A whirling magnet/coil gizmo can be a motor or generator.
Motors have a net power input. Generators have a net power output. The exact nature of a thing spinning around is NOT determined by averages of current and voltage, because the power could be positive or negative. That 'Irms' and 'Vrms' can only give a limit of the magnitude of the power, which means the number you get can be correct, or 200% too high. The factor that the averages miss can be anywhere in the (-1, +1) range and (for simple sinewave case) is the sine of a phase angle.
Multiply I times V, and average THAT, and you have power. It is possible to divide power by Irms and Vrms to get sine(phase)
** The OP has a " brushless DC " fan motor - you ass.
" What I'm trying to do: measure the current and voltage going into each winding of a brushless fan motor, each winding is recieving some ugly looking pulses in order to run the motor, I'm intersted in find the power per each winding. "
** There are no sine wave voltages or currents involved.
You don't need to do all that work. Sample the currend and voltage them multiply the instantaneous values, average over a cycle, or hundred. No need to convert to RMS anything, though that might be interesting (VA measurement) too.
If you have access to the voltage and current waveforms, simply sample them at the same time and multiply to get the instantaneous power, then average over some time (a cycle or hundred is good). There is no need to do any ugly math, though VA might be interesting too... BTW, a current transformer may be better than a resistor.
The most popular hand-held meter that gave true rms voltage used to be the Fluke 87, which could cover the audio bandwidth, using an analog rms converter (100KHz limit). It produced rms current measurements bt measuring a shunt resistor voltage, which is the most common current measurement method.Other meters are available that use the same kind of hardware.
The old Fluke 8922a bench powered meter used to be usefull with higher voltage/power circuits. It used a thermal converter, followed by conventional voltage metering and is highly sensitive to calibration. This has a 1MHz or 11MHz bandwidth, depending on input voltage range. There are other similar meters.
More recently, given the availability of digital scopes with good bandwidth and math capability, mathematical derivation is more common.
You can't just multiply an RMS current and RMS voltage measurement to get power. Their phase relationship is important. As an extreme example, an amplifier driving an inductor with 1Vrms will dissipate internal losses equivalent to driving a short circuit, while the load power is minimal.
A digital scope math calculation uses time-coherent multiplication to produce a fairly reliable power indication for related waveforms in its display/memory.
DC-offset will also affect measurements, the ability of the meter to produce meaningful readings and the operator's ability to understand what the reading means.
"You can't just multiply an RMS current and RMS voltage measurement to get power." - RL
So.... I've read that Vrms*Irms = Pavg..... but..... you're saying I need to include the phase shift?! Isn't the phase shift implied in the waveform? I have two meters hooked up, and I'm reading a Vrms and an Irms, if I multiply these together don't I get the Apparent Power? Aren't RMS values scalars? No phaeses involved.... and isn't Vrms*Irms = P apparent?
You really do have to worry about the phase shift. When the voltage and current are in phase, the power is going in the same direction as the current. When they're out of phase, the power is going in the opposite direction. From the wall socket's perspective, that's the difference between a motor and a generator.
If you imagine a load that drew some enormous current right when the voltage was near 0, and much less current near the peaks, you could be off by a factor of 100 or more if you just multiplied RMS voltage times RMS current. The same is true of a purely reactive load, in which the voltage is 1/4 cycle out of phase with the current. For instance, this is what it would look like with a capacitor across the line.
I___ ___ / X \\ V / / \\ \\ /---/---\\---\\---/---------------------- / \\ \\ / / \\ X
-- --- -- | | | | | | | | | | 1 2 3 4
In quarter-cycle 1, I >0 and V < 0, so the VI < 0 : power is going from the capacitor into the wall socket. In QC 2, I>0 and V>0, so VI > 0: the power is going from the wall socket into the capacitor. In QC 3, I0, so VI < 0 again, and In QC 4, I
Hmmmmm..... ok, well my voltages will definitely be out of phase since there's some reactance involved.... and I agree with everything you are saying here, but this is looking at just one cycle of each.... I'm assuming a meter is constantly sampling the signal and so the RMS value is some sort of average that is constantly being updated, and the average is maybe taken over a hundred cycles or somethign like that not just one cycle, so do you think that since the meter is taking an rms average of many cycles that somehow the phase shift would then be included in there when I multiply the values together? I'm fine with the fact that I'm getting the reacitve power (the negative power going back to the source) in my readings, I'm fine with the fact that I'm getting apparent power instead of true/active power... but I only have meters and am trying to avoid the phase issue.... I'm using the two-wattmeter method to do this (thanks Phantom)
yes, completely agree, that a purely reactive circuit has no real/true power component to it. I don't mind that, I know that I am measuring Apparent Power which is not True Power.... but my question is, Do I Have To Worry About The Phase Shift For Making Apparent Power Measurements By Taking RMS Voltage and Current Readings From A DMM... at this point, I do not think that I need to know the phase shift to make a measurement of the apparent power... Vrms*Irms=apparent power, RMS values are scalars and contain no phase information...
Apparent power is what you'd get by doing that. On the other hand, you might just as well throw darts at a piece of graph paper. "Apparent power" is not a term of praise.
If you know the sign of the reactance, you could correct the power factor and measure it afterwards.
What you're doing isn't the two-wattmeter method. You are using two separate meters and two separate meters don't make one wattmeter:
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The essence of an electrodynamometer wattmeter is that the movement has two coils, a current coil and a voltage coil. The torque that deflects the needle is proportional to the instantaneous product of the current and voltage. The mechanical inertia of the movement smooths out the pulsations that might occur due to the fact that the instantaneous power varies from 0 to some magnitude rapidly.
The reactive power is dealt with by the fact that when the power is returning to the source, the needle would ordinarily be deflected backward, so the average of forward and reverse power is what needle indicates.
When you use two separate RMS responding meters, you are in fact measureing apparent power. The real power will always be less than or equal to the apparent power. If the load is purely resistive, they will be equal. If the load is more or less reactive, the real power will be less than the apparent power, and sometimes the real power may be MUCH less than the apparent power.
If you apply, for example, a 24 VAC stimulus (using a transformer for safety) to a 10 microfarad capacitor (not an electrolytic), the apparent power would be
2.17 volt-amperes. I have such a capacitor lying around in my junk box, and it has an ESR of 50 milliohms. The real power drawn in this case would only be 4.5 milliwatts. This is almost a one thousand to one ratio of apparent power to real power.
Furthermore, you have what amounts to a three-phase load. If you let one lead of the motor circuit be reference, and measure the voltage and current in the other two with respect to that reference with separate RMS responding meters, you will not be carrying out the two-wattmeter method. You have to use actual wattmeters, or their equivalent, a two channel scope with trace math.
The three phase nature of your load additionally complicates things. I think that since your load is probably inductance + resistance, the two separate measurements probably add up to the apparent power, but I'd have to think about a little to be sure.
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