I have a Kill-A-Watt meter (you can find them off Amazon), and it measures the AC power draw off the wall socket. Not the same thing, but you know the amp is providing no more than the AC draw. In general class A/B amps are around 50% efficient, so if you draw 100W you are probably getting around 50W amp power.

These devices are cheap and useful to play with. I'm using it with my computer stuff. My computer in "shut down" position is 10W while in "standby" is 12W, so I know that the computer still draws power when "shut down." However, in standby, using only 2W more, I can restart in about 2 seconds. Useful thing to have learned.

Sorry Gonk but I just can't agree with that statement. If anything the breaker will trip at a slightly higher level. Since breakers are designed and rated at a temp of 40C (104F) and usually are not in an environment of such high temps, the tripping amperage would be slightly higher. According to a typical manufacturers Trip Chart a breaker will carry the full current rating indefinitely and will carry additional current for proportionally less time. The chart shows how a typical 15 amp breaker will carry 2 times the rated current (30 amps or 3600 watts) for 10 seconds and even carry 5 times the rated current (75 amps or 9000 watts) for 1 second.

While not directly related, my compressor is rated for 15 amps at 120v and 7.5 at 220v (switchable), I can run the compressor continously on a 15 amp circuit without tripping the breaker. So maybe breakers are not created equally? Another unrelated example is one of my welders is rated at 20amps. I have plugged it in a 15 amp circuit and tacked 1/4" plate, so that tells me that a 15 amp breaker will exceed the current load for very short times.

NoMoney I think your examples are directly related to the operation of all residential thermomagnetic circuit breakers. A circuit breaker is not a current limiting device, it's just a bi-metallic strip conducting electricity..... to much current for to long and the strip heats and bends,tripping the breaker.

"Rated" is the key - does it actually consume 15 amps continuously, or is it rated to be used on a 15A circuit? Nameplate ratings don't necessarily equate to actual consumption.

To accurately measure the power output of an amplifier, all you need is a true RMS voltmeter with sufficient bandwidth and some 8 ohm power resistors of appropriate wattage rating. For a sine wave input/output, it's much easier as any RMS meter will do. Voltage squared divided by the resistance is power.

To be more accurate you could use an oscilloscope to view the waveform. This would allow you to adjust the drive until the amplifier was just below clipping. The oscilloscope will give you the peak value of the output waveform. For a sine wave, the RMS value is the peak value divided by the square root of two (about .7071). Apply the formula above and you have an answer.

McIntosh used to run clinics at various dealers where you could bring in a preamp or power amp and they would measure it and give you harmonic and IM distortion figures as well as the power output of the amplifier (at no charge). Most good repair shops have instrumentation that can measure amplifier output and harmonic distortion. For a fee, they'd probably be happy to measure and give you the numbers.

Steve

Correction to the above, before anyone else calls me on it. Square root two is 1.414; root two divided by two is the .7071 by which you multiply peak voltage to get RMS.

Sorry about that; fingers faster than brain today.

Steve