>> Haha why spend $100 on a true wattmeter when a $5 resistor shunt and a
>> cheapo DMM will do the samething?
>> article mentions
>> It only cost $5, because it doesn't give out any meaningful result >when
>> you're measuring anything more complex than Christmas lights and table >> lamps.
>> Measure computer or refrigerator's power consumption? yeah right. >It's
>> sad TechTV is supporting this non-sense.
>> Some people just don't get that things aren't as simple as basic P=I*V >>
>> First of all, they're totally ignoring power factor and when you're
>> dealing with a load with a PF of 0.5 to 0.6 and decide to ignore it, >you
>> might as well not measure it.
>> You can't measure the current draw of most computer power supplies >with a
>> cheap DMM either in A mode or V mode with a shunt. Except with a true
>> DMM multimeter, something cheapskate is unlikely to have, the readings
>> from measuring a peculiar current waveform of a computer power supply >is
>> Test result for my CRT(which has a switch mode power supply just like >a
>> PC and I'm not going to unplug my PC right now):
>> Line voltage is 123.3V
>> True power meter reports:
>> PF 0.63
>> Regular DMM reports:
>> 123.3*0.67A= 83W
>> Damn talk about way off.
>I had the same problem when I tried to measure the current and power of
>a $5 Intermatic timer, the one with the dial with the pegs on it on the
>front. It measured more than 2.5W with the DMM, but it was probably
>less than 1.8W when the PF was taken into consideration. The motor has
>a lot of reactance.
You could take a simultaneous voltage [Not the voltage drop across the series (shunt) resistor, but the SOURCE voltage] reading, integrate the two instantaneous values over a period of 2*pi radians, and do a vector sum. That would give you true RMS power, no?
-- -john wide-open at throttle dot info
~~~~~~~~ "The first step in intelligent tinkering is to save all the parts." - Aldo Leopold ~~~~~~~~
I'm guessing that you meant a simultaneous voltage and current measurement. You seem to be proposing an incorrect mathematical approach to a real world issue. You can't integrate two instantaneous values of two different things. Could you please re-state this to make it clear.
My guess is you really mean integrate all V readings from 0 to 2 pi radians, and also integrate the I readings for the same 360 degrees. But that mathematical solution is still incomplete - you then need to multiply. Is that what you have in mind?
Now, if you mean this to be a practical solution, what instrumentation do you buy to sample the V and I, how many samples does it take during the cycle, how are the measurements stored, and how do you integrate them?
It's much be cheaper and easier to buy a true RMS DMM and measure the V across a shunt resistor, then compute the power. Hell, I've read even have a power measurement device for about 40 bucks, which is even cheaper than a true RMS DMM, and involves no computation. I don't know if it's any good.
Not vector sum, rather multiply the two, point by point and average over the cycle. BTW, it's average power (RMS power makes no sense). The vector sum would give you the power factor (or more precisely lead/lag angle). Of course the power factor is also the RMS(V) * RMS(I) / average power. Yes, this all is fairly easy to do with a microcontroller and a couple of ADCs.