Given that the line-to-ground voltages of $V_{a g}=280 \angle 0^{\circ}, V_{b g}=250 \angle-110^{\circ}$, and $V_{c g}=290 \angle 130^{\circ}$ volts are applied to a balanced-Y load consisting of $(12+j 16)$ ohms per phase. The load neutral is kept open. Draw the sequence networks and calculate $I_{0}, I_{1}$, and $I_{2}$, the sequence components of the line currents. Then calculate the line currents $I_{a}, I_{b}$, and $I_{c}$.

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The Answer of Given that the line-to-ground voltages of $V_{a g}=280 \angle 0^{\circ}, V_{b g}=250 \angle-110^{\circ}$, and $V_{c g}=290 \angle 130^{\circ}$ volts are applied to a balanced-Y load consisting of $(12+j 16)$ ohms per phase. The load neutral is kept open. Draw the sequence networks and calculate $I_{0}, I_{1}$, and $I_{2}$, the sequence components of the line currents. Then calculate the line currents $I_{a}, I_{b}$, and $I_{c}$.

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∘

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