Let's solve each part **step by step**: --- ### **Given:** - **Source:** Delta-connected, positive phase sequence - \( V_{ab} = 240 \angle 30^\circ \) V - \( Z_L = 5 + j6 \) Ω - \( Z_Y = 36 - j14 \) Ω --- ## **I. Line Currents** ### **Step 1: Find Phase Voltages (Delta)** The given \( V_{ab} \) is the line-to-line voltage of the **delta source**. Since the sequence is **positive**: - \( V_{ab} = 240 \angle 30^\circ \) V - \( V_{bc} = 240 \angle -90^\circ \) V - \( V_{ca} = 240 \angle 150^\circ \) V --- ### **Step 2: Find Line Currents** #### **A. Through \( Z_L \) (Line Impedance):** Each line has series impedance \( Z_L \). The load is connected in Y with neutral. #### **B. Y-Connected Load Analysis** - The phase voltage across each \( Z_Y \) (Y-load) is the voltage at nodes \( A \), \( B \), \( C \) with respect to neutral \( N \). - The total impedance in each line: \( Z_{total} = Z_L + Z_Y \) \[ Z_{total} = (5 + j6) + (36 - j14) = (41 - j8) \ \Omega \] Magnitude and angle: \[ |Z_{total}| = \sqrt{41^2 + (-8)^2} = \sqrt{1681 + 64} = \sqrt{1745} \approx 41.78\ \Omega \] \[ \theta_{Z_{total}} = \arctan\left(\frac{-8}{41}\right) \approx -11.0^\circ \] So, \[ Z_{total} = 41.78 \angle -11^\circ \ \Omega \] #### **C. Phase Voltages for Y-Load** For Y-connected load, the phase voltage is: \[ V_{AN} = \frac{V_{ab}}{\sqrt{3}} \angle -30^\circ \] \[ V_{AN} = \frac{240}{\sqrt{3}} \angle (30^\circ - 30^\circ) = 138.56 \angle 0^\circ \ \text{V} \] Similarly, \[ V_{BN} = 138.56 \angle -120^\circ \ \text{V} \] \[ V_{CN} = 138.56 \angle 120^\circ \ \text{V} \] #### **D. Find Line Currents** \[ I_A = \frac{V_{AN}}{Z_{total}} = \frac{138.56 \angle 0^\circ}{41.78 \angle -11^\circ} = 3.317 \angle 11^\circ \ \text{A} \] \[ I_B = \frac{V_{BN}}{Z_{total}} = \frac{138.56 \angle -120^\circ}{41.78 \angle -11^\circ} = 3.317 \angle (-120^\circ + 11^\circ) = 3.317 \angle -109^\circ \ \text{A} \] \[ I_C = \frac{V_{CN}}{Z_{total}} = \frac{138.56 \angle 120^\circ}{41.78 \angle -11^\circ} = 3.317 \angle (120^\circ + 11^\circ) = 3.317 \angle 131^\circ \ \text{A} \] --- ## **II. Complex Power Absorbed by the Lines** The power absorbed by each line impedance \( Z_L \): Power per phase: \[ S_{L,phase} = I_{line}^2 \cdot Z_L \] Where \( Z_L = 5 + j6 \ \Omega \). \[ I_{line} = 3.317 \ \text{A} \] \[ S_{L,phase} = (3.317)^2 \cdot (5 + j6) = 11.01 \cdot (5 + j6) = 55.05 + j66.06 \ \text{VA (per phase)} \] Total (three phases): \[ S_{L,total} = 3 \cdot (55.05 + j66.06) = 165.15 + j198.18 \ \text{VA} \] --- ## **III. Complex Power Absorbed by the Loads** Power per phase for load \( Z_Y \): \[ S_{Y,phase} = I_{line}^2 \cdot Z_Y \] Where \( Z_Y = 36 - j14 \ \Omega \). \[ S_{Y,phase} = 11.01 \cdot (36 - j14) = 396.36 - j154.14 \ \text{VA (per phase)} \] Total: \[ S_{Y,total} = 3 \cdot (396.36 - j154.14) = 1189.08 - j462.42 \ \text{VA} \] --- ## **IV. Power Factor of the Load (Load Only)** \[ \text{Power Factor} = \cos(\theta_{Z_Y}) \] \[ \theta_{Z_Y} = \arctan\left(\frac{-14}{36}\right) = -21.4^\circ \] So, \[ \text{Power Factor} = \cos(21.4^\circ) = 0.931 \ (\text{leading, since current leads voltage for negative imaginary part}) \] --- ## **Final Answers (Summary Table)** | Part | Answer | |------|--------| | **I. Line currents** | \( I_A = 3.32 \angle 11^\circ\ \text{A} \); \( I_B = 3.32 \angle -109^\circ\ \text{A} \); \( I_C = 3.32 \angle 131^\circ\ \text{A} \) | | **II. Complex power by lines** | \( S_{L, total} = 165.15 + j198.18 \ \text{VA} \) | | **III. Complex power by loads** | \( S_{Y, total} = 1189.08 - j462.42 \ \text{VA} \) | | **IV. Power factor (load)** | \( 0.931 \) leading | If you need further breakdowns or phasor diagrams, let me know!

Let's go through each part step-by-step in a simple way. --- ### Given Data: - The source is a delta-connected source with positive phase sequence. - Line-to-line voltage: \( V_{ab} = 240 \angle 30^\circ \) volts. - Series line impedance \( Z_L = 5 + j6 \) ohms. - Load impedance \( Z_Y = 36 - j14 \) ohms. --- ### **Part 1: Find the line currents** **Step 1:** Convert the line-to-line voltage to phase voltages for the load in the Y-connection. - For a delta source, the phase-to-neutral voltage (assuming Y-connected load) is: \[ V_{ph} = \frac{V_{line}}{\sqrt{3}} = \frac{240}{\sqrt{3}} \approx 138.56 \text{ V} \] - The phase voltages are phase-shifted by 0°, -120°, and +120° for each phase: - \( V_{AN} = 138.56 \angle 0^\circ \) - \( V_{BN} = 138.56 \angle -120^\circ \) - \( V_{CN} = 138.56 \angle 120^\circ \) **Step 2:** Find the total impedance in each line (load + line impedance): \[ Z_{total} = Z_L + Z_Y = (5 + j6) + (36 - j14) = 41 - j8 \] Calculate magnitude and angle: \[ |Z_{total}| = \sqrt{41^2 + (-8)^2} \approx 41.78\, \Omega \] \[ \theta_{Z_{total}} = \arctan\left(\frac{-8}{41}\right) \approx -11^\circ \] **Step 3:** Calculate the line currents: - For phase A: \[ I_A = \frac{V_{AN}}{Z_{total}} = \frac{138.56 \angle 0^\circ}{41.78 \angle -11^\circ} = 3.32 \angle 11^\circ \text{ A} \] - For phase B: \[ I_B = \frac{138.56 \angle -120^\circ}{41.78 \angle -11^\circ} = 3.32 \angle (-120^\circ + 11^\circ) = 3.32 \angle -109^\circ \text{ A} \] - For phase C: \[ I_C = \frac{138.56 \angle 120^\circ}{41.78 \angle -11^\circ} = 3.32 \angle (120^\circ + 11^\circ) = 3.32 \angle 131^\circ \text{ A} \] --- ### **Part 2: Power absorbed by the lines** The power in each line: \[ S_{line} = I_{line}^2 \times Z_L \] Using the current magnitude: \[ I_{line} = 3.32\, \text{A} \] Calculating: \[ S_{line\, per\, phase} = (3.32)^2 \times (5 + j6) = 11.02 \times (5 + j6) = 55.1 + j66.12\, \text{VA} \] Total over three phases: \[ S_{lines} = 3 \times (55.1 + j66.12) = 165.3 + j198.36\, \text{VA} \] --- ### **Part 3: Power absorbed by the loads** Power in each load: \[ S_{load\, per\, phase} = I_{line}^2 \times Z_Y = 11.02 \times (36 - j14) = 396.7 - j154.3\, \text{VA} \] Total load power: \[ S_{loads} = 3 \times (396.7 - j154.3) = 1190.1 - j462.9\, \text{VA} \] --- ### **Part 4: Power factor of the load** The load impedance angle: \[ \theta_{Z_Y} = \arctan\left(\frac{-14}{36}\right) \approx -21.4^\circ \] Since the imaginary part is negative, the current leads the voltage, indicating a **leading power factor**. The magnitude of the power factor: \[ \cos(21.4^\circ) \approx 0.93 \] --- ### **Final summary:** - The line currents are approximately 3.32 A each, with phase angles of 11°, -109°, and 131° respectively. - The power absorbed by the lines totals about 165 VA (real) and 198 VA (reactive). - The power absorbed by the loads totals about 1190 VA (real) and 463 VA (reactive). - The load's power factor is about 0.93 and is leading. Let me know if you'd like more details!