--- title: "Fault current: two 500 kVA alternators" subject: "Power Plant Equipment" chapter: 4 key: "semester_7/power_plant_equipment:fault-two-500kva-alternators" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-two-500kva-alternators site: BCT Engineering --- # Fault current: two 500 kVA alternators Power Plant Equipment · Chapter 4: Substation Equipments - Asked in: 2073 Chaitra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-two-500kva-alternators - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4 ## Question Below figure shows two Alternators operating in parallel. [Figure: G1 500 kVA, X1 = 15%, connected through a breaker to point A; G2 500 kVA, X2 = 15%, connected through a breaker to point C; points A, B, C lie on the 11 kV busbar; the feeder leaves from point B through a breaker and a reactor X = 20 ohms, with a 3 phase to ground fault on the feeder] (i) Calculate the fault current in outgoing feeder and fault current supplied by each alternator. (ii) If a reactor of 10 ohm is connected between point A and point B, calculate new fault current in outgoing feeder and fault current supplied by each alternator. ## Short answer Using ohmic values, each generator reactance is 36.3 Ω; with 20 Ω feeder, total is 38.15 Ω, giving 166.47 A fault current, 83.24 A per alternator. With 10 Ω between A and B, total is 40.347 Ω, giving 157.40 A; G1 supplies 69.17 A and G2 supplies 88.23 A. ## Answer Work in ohms. Phase voltage $V_{ph} = 11000/\sqrt3 = 6350.85$ V, and $I_f = V_{ph}/X_{total}$. $$ \begin{aligned} Z_{base} &= \frac{11^2}{0.5} = 242\ \Omega \\ X_{G1} = X_{G2} &= 0.15\times242 = 36.3\ \Omega \end{aligned} $$ ```text G1 500kVA G2 500kVA 15% 15% | | CB CB | | ===A==[10]===B======C=== 11 kV | ([10] only in part ii) CB | X=20 ohm | F ``` #### (i) Without reactor between A and B $$ \begin{aligned} X_{par} &= \frac{36.3}{2} = 18.15\ \Omega \\ X_{total} &= 18.15 + 20 = 38.15\ \Omega \\ I_f &= \frac{6350.85}{38.15} = 166.47\ \text{A} \end{aligned} $$ Each alternator supplies $166.47/2 = 83.24$ A. #### (ii) With 10 Ω between A and B $$ \begin{aligned} X_{G1,path} &= 36.3 + 10 = 46.3\ \Omega \\ X_{par} &= \frac{46.3\times36.3}{46.3+36.3} = 20.347\ \Omega \\ X_{total} &= 20.347 + 20 = 40.347\ \Omega \\ I_f &= \frac{6350.85}{40.347} = 157.40\ \text{A} \end{aligned} $$ Sharing (inversely as path reactance): $$ \begin{aligned} I_{G1} &= 157.40\times\frac{36.3}{82.6} = 69.17\ \text{A} \\ I_{G2} &= 157.40\times\frac{46.3}{82.6} = 88.23\ \text{A} \end{aligned} $$ **Answer:** (i) $I_f$ = **166.47 A**, each alternator **83.24 A**. (ii) $I_f$ = **157.40 A**; G1 supplies **69.17 A**, G2 supplies **88.23 A**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the alternator reactance in ohms found from 15% on 500 kVA, 11 kV? A. 0.15 x 11^2/0.5 = 36.3 ohm B. 0.15 x 11/0.5 = 3.3 ohm C. 0.15 x 0.5/11^2 = 0.0006 ohm D. 0.15 x 11000/sqrt3 = 952 ohm Answer: A. Z_base = kV^2/MVA = 121/0.5 = 242 ohm, and X = 0.15 x 242 = 36.3 ohm. (Key point: Reactance in ohms from percentage) 2. In part (i), what is the total reactance to the fault and the fault current? A. 72.6 + 20 = 92.6 ohm, 68.6 A B. 36.3 + 20 = 56.3 ohm, 112.8 A C. 18.15 + 20 = 38.15 ohm, 166.47 A D. 18.15 ohm, 349.9 A Answer: C. The two alternators are in parallel (18.15 ohm) and in series with the 20 ohm feeder reactor; I = 6350.85/38.15 = 166.47 A. (Key point: Fault current without reactor) 3. Why does each alternator supply 83.24 A in part (i)? A. Each supplies the full 166.47 A B. The alternators are identical so they share the fault current equally C. G1 supplies more because it is nearer the feeder D. Each supplies the square root of the total Answer: B. Both have the same reactance and the same path, so the total current divides equally. (Key point: Sharing of current in part (i)) 4. With a 10 ohm reactor between A and B, which result is correct? A. G1 and G2 still share equally B. G1 supplies less current than G2 because its path has higher reactance C. G1 supplies more current than G2 because its path has higher reactance D. The feeder fault current rises above 166.47 A Answer: B. G1's path is 46.3 ohm vs 36.3 ohm for G2, and current divides inversely to reactance: 69.17 A vs 88.23 A, and total falls to 157.40 A. (Key point: Effect of 10 ohm reactor between A and B) ## Related in Power Plant Equipment - [Fault current: four 1600 kVA generators](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-four-1600kva-generators) — Power Plant Equipment, chapter 4 - [Fault current: two 50 MW generators, reactor inductance](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-50mw-inductance) — Power Plant Equipment, chapter 4 - [Fault level: three generators, feeder reactor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-three-generators-feeder) — Power Plant Equipment, chapter 4 - [Fault level: adding a third generator](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-third-generator-added) — Power Plant Equipment, chapter 4 - [Fault level: two 25 MVA generators, 50% reduction](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-25mva-50-percent) — Power Plant Equipment, chapter 4 - [Fault current: 50 MW and 40 MW generators](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-50mw-40mw-equal-current) — Power Plant Equipment, chapter 4 ## Related in other subjects - [L-L fault through Zf with two motors](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-motors-zf-new) — Power System Analysis II, chapter 5 - [Fault at receiving end of 220 kV lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-220kv-two-generators) — Power System Analysis II, chapter 3 - [L-G fault at bus 1 through Zf](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-bus1-parallel-lines) — Power System Analysis II, chapter 5 - [L-L fault current in kA at bus 3](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-bus3-ka) — Power System Analysis II, chapter 5