--- title: "Fault current: four 1600 kVA generators" subject: "Power Plant Equipment" chapter: 4 key: "semester_7/power_plant_equipment:fault-four-1600kva-generators" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-four-1600kva-generators site: BCT Engineering --- # Fault current: four 1600 kVA generators Power Plant Equipment · Chapter 4: Substation Equipments - Asked in: 2074 Asoj, 4+4 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-four-1600kva-generators - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4 ## Question Below figure shows 4 Nos of identical generators operating in parallel. Each generator is rated as 1600 kVA, 11kV, Xg = 0.2 pu. [Figure: G1, G2, G3, G4, each through Xg, connected to an 11 kV busbar at points A, B, C, D; an outgoing feeder from the busbar between B and C through a circuit breaker and a reactor X = 10 Ω, with a 3-phase to ground fault at its end] (i) Calculate fault current in outgoing feeder and fault current supplied by each generator. (ii) If reactors of 5 ohm each are connected between point A and B, point C and D respectively. Calculate fault current in outgoing feeder and fault current supplied by each generator. ## Short answer Using 11 kV phase voltage 6350.85 V and generator reactance 15.125 Ω, four parallel generators plus 10 Ω feeder give 13.781 Ω, so fault current is 460.83 A, 115.21 A per generator. With 5 Ω bus reactors between A-B and C-D, total current is 443.57 A; G1/G4 supply 95.16 A each, G2/G3 supply 126.62 A each. ## Answer Work in ohms at 11 kV. Phase voltage $V_{ph} = 11000/\sqrt3 = 6350.85$ V. Fault current $I_f = V_{ph}/X_{total}$. $$ \begin{aligned} Z_{base} &= \frac{11^2}{1.6} = 75.625\ \Omega \\ X_g &= 0.2\times75.625 = 15.125\ \Omega \end{aligned} $$ ```text G1 G2 G3 G4 |Xg |Xg |Xg |Xg A--[5]--B----+---C--[5]--D 11 kV bus | ([5] only in part ii) CB | X=10 ohm | F ``` #### (i) Without bus reactors All four generators are directly in parallel: $$ \begin{aligned} X_{par} &= \frac{15.125}{4} = 3.781\ \Omega \\ X_{total} &= 3.781 + 10 = 13.781\ \Omega \\ I_f &= \frac{6350.85}{13.781} = 460.83\ \text{A} \end{aligned} $$ Each generator supplies $460.83/4 = 115.21$ A. (Fault MVA $=\sqrt3\times11\times0.46083 = 8.78$ MVA.) #### (ii) With 5 Ω reactors between A–B and C–D G1 and G4 now reach the feeder through 5 Ω; G2 and G3 connect directly. $$ \begin{aligned} X_{G1} = X_{G4} &= 15.125 + 5 = 20.125\ \Omega \\ \frac{1}{X_{par}} &= \frac{2}{20.125} + \frac{2}{15.125} = 0.23161 \\ X_{par} &= 4.3176\ \Omega \\ X_{total} &= 4.3176 + 10 = 14.3176\ \Omega \\ I_f &= \frac{6350.85}{14.3176} = 443.57\ \text{A} \end{aligned} $$ The bus voltage behind the reactors is $I_f X_{par} = 443.57\times4.3176 = 1915.2$ V, so $$ \begin{aligned} I_{G1} = I_{G4} &= \frac{1915.2}{20.125} = 95.16\ \text{A} \\ I_{G2} = I_{G3} &= \frac{1915.2}{15.125} = 126.62\ \text{A} \end{aligned} $$ Check: $2(95.16) + 2(126.62) = 443.57$ A. **Answer:** (i) $I_f$ = **460.83 A**, each generator **115.21 A**. (ii) $I_f$ = **443.57 A**; G1 and G4 supply **95.16 A** each, G2 and G3 supply **126.62 A** each. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is Xg = 0.2 pu for a 1.6 MVA, 11 kV generator? A. 15.125 ohm B. 0.2 ohm C. 75.6 ohm D. 7.56 ohm Answer: A. Z_base = 11^2/1.6 = 75.625 ohm; times 0.2 = 15.125 ohm. (Key point: Generator reactance in ohms) 2. With four generators directly in parallel and a 10 ohm feeder, what is the fault current? A. 1843 A B. 460.8 A C. 6350 A D. 230 A Answer: B. X_total = 15.125/4 + 10 = 13.78 ohm; I = 6350.85/13.78 = 460.8 A. (Key point: Part (i) fault current) 3. How much does each generator supply in part (i)? A. 460.8 A each B. 230.4 A each C. 115.2 A, equal because they are identical D. Different for each generator Answer: C. Identical generators in parallel share equally: 460.83/4. (Key point: Part (i) sharing) 4. With 5 ohm reactors between A-B and C-D, which generators supply less current and why? A. G2 and G3, because they have the extra 5 ohm B. All four equal C. G1 and G4 supply more because of the reactors D. G1 and G4, because they reach the feeder through the extra 5 ohm Answer: D. G1 and G4 branches are 20.125 ohm versus 15.125 ohm, so they supply 95.2 A versus 126.6 A. 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