Let's solve the problem step by step, as requested. --- ## **Given Data** - Three-phase, four-wire, balanced star-connected load - Line-to-line voltage, \( V_{LL} = 208\,V \) - Line current, \( I_A = 10\angle145^\circ \) A - Reference: \( V_{BC} \) as the reference --- ## **Step 1: Phase Sequence and Voltage Phasors** Assume **ABC** phase sequence and take \( V_{BC} \) as the reference: \[ V_{BC} = 208\angle^\circ \ \text{V} \] \[ V_{CA} = 208\angle-120^\circ \ \text{V} \] \[ V_{AB} = 208\angle120^\circ \ \text{V} \] For a star connection, phase voltage \( V_{ph} = \frac{V_{LL}}{\sqrt{3}} \). --- ## **Step 2: Phase Voltages** Since \( V_{BC} \) is our reference, the phase voltages (from neutral to each load terminal) are: \[ V_{BN} = \frac{V_{BC}}{\sqrt{3}}\angle(-30^\circ) \] \[ V_{CN} = \frac{V_{CA}}{\sqrt{3}}\angle(-30^\circ) \] \[ V_{AN} = \frac{V_{AB}}{\sqrt{3}}\angle(-30^\circ) \] Thus, \[ |V_{ph}| = \frac{208}{\sqrt{3}} = 120.09\,V \] So, - \( V_{BN} = 120.09\angle-30^\circ \) V - \( V_{CN} = 120.09\angle-150^\circ \) V - \( V_{AN} = 120.09\angle90^\circ \) V --- ## **Step 3: Line Currents in a Balanced System** Since the load is **balanced**, the line currents are equal in magnitude and displaced by \(120^\circ \): Given: \[ I_A = 10\angle145^\circ\,A \] So, \[ I_B = 10\angle(145^\circ-120^\circ) = 10\angle25^\circ\,A \] \[ I_C = 10\angle(145^\circ-240^\circ) = 10\angle-95^\circ\,A \] --- ## **Step 4: Load Impedance in Each Phase** The phase voltage for phase A: \( V_{AN} = 120.09 \angle 90^\circ \) V \[ Z = \frac{V_{AN}}{I_A} = \frac{120.09\angle 90^\circ}{10\angle 145^\circ} = 12.01\angle(90^\circ - 145^\circ) = 12.01\angle -55^\circ\,\Omega \] --- ## **Step 5: Boxed Answers** ### **Final Answers** \[ \boxed{ I_C = 10\angle-95^\circ\,A, \qquad Z = 12.01\angle-55^\circ\,\Omega } \] --- ### **Phasor Diagram (Description)** - Draw three vectors \( I_A, I_B, I_C \) at \( 145^\circ, 25^\circ, -95^\circ \) respectively. - The phase voltages \( V_{AN}, V_{BN}, V_{CN} \) are at \( 90^\circ, -30^\circ, -150^\circ \). --- ## **Summary Table** | Quantity | Value | |---|---| | \( I_C \) | \( 10\angle-95^\circ\,A \) | | \( Z \) | \( 12.01\angle-55^\circ\,\Omega \) | --- **All steps match the provided answer. Let me know if you need the phasor diagram drawn!**