Calculating true RMS current
- Thread startermicrostar
- Start date
There's nothing to calculate. If the ammeter is a true RMS type, take its reading at face value and it will accurately relate to the heating of wires, fuses, and circuit breakers. If the meter isn't true RMS, then the measurement is inaccurate except when the dimmer is a full output. Basically, true RMS means it can properly measure even when the wave form is not sinusoidal.
If the meter isn't a true RMS meter, you are pretty much out of luck, practically speaking. You'd need to know exactly how the meter was measuring things (most likely average responding) and the exact waveform being measured--which, in this case, mostly means the phase angle at which the dimmer is switching for whatever setting you're using--and then doing the mathematical analysis to get an appropriate conversion factor. That is assuming the frequency response of the clamp-on part of the meter is sufficiently high to follow the transients of the current waveform with adequate fidelity; otherwise, that too would have to be taken into account.
If it's a peak-responding meter, which would be quite a surprise these days, the results will be useless for the upper 50% of the conduction range since the peak voltage doesn't change between 50% and 100% on time from the dimmer.
Just in case it's not clear, "average responding" means that the meter is actually measuring the average of the voltage over time--integrating it, in other words, which is pretty easily accomplished with a capacitor circuit--and scaling the output so that, for a sine wave, the number displayed corresponds to a correct RMS value. An RMS responding meter (a "true RMS" meter) averages the square of the input value and then takes the square root of the result, hence root-mean-square or RMS. A "peak responding" meter measures the extreme (peak-to-peak) voltage internally and scales that to display an RMS value that would be correct for a sine-wave input. Peak responding meters are quite uncommon, and have been for decades, partly because transient spikes can lead to wildly inaccurate measurements with them in many practical applications.
If it's a peak-responding meter, which would be quite a surprise these days, the results will be useless for the upper 50% of the conduction range since the peak voltage doesn't change between 50% and 100% on time from the dimmer.
Just in case it's not clear, "average responding" means that the meter is actually measuring the average of the voltage over time--integrating it, in other words, which is pretty easily accomplished with a capacitor circuit--and scaling the output so that, for a sine wave, the number displayed corresponds to a correct RMS value. An RMS responding meter (a "true RMS" meter) averages the square of the input value and then takes the square root of the result, hence root-mean-square or RMS. A "peak responding" meter measures the extreme (peak-to-peak) voltage internally and scales that to display an RMS value that would be correct for a sine-wave input. Peak responding meters are quite uncommon, and have been for decades, partly because transient spikes can lead to wildly inaccurate measurements with them in many practical applications.
STEVETERRY
Well-Known Member
STEVETERRY
Well-Known Member
David Ashton
Well-Known Member
Sorry people but a true RMS meter will NOT give you an accurate reading of dimmer load if there are many dimmers at different levels, it is just too complex a waveform, if you don't believe me call the experts at Fluke, a cheap moving iron meter is actually more accurate, you also have the complication of DC content to stuff things up.
microstar
Well-Known Member
I did say I was measuring the current draw on "a dimmer". So on one dimmer, not many. I can see where you might interpret that to mean "a dimmer rack", but not the case here.
microstar
Well-Known Member
The whole purpose of what I was trying to do was to figure out what wattage PAR64 lamps were on EACH individual dimmer. There were a pair of lamps per dimmer. The fixtures were not accessible at the time.
By measuring the current draw, I could figure out if it was 1kw or 500 watt lamps. The 8 amp figure mystified me because it was two of the brighter lamps, not the weaker ones. The individual dimmer
was set to full output by means of the Test button on the dimmer. The amperage figure did not make sense so that was the basis of my question.
David Ashton
Well-Known Member
microstar
Well-Known Member
120vac
David Ashton
Well-Known Member
OK, now I understand your confusion, you see you are measuring a lamp which is a resistive load so your ordinary meter is giving you accurate data, the 8A at 120V is 960W or in practice 1K, the reason for wanting true RMS meters is that many circuits contain inductive or capacitive currents which are out of phase or have dimmers chopping bits out of the front or back of the waveform or switch mode power supplies sucking their power at the top of the waveform, so the complications which lead to horrendously complex currents which use "true RMS" meters which come close to sorting all that out. However you can ignore all that as it is not relevant to your situation which is to use basic Ohms law for your calculations.
Some/much pertinent info in this thread:
Based on the data, it appears that: at FULL, the Fluke 30 is nominally close to the 336. However, with dimmers at 50%, the Fluke 30 UNDER-READS by about 15 to 30%: Consistent across phases in each rack, but varies from rack to rack. Notice that the neutral is approximately 1.33 times higher than any hot phase, hence the reason for the double neutral on Touring Racks, that I personally have never seen used, but might consider it if I exceeded 300A on 4/0 feeder.
...
So as much as it kills me to admit it, STEVETERRY is correct.One must use a trueRMS meter to get accurate readings with a less-than-100% level, but the Fluke 30 is accurate enough at measuring total load. So I still have mixed emotions as to whether it was worth it for me to buy a new meter. I still think I should have bought the model 337.
STEVETERRY
Well-Known Member
jhochb
Active Member
Good Morning
A 1K calculates
W/V=A
1000/120= 8.33333
A rough guild
convert
RMS to Peak is 1.414
peak to RMS is .707
Ohms Law is for DC calculations
It work for AC for simple calculations
Frequencies and harmonics complicate things
yes
you can get up to 3 or more times the current on the neutral when the loads are in/out of balance and between 30 - 70%
A 1K calculates
W/V=A
1000/120= 8.33333
A rough guild
convert
RMS to Peak is 1.414
peak to RMS is .707
Ohms Law is for DC calculations
It work for AC for simple calculations
Frequencies and harmonics complicate things
yes
you can get up to 3 or more times the current on the neutral when the loads are in/out of balance and between 30 - 70%
Last edited:
Um, no. Were that true, ETC would have been building touring racks with THREE rather than double neutral.
See the thread https://www.controlbooth.com/threads/why-two-neutral-camlok-inputs.8986/
and
