Question 9.3: A balanced wye-connected generator delivers power to the bal......
A balanced wye-connected generator delivers power to the balanced wye-connected load as shown in Fig. 9.11. The line voltage of the generator is 200 V rms. Determine the line currents for abc phase sequence.
For abc phase sequence, the phase voltages are,
V_{\mathrm{an}}={\frac{V_{\mathrm{L}}}{\sqrt{3}}}|\underline{-30^{\circ}}={\frac{200}{\sqrt{3}}}|\underline{-30^{\circ}}=115.47|\underline{-30^{\circ}} V (9.39)
V_{\mathrm{bn}}=115.47[{\underline{{-150}}}^{\circ}\mathbf{V} (9.40)
V_{\mathrm{cn}}=115.47|\underline{90^{\circ}}V (9.41)
The total impedance is,
Z_{t}=0.03+j5+5|\underline{30^{\circ}}=8.68|\underline{59.83^{\circ}}\Omega (9.42)
The line currents are calculated as,
I_{\mathrm{aA}}={\frac{V_{\mathrm{an}}}{Z_{t}}}={\frac{115.47|\underline{-30^{\circ}}}{8.68|\underline{59.83^{\circ}}}}=13.30|\underline{-89.83^{\circ}}\,\mathrm{A} (9.43)
I_{\mathrm{bB}}={\frac{V_{\mathrm{bn}}}{Z_{t}}}={\frac{115.47|\underline{-150^{\circ}}}{8.68|\underline{59.83^{\circ}}}}=13.30|\underline{150.17^{\circ}}\,\mathrm{A} (9.44)
I_{\mathrm{bB}}={\frac{V_{\mathrm{bn}}}{Z_{t}}}={\frac{115.47|\underline{90^{\circ}}}{8.68|\underline{59.83^{\circ}}}}=13.30|\underline{30.17^{\circ}}\,\mathrm{A} (9.45)