Question 23.1: The phase sequence of the Y-connected generator in Fig. 23.1...
The phase sequence of the Y-connected generator in Fig. 23.13 is ABC.
a. Find the phase angles \theta_{2} \text { and } \theta_{3} .
b. Find the magnitude of the line voltages.
c. Find the line currents.
d. Verify that, since the load is balanced, I _{N}=0.
a. For an ABC phase sequence,
\theta_{2}=-120^{\circ} \quad \text { and } \quad \theta_{3}=+120^{\circ}.
\text { b. } E_{L}=\sqrt{3} E_{\phi}=(1.73)(120 V )=208 V \text {. Therefore, }E_{A B}=E_{B C}=E_{C A}=208 V.
\text { c. } V _{\phi}= E _{\phi} . \text { Therefore, }V _{a n}= E _{A N} \quad \quad \quad V _{b n}= E _{B N} \quad \quad \quad V _{c n}= E _{C N}.
I _{\phi L}= I _{a n}=\frac{ V _{a n}}{ Z _{a n}}=\frac{120 V \angle 0^{\circ}}{3 \Omega+j 4 \Omega}=\frac{120 V \angle 0^{\circ}}{5 \Omega \angle 53.13^{\circ}}.
=24 A \angle-53.13^{\circ}.
I _{b n}=\frac{ V _{b n}}{ Z _{b n}}=\frac{120 V \angle-120^{\circ}}{5 \Omega \angle 53.13^{\circ}}=24 A \angle-173.13^{\circ}.
I _{c n}=\frac{ V _{c n}}{ Z _{c n}}=\frac{120 V \angle+120^{\circ}}{5 \Omega \angle 53.13^{\circ}}=24 A \angle 66.87^{\circ}.
\text { and, since } I _{L}= I _{\phi L},I _{A a}= I _{a n}=24 A \angle-53.13^{\circ}.
I _{B b}= I _{b n}=24 A \angle-173.13^{\circ}.
I _{C c}= I _{c n}=24 A \angle 66.87^{\circ}.
d. Applying Kirchhoff’s current law, we have
I _{N}= I _{A a}+ I _{B b}+ I _{C c}.
In rectangular form,
I _{A a}=24 A \angle-53.13^{\circ} \ = \ 14.40 A -j 19.20 A.
I _{B b}=24 A \angle-173.13^{\circ}=-22.83 A -j 2.87 A.
I _{C c}=24 A \angle 66.87^{\circ} \quad = \quad \underline{ 9.43 A +j 22.07 A } .
\sum\left( I _{A a}+ I _{B b}+ I _{C c}\right) \quad=\quad 0+j 0.
\text { and } I _{N} \text { is in fact equals to zero, as required for a balanced load. }