Vbc=Vb-Va = Vb-Vc+ (Vn-Vn) = (Vb-Vn)-(Vc-Vn) = Vbn-Vcn
(-120)-(120)= -240

Vca = Vc-Va = Vc-Va + (Vn-Vn) = (Vc-Vn)-(Va-Vn) = Vcn-Van
120 - 0 = 120

 

Lul I actually used your method and got his answer
Apparent power (q) Vab = 5760Var
Power = 5760W

Vbc
q= 5760
Power = 4988.31

Vca
q = 3840
p = 3325.54

ptot = 14073.85w

Now so do I. I must have transcribed something wrong on the scrap paper I was using, which has since been thrown away.

 

There is no "N", not is there any 120V in your original question and sketch, I don't know why you've concocted them now. Just solve the problem, don't create others.

 

There is no "N", not is there any 120V in your original question and sketch, I don't know why you've concocted them now. Just solve the problem, don't create others.

The 'n' came about by trying to understand the line-line phase relationships given the phase voltage sequence polarity.