Question 21.4: In the Y-Y system of Figure 21–21, determine the following: ...

This system has a balanced load Z _{a}= Z _{b}= Z _{c}=22.4 \angle 26.6^{\circ}  \Omega.

(a) The load currents are

I _{Z a}=\frac{ V _{\theta a}}{ Z _{a}}=\frac{120 \angle 0^{\circ}  V }{22.4 \angle 26.6^{\circ}  \Omega}=5.36 \angle-26.6^{\circ} A

I _{Z b}=\frac{ V _{\theta b}}{ Z _{b}}=\frac{120 \angle 120^{\circ}  V }{22.4 \angle 26.6^{\circ}  \Omega}= 5 . 3 6 \angle 9 3 . 4 ^ { \circ } A

I _{Z c}=\frac{ V _{\theta c}}{ Z _{c}}=\frac{120 \angle-120^{\circ}  V }{22.4 \angle 26.6^{\circ}  \Omega}=5.36 \angle-147^{\circ} A

(b) The line currents are

I _{L 1}=5.36 \angle-26.6^{\circ} A

I _{L 2}= 5 .36 \angle 9 3 . 4 { }^{\circ} A

I _{L 3}= 5 . 3 6 \angle-147^{\circ} A

(c) The phase currents are

I _{\theta a}=5.36 \angle-26.6^{\circ} A

I _{\theta b}= 5 . 3 6 \angle 9 3 . 4 ^ { \circ } A

I _{\theta c}=5.36 \angle-147^{\circ} A

(d) I _{\text {neut }}= I _{Z a}+ I _{Z b}+ I _{Z c}

= 5.36 \angle-26.6^{\circ} A +5.36 \angle 93.4^{\circ} A +5.36 \angle-147^{\circ}  A

= (4.80 A -j 2.40 A )+(-0.33 A +j 5.35 A )+(-4.47 A -j 2.95 A )= 0  A

If the load impedances were not equal (unbalanced load), the neutral current would have a nonzero value.

(e) The load voltages are equal to the corresponding source phase voltages.

V_{Z a}=120 \angle 0^{\circ} V

V_{Z b}=120 \angle 120^{\circ} V

V_{Z c}=120 \angle-120^{\circ} V