AC/DC question

You would be well advised to completely forget the concept "RMS power". While it can be defined mathematically, it has no physical significance whatsoever. It was a term cooked up by idiots in the hi-fi marketplace of the '60s. These guys simply did not understand the reason for existance of the concepts rmms voltage, rms current and _average_ power.
RMS voltage and current are, however, entirely meaningful. If you compare half- and full-wave _voltage_, you will find the rms voltage of the half wave to approx .707 times the rms voltage of the full wave. With a resistor load, Power=(voltage squared)/resistance thus when you square .707 (actually 1/SQRT(2)), you get 1/2 so the half wave rectified wave delivers exactly half the power to the load.
Ted
Reply to
Ted Edwards
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It isn't and thereby hangs the difficulty. "Ideal DC ammeter" is well defined. "Ideal AC ammeter" is not. Making the usual steady state assumptions, DC current is fundamentally electrons/second in a more convenient unit called coulombs/second or Ampere's. AC is fundamentally time dependent. Does the "ideal" meter read average, rms, peak or some other measure? Is the waveform sinusoidal or something else? e.g. for a square wave, average=peak=rms while for a sine wave, rms=(2^-.5)peak, average=0 and average of the absolute value = 2/pi. Other waveforms, other values.
Ted
Reply to
Ted Edwards
The ultimate true rms meter. :-) I think I still have one of those.
Ted
Reply to
Ted Edwards
Grants original premise did not include everything being perfect. It may be what he intended, but not what he said.
In what I said, I was pointing out that even with no power losses ( conservation of energy and all that ), the AC Power is not AC volts times AC amps. In the real world with losses small enough to be ignored, the AC current will be bigger than the DC current, because DC volts times DC amps does equal Power.
Dan
Reply to
dcaster
I meant one that reads RMS values in an ideal sense, and I also meant the source was a perfect sinusoid.
Grant
Reply to
Grant Erwin
Dan, You got close but missed the mark again. Grant's original statement supposedly dealt with any and all losses. The product of AC volts and AC amps would be exactly the same as DC volts and amps of the same value.
Bob Swinney
Reply to
Robert Swinney
Ted sez (about Grant's premise): " It isn't and thereby hangs the difficulty..."
Assume Grant's ideal meter to be a thermocouple whereby resulting temperature, not voltage, is the measured entity. Elimination of meter ballistics, inequalities of time, etc. must be assumed if we are to believe Grant's perfect meter. In which case, AC power = DC power, if fundamental precepts are considered. RMS AC is such a precept.
Bob Swinney
Reply to
Robert Swinney
I see him mentioning perfect AC ammeter and a prefect DC ammeter. I don't see him mentioning anything about losses. But as said, the transformer could be essentially lossless and still have a power factor that is not close to one.
Try this at home, measure the AC current to a transformer with the secondary open circuit. The AC current to the transformer will not be zero. The product of AC volts and AC amps will not be exactly the same as the product of the DC volts and Amps unless the open circuit Inductance is infinite. It has nothing to do with losses.
Dan
Reply to
dcaster
Grant sez:
"I meant one that reads RMS values in an ideal sense, and I also meant the source was a perfect sinusoid."
Grant approached the problem from the basis of pure physics. Most everyone jumped in to prove that bumblebee can't fly. It's natural. Engineers only measure Reality; physicists explain it.
Bob Swinney
Reply to
Robert Swinney
"Ted Edwards" wrote: You would be well advised to completely forget the concept "RMS power". (clip) ^^^^^^^^^^^^^ Thanks, Ted. "RMS power" is one of my pet gripes. I don't know how I missed it here. In the last thread I recall on the topic, there were people who didn't get it, even after it had been fully explained from several directions. Calculating RMS power is as meaningful as calculating miles per decibel on your car.
Reply to
Leo Lichtman
"Robert Swinney" wrote: Assume Grant's ideal meter to be a thermocouple whereby resulting temperature, not voltage, is the measured entity. (clip) ^^^^^^^^^^^^ A thermo-based ammeter would measure RMS amps, by definition, since it sees the current as having the same heating value as DC. But, if you think of current as coulombs/second, (relevant to electroplating) you want to know the AVERAGE current, which is different.
Reply to
Leo Lichtman
Yes but realize that the output will not be smooth DC but will be rectified half-sinusoid pulses. I doubt a DC meter has any guaranteed averaging so in such a situation it will simply read somewhere between the minimum and maximum instantaneous values. Also the type of loading is undefined and the type of load can influence the current waveshape.
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Reply to
Dave

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