1. **Problem Statement:** We have a 4-wire three-phase system with currents $I_1=8A$, $I_2=5A$, and $I_3=4A$. We need to: a. Draw the phasor diagram and find the neutral current. b. Prove the neutral current using trigonometry. c. Calculate real, reactive, and apparent power for a 400V balanced three-phase generator providing 200A at power factor 0.92 for star and delta connections. 2. **Phasor Diagram and Neutral Current:** - The three currents are 120° apart in phase: $I_1$ at 0°, $I_2$ at -120°, and $I_3$ at 120°. - Represent each current as a vector: $$I_1 = 8 \angle 0^\circ = 8 + j0$$ $$I_2 = 5 \angle -120^\circ = 5 \cos(-120^\circ) + j5 \sin(-120^\circ) = -2.5 - j4.33$$ $$I_3 = 4 \angle 120^\circ = 4 \cos 120^\circ + j4 \sin 120^\circ = -2 + j3.46$$ - Sum the currents to find neutral current $I_N$: $$I_N = I_1 + I_2 + I_3 = (8 - 2.5 - 2) + j(0 - 4.33 + 3.46) = 3.5 - j0.87$$ - Magnitude of neutral current: $$|I_N| = \sqrt{3.5^2 + (-0.87)^2} = \sqrt{12.25 + 0.76} = \sqrt{13.01} \approx 3.61A$$ 3. **Trigonometric Proof:** - Using cosine and sine components for each current: $$I_1: x=8, y=0$$ $$I_2: x=5 \cos 240^\circ = 5(-0.5) = -2.5, y=5 \sin 240^\circ = 5(-0.866) = -4.33$$ $$I_3: x=4 \cos 120^\circ = 4(-0.5) = -2, y=4 \sin 120^\circ = 4(0.866) = 3.46$$ - Sum x-components: $8 - 2.5 - 2 = 3.5$ - Sum y-components: $0 - 4.33 + 3.46 = -0.87$ - Neutral current magnitude matches previous calculation: $3.61A$ 4. **Power Calculations for Balanced Load:** Given: - Line voltage $V_L = 400V$ - Line current $I_L = 200A$ - Power factor $pf = 0.92$ - Angle $\phi = \cos^{-1}(0.92) \approx 23.07^\circ$ **Star Connection:** - Phase voltage $V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{400}{1.732} \approx 230.94V$ - Phase current $I_{ph} = I_L = 200A$ - Real power: $$P = \sqrt{3} V_L I_L pf = 1.732 \times 400 \times 200 \times 0.92 = 127,475W$$ - Reactive power: $$Q = \sqrt{3} V_L I_L \sin \phi = 1.732 \times 400 \times 200 \times \sin 23.07^\circ = 1.732 \times 400 \times 200 \times 0.392 = 54,352 VAR$$ - Apparent power: $$S = \sqrt{3} V_L I_L = 1.732 \times 400 \times 200 = 138,560 VA$$ **Delta Connection:** - Phase voltage $V_{ph} = V_L = 400V$ - Phase current $I_{ph} = \frac{I_L}{\sqrt{3}} = \frac{200}{1.732} \approx 115.47A$ - Real power: $$P = 3 V_{ph} I_{ph} pf = 3 \times 400 \times 115.47 \times 0.92 = 127,475W$$ - Reactive power: $$Q = 3 V_{ph} I_{ph} \sin \phi = 3 \times 400 \times 115.47 \times 0.392 = 54,352 VAR$$ - Apparent power: $$S = 3 V_{ph} I_{ph} = 3 \times 400 \times 115.47 = 138,560 VA$$ **Summary:** - Neutral current magnitude is approximately $3.61A$. - Power values are the same for star and delta connections as expected for balanced loads.