A Balanced Y-Connected Generator with Terminal Voltage$V_{b c}=480 \angle 90^{\circ}$

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The Answer is $\begin{aligned} \bar{I}_{ ext {Line } 0} & =\bar{I}_{ ext {Line } 2}=0 \ \bar{I}_{ ext {Line } 1} & =\frac{V_{g 1}}{\bar{Ƶ}_{ ext {Line } 1}+\frac{\bar{Ƶ}_{\Delta}}{3}}=\frac{\frac{480}{\sqrt{3}} \angle 180^{\circ}}{0.5 \angle 80^{\circ}+\frac{20}{3} \angle 40^{\circ}} \ & =\frac{277.14 \angle 180^{\circ}}{5.194+j 4.778}=\frac{277.14 \angle 180^{\circ}}{7.057 \angle 42.61^{\circ}}=39.27 \angle 137.4^{\circ} \mathrm{A} \end{aligned}$

\begin{aligned}\bar{I}_{ ext {Line } 0} & =\bar{I}_{ ext {Line } 2}=0 \\bar{I}_{ ext {Line } 1} & =\frac{V_{g 1}}{\bar{Ƶ}_{ ext {Line } 1}+\frac{\bar{Ƶ}_{\Delta}}{3}}=\frac{\frac{480}{\sqrt{3}} \angle 180^{\circ}}{0.5 \angle 80^{\circ}+\frac{20}{3} \angle 40^{\circ}} \& =\frac{277.14 \angle 180^{\circ}}{5.194+j 4.778}=\frac{277.14 \angle 180^{\circ}}{7.057 \angle 42.61^{\circ}}=39.27 \angle 137.4^{\circ} \mathrm{A}\end{aligned}

A Balanced Y-Connected Generator with Terminal Voltage $V_{b c}=480 \angle 90^{\circ}$

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A Balanced Y-Connected Generator with Terminal Voltage Vbc=480∠90∘V_{b c}=480 \angle 90^{\circ}Vbc​=480∠90∘

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A Balanced Y-Connected Generator with Terminal Voltage $V_{b c}=480 \angle 90^{\circ}$

Correct Answer:
Verified