--- title: "Reactive power flow between machines via j10 Ω" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:two-machines-reactive-j10" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-machines-reactive-j10 site: BCT Engineering --- # Reactive power flow between machines via j10 Ω Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2068 Bhadra, 6 marks - Video: [Active & Reactive Power Flow in Power Systems](https://www.youtube.com/watch?v=TIud3FoQxhI) — PQES - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-machines-reactive-j10 - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Two 3-phase synchronous machines are connected by an inductive link of j10 Ohms/phase. If the emfs of machine A and machine B are VA = 200∠10° V, VB = 200∠30° V, determine the direction of the reactive power flow and reactive power loss in the link. Neglect the voltage drops at both the machines. ## Short answer With equal voltage magnitudes, reactive power flows from both machines into the inductive link: each supplies 241.2 var per phase, and the link loses 482.5 var per phase (1447.4 var three-phase). Real power flows from machine B to machine A at 1368.1 W per phase. ## Answer Per phase, with I flowing from A to B through $Z = j10\ \Omega$. Then $S_A = V_AI^*$ is the power supplied by A and $S_B = V_BI^*$ is the power received by B. #### Current $$ \begin{aligned} V_A &= 200\angle10^\circ = 196.96 + j34.73\ \text{V}\\ V_B &= 200\angle30^\circ = 173.21 + j100.00\ \text{V}\\ I &= \frac{V_A - V_B}{j10} = \frac{23.76 - j65.27}{j10} = -6.527 - j2.376\\ &= 6.946\angle-160^\circ\ \text{A} \end{aligned} $$ #### Powers $$ \begin{aligned} S_A &= V_AI^* = (200\angle10^\circ)(6.946\angle160^\circ) = -1368.08 + j241.23\ \text{VA}\\ S_B &= V_BI^* = (200\angle30^\circ)(6.946\angle160^\circ) = -1368.08 - j241.23\ \text{VA}\\ Q_{loss} &= |I|^2X = 48.246\times10 = 482.46\ \text{var} \end{aligned} $$ Check: $Q_{loss} = \frac{|V_A - V_B|^2}{X} = \frac{(2\times200\sin10^\circ)^2}{10} = \frac{69.46^2}{10} = 482.46$ var. #### Direction of reactive power - Machine A supplies +241.23 var into the link. - Machine B receives −241.23 var, so it also **supplies** 241.23 var into the link. - So reactive power flows **from both ends into the link**. There is no net transfer of vars from one machine to the other, because the voltage magnitudes are equal (200 V each). Each machine supplies half of the link's $I^2X$. - Real power: 1368.08 W per phase flows from B (leading, 30°) to A (lagging, 10°). | Quantity | Per phase | Three-phase | |---|---|---| | Q supplied by A | 241.23 var | 723.69 var | | Q supplied by B | 241.23 var | 723.69 var | | Reactive loss in link | 482.46 var | 1447.38 var | | Real power B → A | 1368.08 W | 4104.24 W | **Answer:** Reactive power flows from both machines into the link, 241.2 var each per phase. The reactive loss in the link is 482.5 var per phase (1447.4 var for three phases). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the direction of reactive power flow? A. From A to B B. From both machines into the link, since magnitudes are equal C. From B to A D. None; the link carries no vars Answer: B. Equal magnitudes: each supplies half of I^2X. (Key point: Direction of Q) 2. How is the loss in the link calculated? A. Q = I^2 X = 48.246 x 10 = 482.46 var per phase B. Q = I^2 R C. Q = V^2/X using 200 V D. Q = 0 Answer: A. Also check |VA - VB|^2/X = 482.46. (Key point: Reactive loss) 3. Which way does real power flow? A. From A to B B. None C. Cannot be determined D. From B to A, since VB leads VA by 20° Answer: D. Leading voltage is B at 30°. (Key point: Real power direction) 4. How much Q does each machine supply per phase? A. 482.46 var B. 0 var C. 241.23 var D. 723.69 var Answer: C. Half of the 482.46 var loss each; 723.69 var is the three-phase value. 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