Pape, welcome to Axiom. Assuming that what you're asking about is the current flow required by the EP500 during its operation, the relevant form of Ohm's Law is current equals [square root]power/impedance. Using the maximum continuous power capability of the EP500 amplifier in the specs of 500 watts, the square root of 500/4 is 11.2 amperes. Note however that 500 watts will rarely if ever be actually used. At a more typical maximum power usage of 100 watts the number is 5 amperes, and at less than maximum output using a few watts about 1 ampere of current flow is taking place.
These are what might seem to some to be relatively small numbers if they've been impressed by ridiculous "high current" claims of some amplifiers(e.g., 50 or more amperes), but current requirements are controlled by Ohm's Law, not manufacturer hype. So, there would be no specific "amperage recommendations" and typical home electrical wiring is more than capable of meeting any possible requirements.
Close. Your formula is derived from P=I^2R which is tecnically not pertinent to this situation. This formula is used to calculate power dissapation in a purely resistve cicuit where the voltage and current are in phase. Losses.
For peak wall current, without the circuit power factor, the best we can do is P= E x I. Where I = P/E. We can assume the voltage is ~120v (canada), so 500w/120v = 4.167A
But this is also incorrect, as 500w is the RMS output of the amp under very specific conditions, such as input signal level and gain settings. The actual AC draw of the amplifier must be a combination of the in phase (watts) and out of phase components (Vars) of the power supply circuit.
The power supply section is what is being referenced by current input here. Keepng the DC cap filter full after rectification, plus transformer losses, plus circuit component losses, is what the branch circuit sees as AC current draw. The DC current rushing in to fill the empty reserve cap and filter the voltage across the bridge rectifier can be SUBstantial. At startup and prolonged peaks the cap needs to be recharged and draws on the circuit to reach peak voltage again.
The output ampere rating quoted by manufacturers is at certain frequencies under max gain and signal input in the output circuit. The output transistors are saturated. These are theoretical numbers for actual calculations, as under normal music or soundtrack situations they likely would not be present. What the large numbers can tell us is how well sized the power supply circuit is matched to the output section of the amplifier, or if there is active current limiting designed into the circuit. This is also why the "doubling down" of amplifiers is a relevant figure. Floyd Toole uses the "power cube measurement" which follows the same logic.
Unless you listen to sine waves or tones from a frequency generator, a calculator wont give you the answer. An ammeter with peak current readout will.
What a nerd, eh?
