significantly higher than 15W, say 20W (if it was exactly half) but I thought it would absorb more of the brunt if it came first in the chain.
Again, I'm probably wrong, just trying to understand the science behind it, and how the power is perfectly balanced, if at all etc. irrespective of the parallel vs series.
Another mathematical idea here, would be that a resistor (or impedance) causes a voltage drop no? So wouldn't the first speaker drop the ac voltage coming at it, and the second then get 'less voltage', and theoretically couldn't that second one due to less voltage, have less power dissipated?
~Phil
) and L'il Texas $110, it's not much of a stretch to pop for the neodymium choice. If they're similar tonally, so much the better. Only $30 to save a heap of weight in a 2x12 combo or cab, worth the price in aspirin & liniment. 
