Question 23.3: For the Δ-Δ system shown in Fig. 23.20: a. Find the phase an...
For the Δ-Δ system shown in Fig. 23.20:
a. Find the phase angles \theta_{2} \text { and } \theta_{3} for the specified phase sequence.
b. Find the current in each phase of the load.
c. Find the magnitude of the line currents.
a. For an ACB phase 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{120 V \angle 0^{\circ}}{\frac{\left(5 \Omega \angle 0^{\circ}\right)\left(5 \Omega \angle-90^{\circ}\right)}{5 \Omega-j 5 \Omega}}=\frac{120 V \angle 0^{\circ}}{\frac{25 \Omega \angle-90^{\circ}}{7.071 \angle-45^{\circ}}}.
=\frac{120 V \angle 0^{\circ}}{3.54 \Omega \angle-45^{\circ}}=33.9 A \angle 45^{\circ}.
I _{b c}=\frac{ V _{b c}}{ Z _{b c}}=\frac{120 V \angle 120^{\circ}}{3.54 \Omega \angle-45^{\circ}}=33.9 A \angle 165^{\circ}.
I _{ ca}=\frac{ V _{ ca}}{ Z _{ ca}}=\frac{120 V \angle -120^{\circ}}{3.54 \Omega \angle-45^{\circ}}=33.9 A \angle -75^{\circ}.
\text { c. } I_{L}=\sqrt{3} I_{\phi}=(1.73)(34 A )=58.82 A . \text { Therefore, }I_{A a}=I_{B b}=I_{C c}=58.82 A.