What does Current mean in a power amp???

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 ?

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,