CI Volume 4, problem 14

[Zack J.]

So the question states:

A 34.5kv, 3-phase, 4-wire, wye-connected distribution planel is experiencing an unbalanced load situation. On phase A (19.9<0) there is a 120+j65 kva load, on phase C (19.9<0) there is a 65+j20 kva load. Calculate the magnitude of the neutral current.

Ok, so I get the fact that V phase A = 19.9<0 and V phase C = 19.9<240 (assuming ABC sequence) and that Ineutral will equal the sum of the current in phase A and C or Ia+Ic-In=0.

Why is the solution then the impedance of the load divided by the voltage equals the current?? Thought V=I * Z. CI in the solution also shows the currents to be the conejugate of the current? Why is this?

Thanks in advance for your help.

 

[Wael]

It is the Apparent Power (KVA) which is divided by the voltage to get the current. S= 120+j65 kva is S not Z

S=VI*

All the best

 

[Zack J.]

Thanks Wael