"Watts" Versus "Current"


Can someone discuss, in layman's terms, the differences here? I've seen some high wattage amps that do not produce much peak current and some low wattage amps that produce a lot of current. Which stat is more important--watts or current--in terms of deciding on a match with speakers? If current is more significant why to we (and manufacturers) talk mostly in watts?
128x128dodgealum
Great analogy from Danmyers. Most important part of his post, IMO is this "vastly different design parameters required for flow/current and pressure/voltage. Which is better? Well, it obviously depends on what you want to do with the water" IOW, depends on your speakers. Read Atmasphere's posts and white paper. Amp choice highly dependent on whether your speakers are current optimized or voltage optimized. Which strongly suggested that at least a part of "system synergy" can be predicted.
Swamp, not to take away from Dan's similar explanation, but if you read my post at the beginning of this thread, you'll realize I said the exact same thing:
Therefore, all watts are not created equal. 1 amp (of current) X 10 volts = 10 watts of power. But 10 amps X 1 volt also = 10 watts. Further, the basic W =A x V really works for direct current (DC) only. The formula is a bit more complicated for AC (as in: music signal). But leaving that aside for now, how the power is created will either be better or less good for certain kinds of speakers.

Why? Because every kind of driver generates a reactive signal BACK TO THE AMP. When the signal from the amp moves a cone back and forth, for example, the action of the cone's voicecoil in the magnetic field actually GENERATES a reverse electric current BACK to the amp! This reverse current adds to the impedance (load) the amp "sees".

What's CRITICAL (in amp selection) IMO, is HOW that (let's call it 'phantom') impedance is created. If the amp sees a capacitive type of load (an electrostatic speaker) it needs for its available watts to consist of lots of amps (current). If it sees a resistive type load (like cones, domes, ribbons and planar magnetics) it needs for its available watts to consist of lots of volts (to overcome the reverse voltage created by those kinds of drivers.)
.
Ns- you bet. I've been watching this thread for a few net sessions and quite honestly had meant to mention it as well. I was mostly referring to the analogy which for non-tech types of math-phobes (I'm not one, but many are) its easier to grasp the concept that way and then on to the math and tech detail. I know I'm right about at the edge of my tech expertise w impedance. I know its the equivalence of resistance and I understand its important in matching amp and speaker. I know (or at least I think I do) that most amps operate better into higher impedance loads and that lower impedance loads will require/draw more "wattage" from an amplifier. I do get confused with input impedance and output impedance and matching source/SUT/phono stage/pre-amp/amp with respect to those parameters. And I have an advanced degree in a science (not a math oriented one, admittedly). A analogies help me make the conceptual connections that allow me to go from trying to remember "rules of thumb" to actual understanding. Then I can make judgments for myself.
Not to throw water (pun intended) on danyers' analogy but the water pressure at the hose nozzle is independent of the size of the hose (WHEN THERE IS NO FLOW), but will actually be LOWER at the end of a smaller hose because of the extra resistance to the water flow offered by the smaller hose.
The rservoir at a certain height is the Voltage and the amount of water flowing through the hose is the current and the size of the hose (pipe) can be considered the load. If the load is high, the current is actually lower.
In danyers' analogy, the reason that the water is 'thrown' further from the shorter hose is not because the pressure is higher, but because less water is being supplied at the nozzle at a higher speed thus the water can travel a further horizontal distance before falling to the ground. The trouble with using water analogies (which is commonly done for electricity) is that one must also understand the mechanics of fluid flow for the analogies to make sense. In this case the analogy is not very usefull, except to misinform.
Gs5556's reply is, by far the best analogy, because it is not an analogy but an explaination using an example.

Salut, Bob P.