--- title: "L-L fault current in kA at bus 3" subject: "Power System Analysis II" chapter: 5 key: "semester_5/power_system_analysis_ii:ll-fault-bus3-ka" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ll-fault-bus3-ka site: BCT Engineering --- # L-L fault current in kA at bus 3 Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems - Asked in: 2073 Shrawan, 10 marks - Video: [Example 3.4. Line to line short circuit](https://www.youtube.com/watch?v=vx_TPt5qKWU) — Francisco Gonzalez-Longatt - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ll-fault-bus3-ka - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5 ## Question Figure below shows the power system network. i) Draw positive, negative and zero sequence networks ii) Determine fault current in kA if line to line fault occurs at Bus 3. System Data: | Equipment | MVA rating | Voltage rating | X1 pu | X2 pu | X0 pu | |---|---|---|---|---|---| | Generator, G1 | 50 | 11 kV | 0.2 | 0.2 | 0.05 | | Generator, G2 | 50 | 11 kV | 0.15 | 0.15 | 0.03 | | Transformer, T1 | 50 | 11/220 kV | 0.1 | 0.1 | 0.1 | | Transformer, T2 | 50 | 11/220 kV | 0.075 | 0.075 | 0.075 | | Line-1 | 50 | 220 kV | 0.12 | 0.12 | 0.42 | | Line-2 | 50 | 220 kV | 0.12 | 0.12 | 0.42 | The pre-fault voltage at Bus-3 is 0.95 p.u at common base of G1. [Figure: G1 (Y, solidly grounded) - bus 1 - T1 (Δ/Y grounded) - bus 2 - Line 1 and Line 2 in parallel - bus 3 - T2 (Y grounded through j0.03 on bus 3 side / Δ) - bus 4 - G2 (Y, grounded through j0.024)] ## Short answer Pre-fault voltage 0.95 pu, Thevenin Z1=Z2=j0.1385 pu (50 MVA, 220 kV base). Sequence current I_a1=0.95/(j0.277)= -j3.431 pu; L-L fault current = √3×3.431=5.94 pu = 0.78 kA. ## Answer **Base:** 50 MVA, 11 kV (generators), 220 kV (lines); all data are on this base. Pre-fault voltage at bus 3: $V_f = 0.95$ pu, no load. #### Single line diagram ```text G1 T1 Line-1 T2 G2 (~)-(1)-)(-(2)======================(3)-)(-(4)-(~) Yg D Yg Line-2 (parallel) Yg D Yg | | j0.03 j0.024 ``` #### (i) Sequence networks Lines in parallel: $X_1 = X_2 = 0.12/2 = 0.06$, $X_0 = 0.42/2 = 0.21$ pu. **Positive sequence** (emfs $E_1$, $E_2$): ```text E1-j0.2-(1)-j0.1-(2)-j0.06-(3)-j0.075-(4)-j0.15-E2 | F ``` **Negative sequence:** same reactances (0.2, 0.1, 0.06, 0.075, 0.15), no emfs, all ends to reference. **Zero sequence:** - G1 is behind the Δ of T1, so it is isolated; T1 connects bus 2 to reference through $j0.1$. - T2 star side (bus 3) grounded through $j0.03$: bus 3 to reference through $j(0.075 + 3\times0.03) = j0.165$; G2 is behind the Δ, so isolated. ```text ref-j0.1-(2)-j0.21-(3)-j0.165-ref | F G1 zero seq: j0.05 isolated; G2: j0.03+3(j0.024) isolated ``` $Z_0 = j0.31 \| j0.165 = j0.1077$ pu (not needed for L-L fault). #### (ii) L-L fault at bus 3 Thevenin impedances: $$ \begin{aligned} Z_1 &= j(0.2+0.1+0.06)\ \|\ j(0.075+0.15) = j0.36\ \|\ j0.225 = j0.1385\ \text{pu} \\ Z_2 &= Z_1 = j0.1385\ \text{pu} \end{aligned} $$ Sequence currents: $$ I_{a1} = -I_{a2} = \frac{V_f}{Z_1+Z_2} = \frac{0.95}{j0.2769} = -j3.431\ \text{pu} $$ Fault current: $$ |I_f| = |I_b| = |I_c| = \sqrt3 \times 3.431 = 5.942\ \text{pu} $$ Base current at 220 kV: $$ I_{base} = \frac{50\times10^6}{\sqrt3 \times 220\times10^3} = 131.2\ \text{A} $$ $$ I_f = 5.942 \times 131.2 = 779.7\ \text{A} = 0.780\ \text{kA} $$ **Answer: L-L fault current at bus 3 ≈ 5.94 pu ≈ 0.78 kA.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Is the zero-sequence network needed to find the L-L fault current? A. Yes, in series B. Yes, in parallel C. No, Ia0 = 0 D. Only if grounded Answer: C. For an L-L fault Ia0 = 0, so the zero network is not involved. (Key point: Zero network in L-L fault) 2. Z1 at bus 3 is found as: A. 0.36 + 0.225 in series B. 0.36 in parallel with 0.225 = 0.1385 pu C. 0.06 only D. 0.36 in parallel with 0.21 Answer: B. Looking from bus 3, the G1 side (0.2+0.1+0.06) and G2 side (0.075+0.15) are in parallel. (Key point: Thevenin Z1) 3. The L-L fault current magnitude is: A. Vf/(Z1+Z2) B. 3Vf/(Z1+Z2) C. sqrt3 Vf/(Z1+Z2) D. Vf/Z1 Answer: C. Ia1 = -Ia2 = Vf/(Z1+Z2), and |Ib| = sqrt3 x Ia1 = 5.94 pu. (Key point: L-L current formula) 4. The per-unit current is converted using the base current at: A. 11 kV B. The faulted bus voltage, 220 kV (131.2 A) C. The generator rating D. Average of 11 and 220 kV Answer: B. Bus 3 is on the 220 kV side, so Ibase = 50 MVA/(sqrt3 x 220 kV) = 131.2 A, giving 0.78 kA. 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