Let's solve each part step by step as per the given tasks: --- ## **Given Data** - **Load Power (\( P \))** = 5 kW = 500 W - **Power factor (pf)** = .8 lagging - **Load Voltage (\( V_r \))** = 230 V - **Line Impedance (\( Z \))** = \( 2 + j3 \) Ω --- ## **1. Calculate the Load Current (\( I \)):** \[ I = \frac{P}{V_r \cdot \text{pf}} \] \[ I = \frac{500}{230 \times .8} \] \[ I = \frac{500}{184} = 27.17 \text{ A} \] --- ## **2. Determine the Sending-End Voltage (\( V_s \)):** The current lags the voltage by \( \cos^{-1}(.8) = 36.87^\circ \). So, \[ I = 27.17 \angle -36.87^\circ \text{ A} \] \[ Z = 2 + j3 = 3.61 \angle 56.31^\circ \, \Omega \] Calculate voltage drop across the line: \[ IZ = (27.17 \angle -36.87^\circ) \times (3.61 \angle 56.31^\circ) \] \[ = 27.17 \times 3.61 \angle (56.31^\circ - 36.87^\circ) \] \[ = 98.13 \angle 19.44^\circ \text{ V} \] Now, express \( V_r \) as a phasor (\( 230 \angle ^\circ \)): \[ V_s = V_r + IZ \] Convert \( IZ \) to rectangular form: \[ 98.13 \cos 19.44^\circ = 92.42 \] \[ 98.13 \sin 19.44^\circ = 32.63 \] \[ IZ = 92.42 + j32.63 \] Add to \( V_r \): \[ V_s = 230 + 92.42 + j32.63 = (322.42 + j32.63) \] Magnitude: \[ |V_s| = \sqrt{(322.42)^2 + (32.63)^2} = \sqrt{104014.85 + 1065.94} = \sqrt{105080.79} = 324.3 \text{ V} \] --- ## **3. Calculate the Voltage Regulation of the Line:** \[ \% VR = \frac{|V_s| - |V_r|}{|V_r|} \times 100 \] \[ = \frac{324.3 - 230}{230} \times 100 \] \[ = \frac{94.3}{230} \times 100 = 41.\% \] --- ## **4. Calculate the Transmission Efficiency:** ### **Power Loss in Line:** \[ P_{\text{loss}} = I^2 R \] \[ = (27.17)^2 \times 2 = 738.2 \times 2 = 1476.4 \text{ W} \] ### **Efficiency:** \[ \eta = \frac{P_{\text{load}}}{P_{\text{load}} + P_{\text{loss}}} \times 100 \] \[ = \frac{500}{500 + 1476.4} \times 100 \] \[ = \frac{500}{6476.4} \times 100 = 77.2\% \] --- ## **Summary Table** | Quantity | Value | |------------------------------- |-------------------| | Load Current (\( I \)) | 27.17 A | | Sending-end Voltage (\( V_s \))| 324.3 V | | Voltage Regulation | 41.% | | Transmission Efficiency (\( \eta \)) | 77.2% | --- ### **Final Answers** 1. **Load current (\( I \))**: **27.17 A** 2. **Sending-end voltage (\( V_s \))**: **324.3 V** 3. **Voltage regulation**: **41.%** 4. **Transmission efficiency**: **77.2%**