Question 9.4: A balanced three-phase wye–delta system is shown in Fig. 9.1......
A balanced three-phase wye–delta system is shown in Fig. 9.17. Per phase load impedance and generator voltage are Z_{\Delta}=5|\underline{{20}}^{\circ}\Omega and 240 V rms, respectively. For ABC phase sequence, calculate the load phase currents and line currents.
The line voltages are,
V_{\mathrm{AB}}={\sqrt{3}}\times240|\underline{30}^{\circ}=415.69|\underline{30}^{\circ}\mathbf{V} (9.63)
V_{\mathrm{BC}}=\sqrt{3}\times240|\underline{-90}^{\circ}=415.69|\underline{-90}^{\circ}\mathrm{V} (9.64)
V_{\mathrm{CA}}=\sqrt{3}\times240|\underline{150}^{\circ}=415.69|\underline{150}^{\circ}\mathrm{V} (9.65)
The phase currents in the load can be determined as,
I_{\mathrm{ab}}={\frac{V_{\mathrm{AB}}}{Z}}={\frac{415.69|\underline{30}^{\circ}}{25|\underline{20}^{\circ}}}=16.63|{\underline{{10^{\circ}}}}\,\mathbf{A} (9.66)
I_{\mathrm{bc}}={\frac{V_{\mathrm{BC}}}{Z}}={\frac{415.69|\underline{-90}^{\circ}}{25|\underline{20}^{\circ}}}=16.63|\underline{-110}^{\circ}\mathbf{A} (9.67)
I_{\mathrm{ca}}={\frac{V_{\mathrm{CA}}}{Z}}={\frac{415.69|\underline{150}^{\circ}}{25|\underline{20}^{\circ}}}=16.63|\underline{130}^{\circ}\mathbf{A} (9.68)
The line currents can be calculated by applying KCL at nodes a, b and c of the circuit in Fig. 9.17 as,
I_{\mathrm{Aa}}=I_{\mathrm{ab}}-I_{\mathrm{ca}}=16.63|\underline{10}^{\circ}-16.63|\underline{130}^{\circ}=28.80|\underline{-20}^{\circ}\mathbf{A} (9.69)
I_{\mathrm{Bb}}=I_{\mathrm{bc}}-I_{\mathrm{ab}}=16.63|\underline{-110}^{\circ}-16.63|\underline{10}^{\circ}=28.80|\underline{-140}^{\circ}\mathbf{A} (9.70)
I_{\mathrm{Cc}}=I_{\mathrm{ca}}-I_{\mathrm{bc}}=16.63|\underline{130}^{\circ}-16.63|\underline{-110}^{\circ}=28.80|\underline{100}^{\circ}\mathbf{A} (9.71)