What does Current mean in a power amp???

I think you asked the wrong question. The correct question is "What amp will drive XXXXXX speakers for XXXX dollars ? My musical tastes are XXXXX and I like to listen loud/medium/soft. My room size is XXXXX."

After that I'd demo the suggestions. An amp with sufficient current will sound powerful, not loud, and will sound clean, not sibilant, not over-bright. It will drive comfortably at your desired listening levels without producing smoke.
I'm an engineer, and that's the way I'd go about it.

Asking how much current is like asking how long a piece of string. You'll end up with a technical pissing contest of a thread, rather than a straight answer.
It reminds me of that old sketch "I'd like to buy a grammaphone ...." Does anyone remember where that was from ? Was it Python ?

I have to disagree with Sean and say Ohms law, and its derivatives, tells you a lot about
An amp and speakers (of course mnf. specs can be doctored). Ohm's Law for DC circuits is the fundamental relationship between voltage, current, and resistance. It is very relevant. It is usually stated as: E = I*R, or V=I*R, where E or V = voltage (in volts. E stands for "electromotive force" which is the same thing as voltage), and I = current (in amps), and R = resistance (in ohms). The equation can be manipulated to find any one of the three if the other two are known. For instance, if you know the voltage across a resistor, and the current through it, you can calculate the resistance by rearranging the equation to solve for R as follows: R = E/I. Likewise, if you know the resistance and the voltage drop across it, you can calculate the current through the resistor as I = E/R.

A related equation is used to calculate power in a circuit: P = E*I, where P = power (in watts), E = voltage (in volts), and I = current (in amps). For example, if you measure 20V RMS and 2.5A into a load, the power delivered to the load is: P = 20*2.5 = 50W. This equation can also be rearranged to solve for the other two quantities as follows: P = E*I, E = P/I, and I = P/E. You can also combine the power equation with the first Ohm's law equation to derive a set of new equations. Since E = I*R, you can substitute I*R for E in the power equation to obtain: P = (I*R)*I, or P = I2R. You can also find P if you know only E and R by substituting I=E/R into the power equation to obtain: P = E*(E/R), or P = E2/R. These two equations can also be rearranged to solve for any one of the three variables if the other two are known. For example, if you have an amplifier putting out 50W into an 8 ohm load, the voltage across the load will be: E = sqrt(P*R) = sqrt(50*8) = 20V RMS.

There is a form of Ohm's law for AC circuits too. In AC you must deal with capacitance(C) and inductance(L) in addition to resistance(R). Together, any combination of the three is impedance (Z). If you simply substitute Z for R in the Ohm’s law formula they are accurate for AC too. E=IZ, I=E/Z, Z=E/I are the AC Ohm’s law equations. You can tell a great deal about how an amp will drive a speaker with these equations all based on Ohm’s law.

Finding Z is a little more difficult because it involves current out of phase with voltage and it involves complex numbers, vectors, etc…

I think I agree with Scot’s statements above but it is a little confusing on a point that is often confused. He says:

“Imagine a river. The amount of water that is moving downstream is analogous to the voltage -- i.e., it's a measure of the size or quantity of the flow (say, 2500 cubic feet per minute). The current, or force, behind the water (usually due to gravity) is the other measure of actual or potential energy.”

This seems to refer to Voltage as “water that is moving” and the “force, behind the water” at the same time. Well….. is voltage like the stuff that is moving(water) or the force moving it(gravity)? Not good to confuse the two. The amount of water moving downhill seems analogous to Power not Voltage and from the equations above you see that these two are not the same. No amount of gravity is going to make a river flow if there is no water.

This is a very basic and important distinction in electronics too. - between matter (electrons) and force (photons). Voltage does not move. Voltage is like a pressure between two points. You might say current flows through a resistor but it is wrong to think of voltage as doing so. Voltage is “across” a resistor not through it.

A battery (voltage supply) does not supply current. The electric charge preexists in the wire but lacks organization if you will. A wire is a conductor precisely because it has free electrons. If a voltage is applied across a wire, with a battery, it will push the preexisting electrons in one direction (toward the positive). But the electrons are not the “force” either. To think this would be the same as confusing sound waves with the air the sound waves travel in. The difference in voltage potential creates an electric field and the electric field is the force. The Electric force moves very quickly. The drift of electrons is not much faster than ketchup. Alternating current vibrates more than flows. At least I think this is the basic model.

I remain,

Rwd,
Do these repsonces answer your question :)

This has probably progressed well beyond the original question, but it makes for an interesting topic of discussion.

I disagree with the statements that ohm's law is not at all applicable, because with the substitution of the term "impedance" for "resistance", ohm's law is very much true for a single-phase AC circuit. As stated by both Sean and Bigtee, the AC impedance is made up of resistance, capacitive reactance, and inductive reactance. The resistive component is constant at all frequencies, but the capacitive reactance is inversely proportional to frequency, and inductive reactance is proportional to frequency. These properties, along with the phenomenon of mechanical resonance, described above by Sean, explain why any given speakers impedance plot can vary so wildly, both above and below the nominal impedance.

However, in order to look at a "simple" model of a loudspeaker, one must have some way to express the relationship between voltage, current, and impedance, and ohm's law provides that relationship. The other relationship required to model the circuit is a basic power equation, which is also different for an AC circuit. For a single-phase AC circuit, Power=Voltage*Current*Cosine of the phase angle between the voltage and current waveforms. This difference in phase is, of course, a result of the net reactance at any given frequency. To apply the power equation without accounting for phase relationships, you must assume that the reactive component of the equivalent impedance is zero, yielding a purely resistive load. I assume (but don't know for sure) that this is how published amplifier ratings are derived, with a discrete frequency sinusoidal waveform applied to an 8 ohm resistive load.

If you accept the above, then in a very roundabout way, ohm's law does in fact have an effect on whether an amplifier can drive a given speaker. If we could build a "complex" mathematical model for a given type of speaker paired with a certain amplifier (and don't forget the role of our choice of speaker cables in this model) ohm's law could describe, at any particular frequency, how much current our voltage source (amplifier) could supply. We could then look at the phase relationship between the voltage and current, apply our power equation, and we would have our value of power at clipping for any frequency that we cared to look at. Since I didn't do very well in differential equations, I will leave this modeling process to the wonderful people who design audio electronics for a living.

BTW, I noticed that while I was formulating this response, a couple additional posts were added. Thanks to clueless for thinking on the same line as me (kind of scary, huh?), and Seandtaylor99 hit the nail right on the head, although I think we are still just short of a full fledged pissing contest :-)

Perhaps "pissing contest" was a little harsh? :-)
The posts are interesting, though.