I need to measure and log power, and I want to do it mostly correctly. I would greatly appreciate a listing of the errors I have made in my method. I want to measure (and log) the power output of a small 3-phase generator.


200 w, 24 v, 200 rpm - 3 phase delta generator

200 w buck/boost mppt controller

-10 v to 10 v DATAQ DI-149 data logger

Three 2-ohm, 25 w. resistors

I think logging the voltage is not a problem: measure across the 75k of a 75k/91k pair. The DI-149 can internally convert to rms and log same.

The DI-149 measures the 20 v. span in increments of 0.02 v. My plan was to put a 2 ohm resistor between each of the 3 outputs of the generator and the inputs of the MPPT and log the rms voltage drop and convert to current.

I will multiply the rms voltage and calculated current to arrive at power (and then throw in I^2 R to account for the power lost in the resistor?)?

I would be satisfied with +/- 10% accuracy. Is this going to get me there?

  • 1
    \$\begingroup\$ No. Power is not calculated by multiplying voltage with current. Also your current sensing resistor wastes way too much energy. Why not just buy a logger that logs power directly, or a power meter and a logger and connect them together? \$\endgroup\$ Commented May 15, 2017 at 20:21
  • \$\begingroup\$ Multiplying Vrms with Irms does not give you power in Watts - it gives you apparent power in VA. For each sample period you need to multiply the V and I samples to get a power value and then rms that. \$\endgroup\$
    – brhans
    Commented May 15, 2017 at 21:01
  • \$\begingroup\$ @brhans I'm afraid power value samples need to be averaged not RMS'ed. \$\endgroup\$
    – carloc
    Commented May 15, 2017 at 22:01
  • \$\begingroup\$ RMS power requires that each V*I is done in real time or filtered and sampled subcycle time then averaged between readings. \$\endgroup\$ Commented May 15, 2017 at 22:13
  • \$\begingroup\$ but if V is reasonably constant over say 10 cycles then you can compute Rms V and I separately but you must measure the RMS phase difference to extract real power from apparent power. So you see a true RMS meter from Fluke is much better. To my knowledge DATAQ has no builtin way to compute V*I(t) real power only Vrms. \$\endgroup\$ Commented May 15, 2017 at 23:20

2 Answers 2


I will multiply the rms voltage and calculated current to arrive at power (and then throw in I^2 R to account for the power lost in the resistor?)?

AC power is more complicated. If you multiply like DC, P=U*I, then you know S, the apparent power in VA. You have to multiply S with the phase angle φ to get the Real Power P in Watts.

I suggest you read the AC power wiki page, and use those formula's to compute your values.

Since you can sample all required 6 channels (voltage and current per phase) at 10 Khz you should be able to get at least some accurate results. However, your buck/boost might cause an inaccuracy in the measurement due to distortion.

You will be in for some serious software work though.


Since you don't know how true power is measured, and you cannot use a true RMS voltmeter to do this, I would suggest you have the wrong datalogger for this simple job.

Perhaps you can rent a Fluke 1736.


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