--- title: "SLG fault on bus with three alternators" subject: "Power System Analysis II" chapter: 5 key: "semester_5/power_system_analysis_ii:slg-three-alternators-common-bus" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/slg-three-alternators-common-bus site: BCT Engineering --- # SLG fault on bus with three alternators Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems - Asked in: 2075 Asoj, 12 marks - Video: [Single Line to Ground Fault on an Unloaded Generator/LG Fault](https://www.youtube.com/watch?v=WtJvn4aLcTE) — Dr.Nagaraja Kumari - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/slg-three-alternators-common-bus - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5 ## Question Three 6.6 kV, 3-phase, 10 MVA alternators are connected to a common bus. Each alternator has a positive sequence reactance of 0.15 pu. The negative and zero sequence reactances are 75% and 30% of positive sequence reactance. A single line-to-ground fault occurs on the bus. Find the fault current for the following cases: (i) All the alternator neutrals are solidly grounded. (ii) One alternator neutral is grounded through 0.3 ohm resistance and the other two neutrals are isolated. ## Short answer Using 10 MVA, 6.6 kV base, parallel reactances: (i) all neutrals solidly grounded, X0=0.015 pu, If=29.27 pu = 25.6 kA; (ii) one neutral through 0.3 Ω (3Rn=0.2066 pu), others isolated, Z0=0.2066+j0.045, If=12.22 pu = 10.7 kA. ## Answer **Base:** 10 MVA, 6.6 kV (rating of each alternator). $$ I_{base} = \frac{10\times10^6}{\sqrt3 \times 6.6\times10^3} = 874.8\ \text{A}, \quad Z_{base} = \frac{6.6^2}{10} = 4.356\ \Omega $$ Each alternator: $X_1 = 0.15$, $X_2 = 0.75 \times 0.15 = 0.1125$, $X_0 = 0.30 \times 0.15 = 0.045$ pu. Pre-fault voltage 1 pu, no load. The positive and negative networks always have all three machines in parallel: $$ X_1 = \frac{0.15}{3} = 0.05\ \text{pu}, \quad X_2 = \frac{0.1125}{3} = 0.0375\ \text{pu} $$ #### (i) All neutrals solidly grounded All three zero-sequence reactances are in parallel: $$ \begin{aligned} X_0 &= \frac{0.045}{3} = 0.015\ \text{pu} \\ I_f &= \frac{3E}{j(X_1+X_2+X_0)} = \frac{3}{j(0.05+0.0375+0.015)} = \frac{3}{j0.1025} = -j29.27\ \text{pu} \\ |I_f| &= 29.27 \times 874.8 = 25\,603\ \text{A} \approx 25.6\ \text{kA} \end{aligned} $$ #### (ii) One neutral grounded through 0.3 Ω, others isolated Only the grounded machine provides a zero-sequence path; its neutral resistance appears as $3R_n$: $$ \begin{aligned} R_n &= \frac{0.3}{4.356} = 0.06887\ \text{pu}, \quad 3R_n = 0.2066\ \text{pu} \\ Z_0 &= 3R_n + jX_0 = 0.2066 + j0.045\ \text{pu} \\ Z_1+Z_2+Z_0 &= 0.2066 + j(0.05 + 0.0375 + 0.045) = 0.2066 + j0.1325 \\ I_f &= \frac{3}{0.2066 + j0.1325} = 12.22\angle{-32.67^\circ}\ \text{pu} \\ |I_f| &= 12.22 \times 874.8 = 10\,692\ \text{A} \approx 10.7\ \text{kA} \end{aligned} $$ #### Results | Case | Fault current | |---|---| | (i) All neutrals solidly grounded | 29.27 pu = 25.6 kA | | (ii) One neutral through 0.3 Ω | 12.22 pu = 10.7 kA | Grounding only one machine through a resistance cuts the earth-fault current to about 42 % and is why generator neutrals on a common bus are normally grounded through impedance, with only one neutral grounded. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are X1 and X2 of the bus each found by dividing a machine's value by 3? A. The three machines are in parallel in the positive and negative networks B. The three machines are in series in those networks C. Only one machine feeds the fault D. The base MVA triples Answer: A. All three alternators feed the bus, so their positive and negative reactances are in parallel: 0.15/3 and 0.1125/3. (Key point: Sequence networks of parallel machines) 2. With all neutrals solidly grounded, what is the zero-sequence reactance seen from the bus? A. 0.045 pu B. 0.135 pu C. 0.015 pu D. 0.15 pu Answer: C. All three X0 = 0.045 pu paths to ground are in parallel, giving 0.045/3 = 0.015 pu. (Key point: Solidly grounded zero-sequence) 3. For the SLG fault, the fault current in case (i) is: A. E/(X1+X2+X0) = 9.76 pu B. 3E/(X1+X2+X0) = 29.27 pu C. sqrt3 E/(X1+X2) = 20 pu D. 3E/(X1+X2) = 34.3 pu Answer: B. I_f = 3E/(X1+X2+X0) = 3/0.1025 = 29.27 pu, about 25.6 kA. (Key point: Single line-to-ground formula) 4. In case (ii), how does the 0.3 ohm neutral resistance enter the calculation, and why do the isolated machines not matter? A. As Rn in the zero network; isolated machines still conduct B. As 3Rn in series with jX0 of the grounded machine; isolated ones give no zero-sequence path C. As Rn in the positive network; isolated machines add X0 in parallel D. As 3Rn divided by 3 machines; isolated machines share the current Answer: B. Neutral current is 3Ia0, so the zero network sees 3Rn in series with jX0, and only the grounded machine gives a ground path. 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