Let's solve each problem step by step: --- ### 1. **Sending-End Voltage for 3-phase Line** #### **Given:** - **Load:** 20 MW (20,000 kW) at unity power factor - **Impedance per phase:** \( Z = 1 + j3 \,\Omega \) - **Efficiency:** 92% #### **To Find:** Sending-end voltage --- #### **Step 1: Calculate Receiving End Power and Current** - Since efficiency \( \eta = 92\% \): \[ \eta = \frac{P_{R}}{P_S} \] \[ P_S = \frac{P_{R}}{\eta} \] So, \[ P_S = \frac{20,000\,kW}{0.92} = 21,739.13\,kW \] #### **Step 2: Find Receiving End Line-to-Line Voltage** - Not given directly. Let's denote it as \( V_{R} \). - Since power factor is unity, \( \cos\phi = 1 \). - For 3-phase: \[ P_{R} = \sqrt{3} V_{R} I_{R} \cos\phi \] \[ 20,000 = \sqrt{3} V_{R} I_{R} \] (All in kW, so voltage in kV.) #### **Step 3: Express Current in Terms of Voltage** \[ I_{R} = \frac{20,000}{\sqrt{3} V_{R}} \] #### **Step 4: Calculate Voltage Drop Across Line** - Per phase drop: \[ \Delta V_{\text{ph}} = I_{R} Z = I_{R} (1 + j3) \] - Since unity p.f., current is in phase with voltage. #### **Step 5: Find Sending-End Voltage** \[ V_{S,ph} = V_{R,ph} + \Delta V_{\text{ph}} \] \[ V_{S,ph} = V_{R,ph} + I_{R} (1 + j3) \] - But we need either \( V_{R} \) or \( I_{R} \) to proceed. Since not specified, this problem is **missing the receiving end voltage** (or current), and can't proceed numerically without it. --- ### 2. **Short 3-phase Transmission Line** #### **Given:** - **Line impedance per phase:** \( Z = 5 + j10\,\Omega \) - **Sending-end (input) line voltage:** \( V_S = 12.8\,kV \) - **Receiving-end load:** 1200 kW, 60 A at lagging p.f. #### **To Find:** Load voltage between lines (\( V_R \)). --- #### **Step 1: Calculate Power Factor** - Power = \( \sqrt{3} V_R I_R \cos\phi \) - \( I_R = 60\,A \) - \( P = 1200\,kW \) \[ \cos\phi = \frac{P}{\sqrt{3} V_R I_R} \] But \( V_R \) is unknown. #### **Step 2: Find Per-phase Load Voltage** - For star connection: \[ V_{R,ph} = \frac{V_R}{\sqrt{3}} \] - Load current is 60 A per phase. #### **Step 3: Calculate Power Factor:** \[ 1200\times 10^3 = \sqrt{3} V_R \times 60 \times \cos\phi \] \[ V_R \cos\phi = \frac{1200 \times 10^3}{\sqrt{3} \times 60} \] \[ V_R \cos\phi = \frac{1200000}{103.923} = 11,546\,V \] But \( V_R \) and \( \cos\phi \) are multiplied. We need one to find the other. Let's try another way: we know current, so total apparent power: \[ S = \sqrt{3} V_R I_R \] \[ |S| = \sqrt{P^2 + Q^2} \] But not enough info to solve directly unless we assume p.f. (which is not given numerically). **If we assume p.f. = 0.8 lagging (common for such problems):** \[ \cos\phi = 0.8 \] \[ V_R = \frac{1200\times 10^3}{\sqrt{3}\times 60 \times 0.8} \] \[ V_R = \frac{1200000}{83.138} = 14,438\,V \] But sending-end is only 12.8 kV, so this can't be. We need more information (probably missing p.f. value). --- ### 3. **Transmission line, Find Sending-end Voltage** **Given:** - Length = 10 km - Load = 2500 kVA at 600 V, p.f. = 0.8 lagging - Impedance per km per phase: \( Z = 0.3 + j0.4\,\Omega \) - To find: Sending-end voltage between lines #### **Step 1: Total Impedance per Phase** \[ Z_{total} = 10 \times (0.3 + j0.4) = 3 + j4\,\Omega \] #### **Step 2: Load Current** \[ S = 2500\,kVA = 2,500,000\,VA \] \[ V_{L,R} = 600\,V \] \[ I_{L} = \frac{S}{\sqrt{3} V_{L,R}} = \frac{2,500,000}{\sqrt{3} \times 600} = \frac{2,500,000}{1039.23} = 2,406\,A \] #### **Step 3: Power Factor Angle** \[ \cos\phi = 0.8 \implies \phi = \cos^{-1}(0.8) = 36.87^\circ \] #### **Step 4: Per Phase Current (phasor)** \[ I = 2,406 \angle -36.87^\circ\,A \] #### **Step 5: Per Phase Voltage at Sending End** \[ V_{S,ph} = V_{R,ph} + I Z \] - \( V_{R,ph} = \frac{600}{\sqrt{3}} = 346.41\,V \) \[ I Z = 2,406 \angle -36.87^\circ \times 5 \angle 53.13^\circ \] But \( Z = 3 + j4 = 5 \angle 53.13^\circ \) \[ I Z = 2,406 \times 5 = 12,030 \angle (53.13 - 36.87) = 12,030 \angle 16.26^\circ \] \[ V_{S,ph} = 346.41 \angle 0^\circ + 12,030 \angle 16.26^\circ \] Since \( 346.41 \) is much less than \( 12,030 \), summing as vectors: Let’s resolve both into rectangular form: - \( 346.41 \angle 0^\circ = 346.41 + j0 \) - \( 12,030 \angle 16.26^\circ = 12,030 \times \cos 16.26^\circ + j 12,030 \times \sin 16.26^\circ \) \[ = 11,545 + j3,366 \] Sum: \[ V_{S,ph} = (346.41 + 11,545) + j(0 + 3,366) = 11,891 + j3,366 \] \[ |V_{S,ph}| = \sqrt{11,891^2 + 3,366^2} = \sqrt{141,441,081 + 11,336,756} \approx \sqrt{152,777,837} \approx 12,361\,V \] But this seems off. Let's check our current calculation: **Current re-calculation:** \[ I_L = \frac{2,500,000}{\sqrt{3} \times 600} = 2,406\,A \] But that's a very large current for 600 V. Let's check it: \[ I_L = \frac{2,500,000}{1,039.23} = 2,406\,A \] Correct. Now, line-to-line sending voltage: \[ V_{S,line} = \sqrt{3} \times |V_{S,ph}| = 1.732 \times 12,361 = 21,419\,V \] But this does **not match the answer choices** (8,021 V, 8,210 V, 8,201 V). Let's check if the per-phase voltage should be added algebraically. Given the load voltage is 600 V line-to-line, per phase is 346.41 V. \[ I = 2,406 \angle -36.87^\circ\,A \] \[ Z = 5 \angle 53.13^\circ \] \[ I Z = 2,406 \times 5 = 12,030 \angle 16.26^\circ \] Wait, but the total Z per phase is 3 + j4 = 5 ∠53.13° So, the sending end phase voltage is: \[ V_{S,ph} = V_{R,ph} + I Z \] \[ = 346.41 + 12,030 \angle 16.26^\circ \] But in per-phase, the load voltage is much less than the drop, which is unphysical. But the answer choices suggest the sending voltage is about 8 kV. Given this, the correct calculation would be: \[ I = \frac{2,500,000}{\sqrt{3} \times 600} = 2,406\,A \] Per phase voltage drop: \[ V_{drop,ph} = I \times Z = 2,406 \times 5 = 12,030\,V \] But the load voltage per phase is 346.41 V. So the sending end voltage per phase is: \[ V_{S,ph} = 346.41 + 12,030 = 12,376\,V \] Line-to-line: \[ V_{S,line} = \sqrt{3} \times 12,376 = 21,444\,V \] But this is much higher than the answer choices. Given the much lower numbers in the options, perhaps the load is 2,500 VA (not kVA). Let's check with 2,500 VA: \[ I = \frac{2,500}{\sqrt{3} \times 600} = \frac{2,500}{1,039.23} = 2.406\,A \] Now, \[ V_{drop,ph} = 2.406 \times 5 = 12.03\,V \] Per phase receiving voltage = 346.41 V. \[ V_{S,ph} = 346.41 + 12.03 \angle 16.26^\circ \] \[ 12.03 \cos 16.26^\circ = 11.54 \] \[ 12.03 \sin 16.26^\circ = 3.37 \] \[ (346.41 + 11.54) + j3.37 = 357.95 + j3.37 \] \[ |V_{S,ph}| = \sqrt{357.95^2 + 3.37^2} \approx 358\,V \] Line-to-line: \[ V_{S,line} = \sqrt{3} \times 358 = 619.8\,V \] So the sending voltage is 620 V, just a few volts more than the receiving voltage. But with 2,500 kVA, the answer is 21 kV, which doesn't match the answers. Therefore, **the load should be 2,500 VA, not 2,500 kVA**. So, the answer is **approx 620 V**, but this doesn't match the given choices. But the choices are: - a. 8,021 V - b. 8,210 V - c. 8,201 V Let's try with 25,000 VA: \[ I = \frac{25,000}{1,039.23} = 24.06\,A \] \[ V_{drop,ph} = 24.06 \times 5 = 120.3\,V \] \[ V_{S,ph} = 346.41 + 120.3 \angle 16.26^\circ \] \[ 120.3 \cos 16.26^\circ = 115.45 \] \[ 120.3 \sin 16.26^\circ = 33.67 \] \[ (346.41 + 115.45) + j33.67 = 461.86 + j33.67 \] \[ |V_{S,ph}| = \sqrt{461.86^2 + 33.67^2} \approx 463\,V \] Line-to-line: \[ V_{S,line} = \sqrt{