Question 9.8: A balanced three-phase wye–wye system is shown in Fig. 9.31.......
A balanced three-phase wye–wye system is shown in Fig. 9.31. For ABC phase sequence, calculate the line current, power supplied to each phase, power absorbed by each phase and the total complex power supplied by the source.
The line currents are calculated as,
I_{Aa}=\frac{230|\underline{15^{\circ}}}{2+j8}=27.89|\underline{-60.96^{\circ}} A (9.156)
I_{Bb}=I_{Aa}|\underline{-120^{\circ}}=27.89|\underline{-180.96^{\circ}} A (9.157)
I_{Cc}=I_{Aa}|\underline{+120^{\circ}}=27.89|\underline{59.04^{\circ}} A (9.158)
Power supplied to each phase is,
P_{A}=V_{P}I_{p}\cos \theta=V_{AN}I_{Aa}\cos(\theta_{\nu}-\theta_{i})=230\times 27.89\times \cos (15+60.96)=1556.20\mathrm{W} (9.159)
Per phase power absorbed by the load is,
P_{L1\phi}=27.89^{2}\times2=155.70\,\mathrm{W} (9.160)
Total complex power supplied by the source is,
S_{t}=3V_{\mathrm{An}}I_{\mathrm{Aa}}^{*}=3\times230|\underline{15^{\circ}}\times27.89|\underline{60.96^{\circ}}=4668.60+j18669.21 \ mathrm{ V A} (9.161)