Calculating amperage

[N. Abstentia]

My Ohm's Law is a bit rusty, just looking for some backup

I'm trying to figure out how much current my amps will be pulling.

I'm using I = W / V correct?

So an amp that produces 500 watts (I'm talking about a real amp here..like an Anthem or Acurus 5 channel) would draw roughly 2.5 amps. Am I on the right track?

[kexodusc]

mmm...no, not exactly...500/120V = 4.17 amps
Or 500/115V = 4.35 amps,

but only if the unit is drawing all 500 watts at once....

Also, input and output current aren't really the same thing

There's also P= I X I/R
current squared divided by resistance

500 watts/8 ohms = current squared or about 7.9 amps...
So the output amps are different than the draw amps.

[N. Abstentia]

Thanks man. I'm mainly concerned with the power needed from the wall, I'm drawing my outlet layout. Looks like I'll be fine to run two big amps on one outlet.

I got the 2.4 amps from here:
http://www.csgnetwork.com/ohmslaw2.html

120volts/500watts = 2.4 amps

I'm confused now

[PAT.P]

To figure current needed for a divice rated in watts is Watts divided by volts=amperes.A toaster or hair dryer takes around 1200 watts divided by 120volts = 10 amperes.Pat.P

[popolz]

Thanks man. I'm mainly concerned with the power needed from the wall, I'm drawing my outlet layout. Looks like I'll be fine to run two big amps on one outlet.

I got the 2.4 amps from here:
http://www.csgnetwork.com/ohmslaw2.html

120volts/500watts = 2.4 amps

I'm confused now

remove the check from the "3 phase"

[mixadude]

mmm...no, not exactly...500/120V = 4.17 amps
Or 500/115V = 4.35 amps,

but only if the unit is drawing all 500 watts at once....

Also, input and output current aren't really the same thing

There's also P= I X I/R
current squared divided by resistance

500 watts/8 ohms = current squared or about 7.9 amps...
So the output amps are different than the draw amps.

Yeah, the speaker wire would be carying 7.9A if it's 500W into an 8 ohm load.

Most ss audio amps are 50 - 75% efficient. 50% efficient would mean that if you turn it up until it's producing 500W of audio, it would be drawing 1000W from the AC cord. So an amp that is actually producing 500W steady state (never happens unless you like to listen to tone or some steady state signal at max output) will actually be drawing 750W to 1000W from the main AC supply.

Now if you put 2 500W audio amps on a circuit, they can draw say 1500 watts (75% efficient) at full output. Furthermore 12Ga wire has roughly .17 ohms of resistance per 100'. Since there are 2 conductors carrying the AC to your amps, that's .34 ohms per 100'. So if you have 12Ga wire in the walls, and a 12 Ga extension cord and 12 ga power cords on your amps, it could add up to 100' of wire from your load center where the breakers are.

Now if you had a zero impedance source of 117V available to your 2 amps that draw 1500W, they would draw 12.82A, well within the safety margin for the circuit. BUT. But when you draw 12.82A through a .34 ohm resistance, you will have a voltage drop of 4.35V at the AC mains input of the amplifier from the resistance of the wire in the wall plus whatever you hook it up with! And if you ran 3 amplifiers it would be 50% more (6.5V of loss) and so forth. This basically means that your 2 500W amps would now be 2- 481W amps. And the more power amps you load down on a circuit and it's AC wiring, the worse it gets. Might as well get smaller amps!

It's not all that bad in the sizes of amplifiers we use for hifi, and the volumes at which we listen, but when you're hooking up big amps (multiple 4KW and more) you can see where it can become quite problematic.

That's my story and I'm stickin to it

[Smokey]

So an amp that produces 500 watts (I'm talking about a real amp here..like an Anthem or Acurus 5 channel) would draw roughly 2.5 amps. Am I on the right track?

Well, sort of. Doing the calculation, 500W/120v = 4.17 amps.

But there is catch here. As Kexodusc mentioned, input (AC Current) and output current (speaker current) aren't really the same thing. Current out of AC plug is called Real current, while current into speaker load is called Apparent current and they will not equal.

Apparent current-or more appropriate Apparent power is usually much higher than drawn real power. So real current drawn from AC plug will be much less than 4.17 apparent current outputed by amplifier.

[mixadude]

My Ohm's Law is a bit rusty, just looking for some backup

I'm trying to figure out how much current my amps will be pulling.

I'm using I = W / V correct?

So an amp that produces 500 watts (I'm talking about a real amp here..like an Anthem or Acurus 5 channel) would draw roughly 2.5 amps. Am I on the right track?

Here's an easy to use ohm's law calculator.

The best way to see how much AC current an amp can draw is to read the specs.

And the current in the AC wire and the current in the speaker speaker wire are both real... but they flow in different paths, in different wires, to different loads, from different sources. They are in a word ... different.

[Glen B]

My Ohm's Law is a bit rusty, just looking for some backup

I'm trying to figure out how much current my amps will be pulling.

I'm using I = W / V correct?

So an amp that produces 500 watts (I'm talking about a real amp here..like an Anthem or Acurus 5 channel) would draw roughly 2.5 amps. Am I on the right track?

All you need to do is read the total power line consumption printed on back of the amplifier. Some manufacturers give you the figure in watts and others give it in amps. In actual operation most amplifiers will not draw the maximum unless you are driving them to their limits. In normal use, actual current draw is a fraction of the maximum rating. The only way to geat a realistic measure of actual system current draw during use is with a clamp-on meter or other suitable measuring tool.

[N. Abstentia]

Thanks guys! I couldn't find that rating on the back of my speakers (yes they have amps built in..Paradigm Active 40's) but it might be in watts as Glen said. Let me look again.