I have sampled values of voltage (phase to ground) and current (line current). I can calculate instantaneous apparent power using: $$ P(i) = V_1(i)I_1(i) +V_2(i)I_2(i) +V_3(i)I_3(i), i\in [1,n]$$

Question: Is there a method to calculate instantaneous active (or reactive) power from sampled data instead of apparent power?

  • \$\begingroup\$ Are these RMS measurements of voltage and current or are the voltage and current values almost a continuous stream of value at a high sampling rate? \$\endgroup\$
    – Andy aka
    Nov 6, 2015 at 12:08
  • \$\begingroup\$ Yes, if you can determine the peak values of I and V, the time shift between these two peak values, and the periodic time. \$\endgroup\$
    – Chu
    Nov 6, 2015 at 23:49

2 Answers 2


Instantaneous power is simply the instantaneous voltage times instantaneous current. This is true regardless of the power factor, amount of "reactive power", phase angle, or any other way of expressing this same issue.

To determine phase angle (or power factor, "reactive power", etc) takes more than one reading. Average real power is the average of instantaneous power.

real power = ave(volt x amps)

The power factor is the real power divided by the "VA" (volt-amps):

power factor = (real power) / (RMS(volts) * RMS(amps))

From that you can calculate the real and imaginary components of the power, and the phase angle if you know the waveshape.


If I got it correctly,then you are probably asking about measuring actual power(active/reactive) of a design by taking readings(power/voltage/current) at different time instances. well if it sounds fine then you simply minimize the error b/w instantaneous & actual by taking more samples. enter image description here


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