Answer:
The phase current in each line conductor are;
[tex]I_{R} = 100.17 < 0A[/tex]
[tex]I_{Y} = 75.13< - 120A[/tex]
[tex]I_{B} = 50.08 <120A[/tex]
Explanation:
Given the following data;
Red phase = 24kW,
Yellow phase = 18kW
Blue phase = 12kW
Line voltage = 415V
For a star connected system, we have;
[tex]Phase voltage (V_{p} ) = \frac{Line voltage}{\sqrt{3}}[/tex]
[tex]Phase voltage (V_{p} ) = \frac{415}{\sqrt{3}}[/tex]
[tex]Phase voltage (V_{p} ) = 239.6V[/tex]
The phase sequence for RYB is given by;
[tex]V_{R} = 239.6<0\\V_{Y} = 239.6<120\\V_{B} = 239.6<-120[/tex]
[tex]Phase current (I) = \frac{Phase power}{Phase voltage}[/tex]
[tex]Hence, I = \frac{P}{V}[/tex]
For the Red phase;
[tex]I_{R} = \frac{24000}{239.6<0}[/tex]
[tex]I_{R} = 100.17 < 0A[/tex]
For the Yellow phase;
[tex]I_{Y} = \frac{18000}{239.6<120}[/tex]
[tex]I_{Y} = 75.13< - 120A[/tex]
For the Blue phase;
[tex]I_{B} = \frac{12000}{239.6<-120}[/tex]
[tex]I_{B} = 50.08 <120A[/tex]
For the line neutral;
[tex]I_{N} =\sqrt{ (I_{R}^{2} +I_{Y}^{2}+I_{B}^{2}-I_{R}I_{Y}-I_{Y}I_{B}-I_{R}I_{B}[/tex]
Substituting we have, [tex]I_{N} = 43.29A[/tex]
