Question

xample A Δ-connected balanced 3-phase generator with an impedance of 0.6+/0.3 Ω/ph is connected to a Δ-connected balanced load with an impedance of 24 18 Ω/ph. The line joining the generator and the load has an impedance of 0.6 0.7 Ω.ph. Assume a positive sequence for the source voltages and the internal phase voltage for V4,-120 L30 V. Find: The line currents, The phase voltages at the load and the generator terminal line voltages. 0.6+0.7j 1a-309.8387 2-26.57 A la-62298 L-37.69, A 309.83872-26.57 6.2298<-157.69 A 0.2+ 0.1j 8+6j =6.2298 L82.31 I4B 3.5968 2-7.69 A AB=107.9025し29.18 v VBc 107.90252-90.82 V pa=107.9025し149.18 V A -,-1 17.6335し30.23 V bc-117.63352-89.77° T,-67.9157し0.23 V V ca=117.6335 150.23 y

When analyzing a delta-delta connection, we typically convert it to Y-Y connection and then consider the single phase equivalent circuit (shown in the figure) to anaylze it easily. So here, my problems come, how can I related the phase and line voltages in Y-connection to that of the original delta-connection.

If I calculated V_a, then how did I use V_ab = sqrt(3)*V_a*phase(30^o), even though in a delta connection the phase voltage is the same line voltage? Please someone clear this confusion for me.

Answer 1

a. C. oe same tane 疗 al so Ph