--- title: "Power by two generators, 1.05∠5° and 1.05∠20° pu" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:two-generators-1-05-pu" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-generators-1-05-pu site: BCT Engineering --- # Power by two generators, 1.05∠5° and 1.05∠20° pu Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2071 Bhadra, 6 marks - 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-generators-1-05-pu - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Determine the amount of power generated by each of the generators in the power system network below. Also specify the direction of power flow in the network. [Figure: E1 = 1.05∠5° pu behind reactance j0.08 pu, connected through a line of resistance 0.025 pu and reactance j0.1 pu to E2 = 1.05∠20° pu behind reactance j0.08 pu.] ## Short answer Total series impedance is 0.025+j0.26 pu and I from E1 to E2 is 1.0494 angle -162.01 pu. G2 generates 1.1012 pu real power while G1 absorbs 1.0737 pu, with 0.0275 pu line loss. G1 supplies 0.2477 pu reactive and G2 supplies 0.0386 pu reactive; real power flows from G2 to G1. ## Answer The two internal reactances and the line form one series path between the two emfs: $$ Z = j0.08 + (0.025 + j0.1) + j0.08 = 0.025 + j0.26\ \text{pu} $$ Let I flow from $E_1$ to $E_2$. Then $S_1 = E_1I^*$ is the power generated by G1 and $S_2 = E_2I^*$ is the power absorbed by G2. ```text j0.08 0.025+j0.1 j0.08 +-/\/\/-----/\/\/-------/\/\/-+ | I --> | (E1) 1.05/5 deg 1.05/20 deg (E2) | | +-----------------------------+ ``` #### Current $$ \begin{aligned} E_1 &= 1.05\angle5^\circ = 1.0460 + j0.0915\\ E_2 &= 1.05\angle20^\circ = 0.9867 + j0.3591\\ I &= \frac{E_1 - E_2}{Z} = \frac{0.0593 - j0.2676}{0.025 + j0.26}\\ &= -0.9981 - j0.3242 = 1.0494\angle-162.01^\circ\ \text{pu} \end{aligned} $$ #### Powers $$ \begin{aligned} S_1 &= E_1I^* = -1.0737 + j0.2477\ \text{pu}\\ S_2 &= E_2I^* = -1.1012 - j0.0386\ \text{pu}\ \text{(absorbed by G2)}\\ S_{loss} &= |I|^2Z = 1.1013(0.025 + j0.26) = 0.0275 + j0.2863\ \text{pu} \end{aligned} $$ | Generator | Real power | Reactive power | |---|---|---| | G1 | absorbs 1.0737 pu (acts as motor) | generates 0.2477 pu | | G2 | generates 1.1012 pu | generates 0.0386 pu | Check: $1.1012 - 1.0737 = 0.0275$ pu (I²R loss) and $0.2477 + 0.0386 = 0.2863$ pu (I²X). #### Direction of power flow - Real power flows from **G2 to G1**, since $E_2$ leads $E_1$ by 15°. - Both generators supply reactive power to the network, and it is consumed in the series reactances. With equal emf magnitudes, each machine supplies part of the $I^2X$. **Answer:** G2 generates 1.101 pu real power and G1 receives 1.074 pu. The loss is 0.0275 pu. G1 and G2 supply 0.248 pu and 0.039 pu reactive power. P flows G2 → G1. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the total impedance between the two emfs? A. 0.025 + j0.1 pu only B. j0.08 + (0.025 + j0.1) + j0.08 = 0.025 + j0.26 pu C. j0.26 pu only D. 0.05 + j0.26 pu Answer: B. Internal reactances and the line are in series and add. (Key point: Total series impedance) 2. In which direction does real power flow? A. From G2 to G1, since E2 leads E1 by 15° B. From G1 to G2 since E1 is lagging C. From G1 to G2 since both emf magnitudes are equal D. There is no real power flow since magnitudes are equal Answer: A. Real power flows from the leading emf to the lagging one. (Key point: Direction of real power) 3. What does a negative real part of S1 = E1 I* mean? A. G1 generates more than G2 B. The line has no loss C. G1 absorbs reactive power only D. G1 absorbs real power and acts as a motor Answer: D. Negative P of the source power means it takes power in. (Key point: Role of G1) 4. How is the result checked? A. Sum of real powers equals the loss B. Both Q values must be equal C. Difference of real powers equals I^2R loss (0.0275 pu); sum of Q equals I^2X D. Real powers must be equal Answer: C. G2 generates 1.1012, G1 absorbs 1.0737; the difference is the line loss 0.0275 pu. (Key point: Loss check) ## Close variants set in other exams - [Power by two generators, 1.1∠15° and 1.1∠0° pu](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-generators-1-1-pu) — Power System Analysis I, chapter 3, asked 2× ## Related in Power System Analysis I - [Reactance diagram: two generators, two motors, j50 Ω line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-two-gens-two-motors) — Power System Analysis I, chapter 3 - [Reactance diagram: three generators, two 220 kV lines](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-three-gens-two-lines) — Power System Analysis I, chapter 3 - [Two sources via 0 - j5 Ω: power flow](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-sources-0-j5) — Power System Analysis I, chapter 3 - [Reactance diagram: 300 MVA, 20 kV generator, two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-300mva-20kv) — Power System Analysis I, chapter 3 - [Reactance diagram: 50 MVA, 13.8 kV base, three generators](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-50mva-13-8kv) — Power System Analysis I, chapter 3 - [Vs from P+jQ and reactive power flow direction](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Avs-from-pq-reactive-flow) — Power System Analysis I, chapter 6 ## Related in other subjects - [N-R first iteration, line reactance 0.1](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Anr-first-iteration-x0-1) — Power System Analysis II, chapter 2 - [G-S two iterations, lines 0.02+j0.11](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-two-iterations-2080) — Power System Analysis II, chapter 2 - [G-S two iterations, lines 0.026+j0.11](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-two-iterations-2075) — Power System Analysis II, chapter 2 - [Neutral reactance/resistance to limit L-G current](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aneutral-reactance-resistance-limit-lg) — Power System Analysis II, chapter 5