Question 23.2: For the three-phase system in Fig. 23.15: a. Find the phase ...
For the three-phase system in Fig. 23.15:
a. Find the phase angles \theta_{2} \text { and } \theta_{3}.
b. Find the current in each phase of the load.
c. Find the magnitude of the line currents.
a. For an ABC sequence,
\theta_{2}=-120^{\circ} \quad \text { and } \quad \theta_{3}=+120^{\circ}.
\text { b. } V _{\phi}= E _{L} \text {. Therefore, }V _{a b}= E _{A B} \quad V _{c a}= E _{C A} \quad V _{b c}= E _{B C}.
The phase currents are
I _{a b}=\frac{ V _{a b}}{ Z _{a b}}=\frac{150 V \angle 0^{\circ}}{6 \Omega+j 8 \Omega}=\frac{150 V \angle 0^{\circ}}{10 \Omega \angle 53.13^{\circ}}=15 A \angle- 5 3 . 1 3 ^{\circ}.
I _{b c}=\frac{ V _{b c}}{ Z _{b c}}=\frac{150 V \angle-120^{\circ}}{10 \Omega \angle 53.13^{\circ}}= 1 5 A \angle- 1 7 3 . 1 3 ^{\circ}.
I _{c a}=\frac{ V _{c a}}{ Z _{c a}}=\frac{150 V \angle+120^{\circ}}{10 \Omega \angle 53.13^{\circ}}=15 A \angle 6 6 . 8 7 ^{\circ}.
\text { c. } I_{L}=\sqrt{3} I_{\phi}=(1.73)(15 A )=25.95 A \text {. Therefore, }I_{A a}=I_{B b}=I_{C c}=25.95 A.