Electrical – Total power measurement in 3-phase unbalanced delta system

If you have resistors from red to blue, from blue to yellow and from yellow to red (i.e. a delta formation load) then the power is the sum of the individual line voltages (squared) divided by the individual resistance across each line:

$$P = \frac{V_{RB}^2}{R_{RB}} + \frac{V_{BY}^2}{R_{BY}} + \frac{V_{YR}^2}{R_{YR}}$$

If you have resistors in a star formation this is more difficult unless you have a neutral wire commoning the three resistors. If you do then measure the individual phase voltages and do individual power calculations then add the three powers to give total load power.

If you don't have a neutral then you will have to calculate the star-point voltage relative to red, blue and yellow respectively. Then you'll have three voltages and three resistors and power is the sum of the individual powers.

I don't see any way to do this. You need the power factor of the individual phases and that's not available. The equation is incorrect.

\$P_{actL1} = P_{appL1}\cdot\cos\varphi_1\$

\$P_{act,tot}/P_{app,tot}= \$

\$= (U\cdot I_{l1}\cdot \cos\varphi_1 + U\cdot I_{l2}\cdot \cos\varphi_2 + U\cdot I_{l3}\cdot \cos\varphi_3)/(I_{l1}\cdot U + I_{l2}\cdot U + I_{l3}\cdot U)\$

\$= ( I_{l1}\cdot \cos\varphi_1 + I_{l2}\cdot \cos\varphi_2 + I_{l3}\cdot \cos\varphi_3)/(I_{l1} + I_{l2} + I_{l3})\$

and that is not equal to \$\cos\varphi_1\$.