Given That the Line-To-Ground Voltages Of$V_{a g}=280 \angle 0^{\circ}, V_{b g}=250 \angle-110^{\circ}$

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The Answer is $\bar{I}_{0}=0 From Test Bank Problem 8.5, \bar{I}_{1}=27.57 \angle-59.76^{\circ} \mathrm{A} ; \bar{I}_{2}=2.487 \angle 26.3^{\circ} \mathrm{A} \begin{aligned} {\left[\begin{array}{c} \bar{I}_{a} \ \bar{I}_{b} \ \bar{I}_{c} \end{array} ight] } & =\left[\begin{array}{ccc} 1 & 1 & 1 \ 1 & a^{2} & a \ 1 & a & a^{2} \end{array} ight]\left[\begin{array}{c} 0 \ 27.57 \angle-59.76^{\circ} \ 2.487 \angle 26.3^{\circ} \end{array} ight] \ & =\left[\begin{array}{c} 27.57 \angle-59.76^{\circ}+2.487 \angle 26.3^{\circ} \ 27.57 \angle 180.24^{\circ}+2.487 \angle 146.3^{\circ} \ 27.57 \angle 60.24^{\circ}+2.487 \angle 266.3^{\circ} \end{array} ight] \ & =\left[\begin{array}{c} 16.120-j 22.72 \ -29.64+j 1.2650 \ 13.530+j 21.46 \end{array} ight]=\left[\begin{array}{l} 27.86 \angle-54.64^{\circ} \ 29.66 \angle 177.56^{\circ} \ 25.36 \angle 57.77^{\circ} \end{array} ight] \mathrm{A} \end{aligned}$

\bar{I}_{0}=0

From Test Bank Problem 8.5,

\bar{I}_{1}=27.57 \angle-59.76^{\circ} \mathrm{A} ; \bar{I}_{2}=2.487 \angle 26.3^{\circ} \mathrm{A}

\begin{aligned}{\left[\begin{array}{c}\bar{I}_{a} \\bar{I}_{b} \\bar{I}_{c}\end{array} ight] } & =\left[\begin{array}{ccc}1 & 1 & 1 \1 & a^{2} & a \1 & a & a^{2}\end{array} ight]\left[\begin{array}{c}0 \27.57 \angle-59.76^{\circ} \2.487 \angle 26.3^{\circ}\end{array} ight] \& =\left[\begin{array}{c}27.57 \angle-59.76^{\circ}+2.487 \angle 26.3^{\circ} \27.57 \angle 180.24^{\circ}+2.487 \angle 146.3^{\circ} \27.57 \angle 60.24^{\circ}+2.487 \angle 266.3^{\circ}\end{array} ight] \& =\left[\begin{array}{c}16.120-j 22.72 \-29.64+j 1.2650 \13.530+j 21.46\end{array} ight]=\left[\begin{array}{l}27.86 \angle-54.64^{\circ} \29.66 \angle 177.56^{\circ} \25.36 \angle 57.77^{\circ}\end{array} ight] \mathrm{A}\end{aligned}

Given That the Line-To-Ground Voltages Of $V_{a g}=280 \angle 0^{\circ}, V_{b g}=250 \angle-110^{\circ}$

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Given That the Line-To-Ground Voltages Of Vag=280∠0∘,Vbg=250∠−110∘V_{a g}=280 \angle 0^{\circ}, V_{b g}=250 \angle-110^{\circ}Vag​=280∠0∘,Vbg​=250∠−110∘

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Given That the Line-To-Ground Voltages Of $V_{a g}=280 \angle 0^{\circ}, V_{b g}=250 \angle-110^{\circ}$

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