A three-phase source is balanced and positive-sequence

A three-phase source is balanced and positive-sequence. Line-to-neutral RMS phasors, with phase $A$ as the angle reference, are \[ \mathbf V_{AN}=120\angle 0^\circ\ \mathrm{V},\qquad \mathbf V_{BN}=120\angle -120^\circ\ \mathrm{V},\qquad \mathbf V_{CN}=120\angle +120^\circ\ \mathrm{V}. \] A balanced Y-connected load has phase impedance \[ Z_Y=8+j6\ \Omega \] in each of $A$, $B$, and $C$, with the common neutral connected to the source neutral by an ideal conductor. All voltages and currents in this packet are RMS phasors. Passive sign convention: each phase current $\mathbf I_A,\mathbf I_B,\mathbf I_C$ enters the corresponding load impedance from the line terminal. 1. Compute $\mathbf I_A= \mathbf V_{AN}/Z_Y$ in polar form. Then write $\mathbf I_B$ and $\mathbf I_C$ by the positive-sequence shifts $-120^\circ$ and $+120^\circ$. Use the conventional engineering angle $\arctan(3/4)=36.87^\circ$. 2. Form the line-to-line voltage $\mathbf V_{AB}=\mathbf V_{AN}-\mathbf V_{BN}$ and prove \[ \mathbf V_{AB}=120\sqrt{3}\angle 30^\circ\ \mathrm{V}. \] 3. Compute the total complex power $\mathbf S_{3\phi}=3\mathbf V_{AN}\mathbf I_A^\ast$, the apparent power $|\mathbf S_{3\phi}|$, and the power factor, including lag/lead. 4. Give the equivalent balanced delta impedance $Z_\Delta=3Z_Y$ that draws the same line currents from the same line voltages. 5. Audit both sentences: (i) “the $120\,\mathrm{V}$ source is a peak phasor, so every current and the total power must be scaled by $1/\sqrt{2}$ or $1/2$”; (ii) “the equivalent delta is $Z_Y/3=8/3+j2$.” Do not replace the packet by a single-phase series-$RLC$ complex-power balance or a frequency-response resonance argument.

$|Z_Y|=10$ and $\arg(Z_Y)=\arctan(3/4)=36.87^\circ$, so $\mathbf I_A=12\angle-36.87^\circ$ A.