Question 9.9: A balanced three-phase wye-connected generator delivers powe......

Delta load is converted to wye load as,

$Z_{1}={\frac{4|\underline{-15^{\circ}}}{3}}=1.33|\underline{-15^{\circ}}\,\Omega$      (9.165)

The value of the parallel loads is,

$Z_{2}={\frac{1.33|\underline{-15^{\circ}}\times6|\underline{10^{\circ}}}{1.33|\underline{-15^{\circ}}+6|\underline{10^{\circ}}}}=1.10|\underline{-10.54^{\circ}}\Omega$      (9.166)

The line current is calculated as,

$I_{\mathrm{Aa}}={\frac{200|\underline{-20^{\circ}}}{0.02+j5+1.10|\underline{-10.54^{\circ}}}}=40.62{|\underline{{-97.07^{\circ}}}}\mathbf{A}$      (9.167)

Power delivers by the first line is calculated as,

$P_{A}=V_{\mathrm{AN}}I_{\mathrm{Aa}}\cos(\theta_{\nu}-\theta_{i})=200\times40.62\cos(-20^{\circ}+97.07^{\circ})=1817.83\,\mathrm{W}$      (9.168)

Total power delivered by the source is,

$P_{t}=3P_{A}=3\times1817.83=5453.49\,\mathrm{W}$      (9.169)