what is the unbalanced current flowing in the neutral of this three-phase,four wire,208/120-volt system? A phase=52 amperes, B phase=44 amperes, C phase=46 amperes.


A. 7.21 amperes
B. 7.07 amperes
C. 7.34 amperes
D. 6.78 amperes

Please show the computation and thanks for any help and it's highly appreciated al_benigno@comcast.net

 

\(\sqrt{(Ia^{2}+Ib^{2}+Ib^{2})-(Ia*Ib)-(Ib*Ic)-(Ia*Ic)}\)

 

Why not simply

\( I_n=I_a\angle{0^o}+I_b\angle{-120^o}+I_c\angle{-240^o}\)

 

Hi! thanks for help I able to answer problems, more power to you

 

Why not simply

\( I_n=I_a\angle{0^o}+I_b\angle{-120^o}+I_c\angle{-240^o}\)

Now that makes sense.

 

Sorry I don't know the masonic symbols. Not part of the club.

 

Hi! guys thanks it's over the problems was solved and I really appreciate your inputs more power to you all.

 

Sorry I don't know the masonic symbols. Not part of the club.

I think T_N_K's defined the negative sequence set.