Question 9.5: A balanced three-phase wye-connected generator delivers powe......

Converting delta-connected load to a wye-connected load yields,

Z_{y}={\frac{Z_{\Delta}}{3}}={\frac{12|\underline{30^{\circ}}}{3}}=4|\underline{30^{\circ}}\Omega     (9.76)

The line currents can be calculated as,

I_{\mathrm{Aa}}={\frac{\nu_{\mathrm{an}}}{Z_{\mathrm{y}}+Z_{L}}}={\frac{230|\underline{0^{\circ}}}{2|\underline{15^{\circ}}+4{|\underline{30^{\circ}}}}}=38.63|\underline{-25.01^{\circ}}\mathrm{A}     (9.77)

I_{\mathrm{Bb}}={\frac{\nu_{\mathrm{bn}}}{Z_{\mathrm{y}}+Z_{L}}}={\frac{230|\underline{-120^{\circ}}}{2|\underline{15^{\circ}}+4{|\underline{30^{\circ}}}}}=38.63|\underline{-145.01^{\circ}}\mathrm{A}     (9.78)

I_{\mathrm{Cc}}={\frac{\nu_{\mathrm{cn}}}{Z_{\mathrm{y}}+Z_{L}}}={\frac{230|\underline{120^{\circ}}}{2|\underline{15^{\circ}}+4{|\underline{30^{\circ}}}}}=38.63|\underline{94.99^{\circ}}\mathrm{A}    (9.79)

The line voltage is calculated as,

V_{\mathrm{AB}}=\sqrt{3}\nu_{\mathrm{an}}\big\vert\underline{{{\mathrm{30}}}}^{\circ}=\sqrt{3}\times230|\underline{30^{\circ}}=398.37|\underline{30^{\circ}}\mathrm{{V}}    (9.80)

Applying KVL between lines A and B of the circuit in Fig. 9.21 yields,

V_{\mathrm{ab}}=V_{\mathrm{AB}}-I_{\mathrm{Aa}}Z_{L}+I_{\mathrm{Bb}}Z_{L}    (9.81)

V_{\mathrm{ab}}=398.37|\underline{30^{\circ}}+2|\underline{15^{\circ}}(38.63|\underline{-145.01^{\circ}}-38.63|\underline{-25.01^{\circ}})=267.60|\underline{35}^{\circ}V    (9.82)

The phase current is calculated as,

I_{\mathrm{ab}}=\frac{V_{\mathrm{ab}}}{Z_{Δ}}=\frac{267.60|\underline{35^{\circ}}}{12|\underline{30^{\circ}}}=22.3|\underline{5^{\circ}}\,{\mathrm{A}}    (9.83)

Other phase currents can be written as,

I_{bc}=I_{ab}|\underline{-120}^{\circ}=22.3|\underline{5^{\circ}-120^{\circ}}=22.33|\underline{-155}^{\circ}A    (9.84)

I_{ca}=I_{ab}|\underline{120}^{\circ}=22.3|\underline{5^{\circ}+120^{\circ}}=22.33|\underline{125}^{\circ}A    (9.85)