I am not an electrical engineer (I did work as an automation technologist a long time ago, but please forgive me my electricity is very rusty). I've been given some data that was measured from a 3-phase delta system using current transformers and voltage leads on a pump system.

The person who gave me it claims that the software for the logger was configured incorrectly, and that the power consumed is half of what it's reading. He insists that the correct way to calculate the power in this situation is to use the averages of the 3 phases (\$V_{avg} \times I_{avg} \times pf_{avg} \times \sqrt{3}\$).

The meter however is using the summation of the three individual phase powers (\$V_1 \times I_1 \times pf_1 + ... \$).

My question is: should these values not come out to be the same in the end? Should we not be able to calculate the power using both methods?

ETA: Here is a sample row of the metering data I have been given, to try and make what I'm working with more clear:

Sample Data

Each of the 3 real channels are measuring a phase voltage, current and power factor - so my approach was to just calculate power using each phase measurement and summing them (much like the meter did). This should remain the same regardless of the way the system is wired right? (same calculation for Wye / Delta?)

  • \$\begingroup\$ Some other sources of error are lack of RMS current vs peak or average current and phase. Realtime VIcos(theta) or 3phi watt meter is best. Others measure avg I over x cycles and avg V which may be ok or not if high pulse unbalanced currents. \$\endgroup\$ Commented May 26, 2017 at 4:13
  • \$\begingroup\$ Yes. I assume Power 5 is your line numbers. \$\endgroup\$ Commented May 26, 2017 at 13:12
  • 1
    \$\begingroup\$ You cannot average power factor. Imagine if one phase had 1000kVA but power factor of 0.01 giving real power of 10kW Another phase had 10kVA with a power factor of 1.00 also giving 10kW. With the third phase having no load the total power would be 20kW. If you calculate the average first you would have a power factor of .505 (ignoring the third phase) with 1010kVA giving an incorrect power of 510.05kW instead of 20kW. The power factor is related to one phase only if the phases are unbalanced in any way. With a balanced load all three power factors should be the same and it will not matter. \$\endgroup\$
    – KalleMP
    Commented Feb 20, 2020 at 19:58

3 Answers 3


The average of three line-to-line voltages multiplied by the average of three line currents multiplied by pf and sqrt(3) will give a reasonably accurate results if the phases are reasonably well balanced.

The sum of three wattmeter reading will only be correct if the measurements are based of the line-to-neutral voltages and the line currents or the line-to-line voltages and the delta phase currents.

If two wattmeters measure the power based on any two line currents and the line-to-line voltage between the iines in which the current is measured and the third line, the power is the sum of the two wattmeter readings. The system does not need to be balanced for the two-wattmeter method to be accurate.


1, 2 and 3 are delta phase voltages. \$cos\theta\$ is fancy electrotech speak for power factor.

$$ P_T = P_1 + P_2 + P_3$$ $$ P_T = V_1 I_1 cos {\theta}_1 + V_2 I_2 cos {\theta}_2 + V_3 I_3 cos {\theta}_3$$

Three-phase balanced delta, so phase voltage, currents and pf are the same. $$\color {red} {P_T = 3 V_{Phase} I_{Phase} cos {\theta}_{Phase}}$$

For a delta, \$ V_{Line} = V_{Phase}\$ and \$ I_{Line} = \sqrt {3}I_{Phase}\$, so using line values.

$$P_T = 3 V_{Line} \frac {I_{Line}} {\sqrt {3}} cos {\theta}_{Line}$$ $$\color {red} {P_T = \sqrt {3} V_{Line} I_{Line} cos {\theta}_{Line}}$$

So you are correct. Whether you use line quantities or phase quantities, the answers will be the same. This works for the load being connected up in wye or delta.

You can use averages for your line quantities. Not sure if you have access to power factor.

Now, you have to reconcile the math and measurements with your co-worker.


If your device measures phase voltages \$V_1\$, \$V_2\$ and \$V_3\$, and phase currents \$I_1\$, \$I_2\$ and \$I_3\$, you have to add the active power \$P = P_1 + P_2 +P_3\$ in [kW] and the apparent power \$S = S_1 + S_2 +S_3\$ in [kVA].

The power factor of the total load is then $$pf = \frac{P}{S}$$ irrespective if the load is connected in star or delta.

Due to the fact that the power factor of each phase is not the same, you will not get the correct answer if you use any other method.

As a side note:

If you have any distortion powers, the power factor $$pf \ne \frac{P}{\sqrt{P^2+Q^2}} $$ with \$P\$ in [kW] and \$Q\$ in [kvar] due to the distortion power \$D\$, also in [kvar].

In such a case \$S^2 = P^2 + Q^2 + D^2\$. However, the average power factor \$pf\$ will still be correct if the formula in the answer is used.


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