I think the criteria for no unique solution is if the delta load currents do not sum to zero (zero-sequence current <> 0).If the phase currents are not separated by 120 degrees, there is no unique solution if we have only the line currents.
I think the criteria for no unique solution is if the delta load currents do not sum to zero (zero-sequence current <> 0).If the phase currents are not separated by 120 degrees, there is no unique solution if we have only the line currents.
Well I didn't double check but your math appears good (I'm not getting any bad vibes ). But this makes my [disguised] point... Doesn't the circuit described have circulating current? i.e. there will be energy (which requires current) passed back and forth between the capacitive and inductive elements as long as the load is energized, and that energy is only dissipated upon deenergization of the circuit. That energy comes in on the lines during the first [few?] cycleWell, double check me on this...but if I had a system with Vab=208<-30, Vbc=208<-150 and Vca=208<90 and between A-B, I connect an inductive impedance of 15.969<6.9, between B-C, I connect a capacitive impedance of 19.941<-18.56, and between C-A, I connect an inductive impedance of 28.139<13.74. My phase currents should be:
Iab=13.025<-36.9
Ibc=10.431<-131.44
Ica=7.392<76.26
In rectangular form:
Iab=10.416-j7.820
Ibc=-6.904-j7.820
Ica= 1.756+j7.180
Iab+Ibc+Ica = 5.268-j8.460 = 9.97<-58.09
Since Ia=Iab-Ica, Ib=Ibc-Iab & Ic=Ica-Ibc, we get
Ia=8.660-j15.000
Ib=-17.320+j0.00
Ic=8.660+j15.000
Ia+Ib+Ic= 0+j0 = 0
This is an unbalanced (and unequal power factor) delta connected loads, where the sum of the line currents equals zero, but the sum of the phase currents does not.
Or, a simpler example:
Vab=208<-30, Vbc=208<-150 and Vca=208<90 and between A-B, I connect a 10 ohm resistor. B-C and C-A are open. Iab = 20.8<-30. Ibc=Ica=0
Iab+Ibc+Ica=Iab=20.8<-30 not =0
Ia=20.8<-30, Ib=20.8<150, Ic=0, Ia+Ib+Ic=0
Or, a simpler example:
Vab=208<-30, Vbc=208<-150 and Vca=208<90 and between A-B, I connect a 10 ohm resistor. B-C and C-A are open. Iab = 20.8<-30. Ibc=Ica=0
Iab+Ibc+Ica=Iab=20.8<-30 not =0
Ia=20.8<-30, Ib=20.8<150, Ic=0, Ia+Ib+Ic=0
And to answer the question about load currents being 120? apart, just add 10 kohm resistors in B-C and C-A. Then these load currents will be 20.8 mA, all power factors the same, all currents 120? apart, but the sum of the currents is still very close to 20.8<-30.Or, a simpler example:
Vab=208<-30, Vbc=208<-150 and Vca=208<90 and between A-B, I connect a 10 ohm resistor. B-C and C-A are open. Iab = 20.8<-30. Ibc=Ica=0
Iab+Ibc+Ica=Iab=20.8<-30 not =0
Ia=20.8<-30, Ib=20.8<150, Ic=0, Ia+Ib+Ic=0