Practical Physics question

I'm not asking for help on homework here, this is side reading, you see.

My friend recently took his bass amp to a specialist to see what was wrong with it. He keeps blowing his driver up, you see. Turns out that he was using a driver rated at 400W, and his head was actually outputting 500W to it (it was labeled wrong).

What I'm interested in is how exactly did this specialist calculate the power output of the amp?

I'm familiar of working out the emf, internal resistance and therefore theoretical output of a DC power source (like a dry cell), but I'm not sure how this would equate to an AC power source like a bass amp.

I have had a thought, though. I suppose you might be able to use the same procedure, but instead of using a simple DC current, use a sine wave generator, or perhaps two, have a known power supply (or in this case amplifier) output an AC wave in antiphase to that which the unknown amplifier is outputting. This would mean that the current would be directly opposing the other wave at all points, and so i suppose you could treat it the same as a DC current.

But, since its AC, the circuit you use to give the output power would need to be symetrical (since current flows both ways, or you would get an uneven wave).

Don't spose there any physisists here that could point me in the right direction?

 

http://encyclopedia.thefreedictionary.com/Internal resistance

Seems its a bit more simple than I thought

But thats just for audio amplifiers, what about other AC sources? Hmmm...

 

But I bet they have changed the AC to DC, Due to power availiblility,,Dc works better.

 

ha AC/DC, just thought i would share that.

eric

 

OK Power is, for DC I*I*R where I is the load current and R is the load resistance

For AC the Power is VICos(phi)/Sq Rt 2 where V is the line voltage, say 120, I is the line current (can also be taken as V/R) and Cos (phi) is the phase angle usually 0.86 for the mains system Sq Rt 2 is 1.414 as the voltage is usually the peak value and the Sq Rt gives the effective DC value

Easy really

 

hehehehe... blowing up amps again!

(hello people..)