1. **Problem Statement:** Calculate the line currents magnitude and phase angles, neutral current, total active, reactive, and apparent power, and system power factor for a 400V three-phase four-wire system supplying an unbalanced star-connected load with: - Phase A: 10kW at unity power factor (PF) - Phase B: 15kW at 0.8 PF lagging - Phase C: 8kW at 0.9 PF leading 2. **Given Data:** - Line-to-neutral voltage $V_{ph} = \frac{400}{\sqrt{3}} \approx 230.94$ V - Phase A: $P_A = 10$ kW, $PF_A = 1$ (unity, so angle $\theta_A = 0^\circ$) - Phase B: $P_B = 15$ kW, $PF_B = 0.8$ lagging (angle $\theta_B = \cos^{-1}(0.8) = 36.87^\circ$) - Phase C: $P_C = 8$ kW, $PF_C = 0.9$ leading (angle $\theta_C = -\cos^{-1}(0.9) = -25.84^\circ$) 3. **Formulas:** - Apparent power $S = \frac{P}{PF}$ - Reactive power $Q = P \tan \theta$ - Current magnitude $I = \frac{S}{V_{ph}}$ - Phase angle of current $\theta$ same as load power factor angle - Neutral current $I_N = \sqrt{I_A^2 + I_B^2 + I_C^2 - I_A I_B - I_B I_C - I_C I_A}$ (vector sum) - Total active power $P_{total} = P_A + P_B + P_C$ - Total reactive power $Q_{total} = Q_A + Q_B + Q_C$ - Total apparent power $S_{total} = \sqrt{P_{total}^2 + Q_{total}^2}$ - System power factor $PF_{system} = \frac{P_{total}}{S_{total}}$ 4. **Step 1: Calculate individual phase apparent and reactive powers** $$ S_A = \frac{10}{1} = 10 \text{ kVA}, \quad Q_A = 10 \times \tan 0^\circ = 0 \text{ kVAR} $$ $$ S_B = \frac{15}{0.8} = 18.75 \text{ kVA}, \quad Q_B = 15 \times \tan 36.87^\circ = 15 \times 0.75 = 11.25 \text{ kVAR} $$ $$ S_C = \frac{8}{0.9} \approx 8.89 \text{ kVA}, \quad Q_C = 8 \times \tan (-25.84^\circ) = 8 \times (-0.484) = -3.87 \text{ kVAR} $$ 5. **Step 2: Calculate phase currents magnitude** $$ I_A = \frac{S_A \times 1000}{V_{ph}} = \frac{10 \times 1000}{230.94} \approx 43.28 \text{ A} $$ $$ I_B = \frac{18.75 \times 1000}{230.94} \approx 81.17 \text{ A} $$ $$ I_C = \frac{8.89 \times 1000}{230.94} \approx 38.47 \text{ A} $$ 6. **Step 3: Express phase currents as complex numbers (phasors)** $$ I_A = 43.28 \angle 0^\circ = 43.28 + j0 $$ $$ I_B = 81.17 \angle -36.87^\circ = 81.17 \cos(-36.87^\circ) + j81.17 \sin(-36.87^\circ) = 64.89 - j48.88 $$ $$ I_C = 38.47 \angle 25.84^\circ = 38.47 \cos 25.84^\circ + j38.47 \sin 25.84^\circ = 34.62 + j16.77 $$ 7. **Step 4: Calculate neutral current as vector sum** $$ I_N = |I_A + I_B + I_C| = |(43.28 + 64.89 + 34.62) + j(0 - 48.88 + 16.77)| = |142.79 - j32.11| $$ $$ I_N = \sqrt{142.79^2 + (-32.11)^2} = \sqrt{20400 + 1031} = \sqrt{21431} \approx 146.4 \text{ A} $$ 8. **Step 5: Calculate total active and reactive power** $$ P_{total} = 10 + 15 + 8 = 33 \text{ kW} $$ $$ Q_{total} = 0 + 11.25 - 3.87 = 7.38 \text{ kVAR} $$ 9. **Step 6: Calculate total apparent power and system power factor** $$ S_{total} = \sqrt{33^2 + 7.38^2} = \sqrt{1089 + 54.44} = \sqrt{1143.44} \approx 33.81 \text{ kVA} $$ $$ PF_{system} = \frac{33}{33.81} \approx 0.976 \text{ lagging} $$ 10. **Implications:** - The neutral current is very high (146.4 A), much higher than individual phase currents, due to unbalance and phase angle differences. - Neutral conductor must be sized to carry this high current safely. - Protective devices must be selected to handle high neutral currents and unbalanced load conditions to prevent overheating and ensure safety. **Final answers:** - Phase currents magnitudes and angles: $I_A = 43.28 \angle 0^\circ$ A, $I_B = 81.17 \angle -36.87^\circ$ A, $I_C = 38.47 \angle 25.84^\circ$ A - Neutral current magnitude: $I_N \approx 146.4$ A - Total active power: 33 kW - Total reactive power: 7.38 kVAR - Total apparent power: 33.81 kVA - System power factor: 0.976 lagging