What does Current mean in a power amp???

All of the explanations above are technically accurate, but here's a very simple way for those not versed in science or engineering to think of current vs. voltage.

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.

A river or stream with little current does not yield much energy (or force). Similarly, you can also have a relatively wide river that is very slow moving, and it does not exert much force. (If you have ever waded a river, you know from experience that it's a lot easier to cross a very slow moving flow than one moving rapidly.)

Conversely, you can have a narrow stream of water moving at very high current, and it will produce quite a bit of energy. To carry this analogy to an extreme, think of the cutting tool known as the "Water Knife" -- it forces a very small stream of water (with an abrasive added) at high pressure (up to 50,000 psi) through a small nozzle, and the stream is capable of cutting through a variety of very hard materials (such as steel).

So, to stretch this analogy, there are two ways for an amp to drive a speaker load: high current (the force), or high voltage (the amount of electricity).

Yes, yes, I know this is a rather unsophisticated explanation from an engineering standpoint, but it has the virtue of being conceptually simple for those who didn't take physics or EE courses in school.

Nikki...,

Let me be straight forward on how good watts are different from bad ones...

If you for example take a look on professional power amps such as Carver highly regarded by DJs available at RadioShack stores, you can see and feel that it has a plenty of boost and the power(~300W/side) to any impedance load and pretty darn cheap(aka $250).

The Ohm's law mentioned up above states that ALL watts are "created" equal. The double wattage of the good amps into the lower impedance loads cannot be rated as a CONTINUES power and needs lots of lines to explain. Shortly I can say that price of an amp is not an explaination why one has a reserve power and the other one hasn't.

To increase the power you either have to increase the current or the voltage.

The main problem in this issue is our 110V wall outlet power that realy limits engineers to work on high-power-quality amps where designing a proper power supply is the most essential issue.

So to correct you in both threads I must state that the main design difference between "bad watts" and "good watts" is that the power supply is OK to handle the large current.

The quality of sonics (I must say here that it's completely different issue from wattage) is the quality of an active amplification elements(tubes, transistors, diodes) and also passive elements. The working area of an amplification elements in high quality equipment is selected so that it covers the widest-possible freequency bandwidth rather than working in the peak values. Thus more transistors or tubes is required to deliver the signal to desirable level with good output characteristics.

Ohms law is not really a good way to figure current output of an amplifier. Ohm's law works in a purely resistive DC circuit. The output of an amplifier is AC. You generally will be required to factor in inductance, capacitive reactance and pure resistance (this makes up impedence)to come up with the current output into any given load. Amps that "Double down" are always better for driving low impedence loads. However, things can happen to an amps output when the load has steep phase angles and high reactance. Look at the dbw that Stereophile uses in rating amps for some reference. The bottom line here is that each amp will behave differently into the reactive load of different speakers. A resistor on the test bench doesn't mean a lot.
With this said, current is the ability of the amp to deliver actual power expressed in watts into a given load. Power(in watts)= volts x amperage This is a simplification of a more complex calculation.

Marakantz

Please reread my earlier post. In that post I was talking about the power supplies, not the amplification circuits. I tossed in some statements about assuming quality amp circuits and the like but the big point was power supplies. When I refer to amps that can double their power rating I am referring to continuous power. For example a Bryston 2B-LP is ~60wpc into 8 ohm but 100 into 4 ohm. This is continuous power not a peak or shot term power rating. I assume Bryston added enough heatsinking for long term use.
I my previous post I looked amplifier output voltage and current.
Output current = current from power supply – small amount to drive associated circuits.
Voltage at the speaker terminal = Voltage at power supply – voltage drop in amp circuit (I don’t know exactly how much this is).
Rather than talk about voltage and current into the amp circuits I was talking about power and voltage out of the amp circuit. Account for a few losses and you have the power and voltage demands of the power supply.
We are both making the same point.

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 ?