--- title: "LLG fault sequence currents at bus 3" subject: "Power System Analysis II" chapter: 5 key: "semester_5/power_system_analysis_ii:llg-fault-bus3-sequence-currents" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/llg-fault-bus3-sequence-currents site: BCT Engineering --- # LLG fault sequence currents at bus 3 Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems - Asked in: 2076 Chaitra, 10 marks - Video: [Lecture 10 unsymmetrical fault example 3](https://www.youtube.com/watch?v=n5ELfExB3Z4) — YanbuCollege - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/llg-fault-bus3-sequence-currents - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5 ## Question Compute the sequence currents for a LLG fault at bus 3 of the following network. The fault impedance is j0.1 pu. All the parameters are in pu. [Figure: G1 (Y grounded) - T1 (Y grounded/Y grounded) - bus 1; G2 (Y grounded) - T2 (Δ on generator side/Y grounded) - bus 2; L1 between buses 1 and 2, L2 between buses 1 and 3, L3 between buses 2 and 3] | Item | Base MVA | X1 | X2 | X0 | |---|---|---|---|---| | G1 | 100 | 0.15 | 0.15 | 0.05 | | G2 | 100 | 0.15 | 0.15 | 0.05 | | T1 | 100 | 0.1 | 0.1 | 0.1 | | T2 | 100 | 0.1 | 0.1 | 0.1 | | L1 | 100 | 0.12 | 0.12 | 0.3 | | L2 | 100 | 0.15 | 0.15 | 0.35 | | L3 | 100 | 0.25 | 0.25 | 0.71 | ## Short answer Thevenin Z1=j0.2200 pu, Z0=j0.3032 pu at bus 3. With Zf=j0.1 pu, sequence currents: I_a1 = -j2.624 pu, I_a2 = j1.923 pu, I_a0 = j0.701 pu. Ground fault current = 3I_a0 = j2.103 pu. ## Answer **Approach:** find the Thevenin impedances at bus 3 from the bus impedance matrices of the positive and zero sequence networks ($Z_2 = Z_1$ since all $X_1 = X_2$), then connect the networks for an LLG fault. Pre-fault voltage $V_f = 1\angle0^\circ$ pu, no load; $Z_f = j0.1$ pu is the impedance from the faulted phases' common point to ground. #### Positive sequence network Generator + transformer branches to reference: bus 1: $j(0.15+0.1) = j0.25$; bus 2: $j0.25$. Lines: 1-2 $j0.12$, 1-3 $j0.15$, 2-3 $j0.25$. ```text E1 E2 | | j0.25 j0.25 | | (1)--j0.12--(2) \ / j0.15 j0.25 \ / (3) F ``` Forming $Y_{bus}$ and inverting gives $$ Z_{bus}^{(1)} = j\begin{bmatrix} 0.1445 & 0.1055 & 0.1299 \\ 0.1055 & 0.1445 & 0.1201 \\ 0.1299 & 0.1201 & 0.2200 \end{bmatrix} $$ So $Z_1 = Z_2 = Z_{33}^{(1)} = j0.2200$ pu. #### Zero sequence network - G1 (Yg) + T1 (Yg/Yg): zero-seq path passes, bus 1 to reference $j(0.05+0.1) = j0.15$ - G2 behind T2 (Δ on generator side, Yg on bus side): bus 2 to reference through $j0.1$ only; G2 is cut off - Lines: 1-2 $j0.3$, 1-3 $j0.35$, 2-3 $j0.71$ $$ Z_{bus}^{(0)} = j\begin{bmatrix} 0.1035 & 0.0310 & 0.0796 \\ 0.0310 & 0.0793 & 0.0470 \\ 0.0796 & 0.0470 & 0.3032 \end{bmatrix} $$ So $Z_0 = j0.3032$ pu. #### Sequence currents for LLG fault (phases b, c to ground through $Z_f$) The negative network is in parallel with the zero network plus $3Z_f$: $$ \begin{aligned} Z_0 + 3Z_f &= j0.3032 + j0.3 = j0.6032 \\ I_{a1} &= \frac{V_f}{Z_1 + \dfrac{Z_2(Z_0+3Z_f)}{Z_2 + Z_0 + 3Z_f}} = \frac{1}{j0.2200 + j\dfrac{0.2200 \times 0.6032}{0.8232}} = \frac{1}{j0.3812} = -j2.6236\ \text{pu} \\ I_{a2} &= -I_{a1}\frac{Z_0+3Z_f}{Z_2+Z_0+3Z_f} = j2.6236 \times \frac{0.6032}{0.8232} = j1.9225\ \text{pu} \\ I_{a0} &= -I_{a1}\frac{Z_2}{Z_2+Z_0+3Z_f} = j2.6236 \times \frac{0.2200}{0.8232} = j0.7011\ \text{pu} \end{aligned} $$ Check: $I_{a1} + I_{a2} + I_{a0} = -j2.6236 + j1.9225 + j0.7011 = 0 = I_a$. #### Phase and fault currents $$ \begin{aligned} I_b &= I_{a0} + a^2 I_{a1} + a I_{a2} = 4.075\angle{165.05^\circ}\ \text{pu} \\ I_c &= I_{a0} + a I_{a1} + a^2 I_{a2} = 4.075\angle{14.95^\circ}\ \text{pu} \\ I_f &= I_b + I_c = 3I_{a0} = j2.103\ \text{pu} \end{aligned} $$ **Answer:** $I_{a1} = -j2.624$ pu, $I_{a2} = j1.923$ pu, $I_{a0} = j0.701$ pu; fault (ground) current $= 2.103$ pu; $|I_b| = |I_c| = 4.075$ pu. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is G2 not part of the zero sequence network? A. G2 has X0 = 0 B. T2 has a delta on the generator side, blocking zero sequence flow C. T2 is Yg-Yg D. L3 has high X0 Answer: B. The delta on the generator side isolates G2, so bus 2 connects to reference through T2's j0.1 only. (Key point: Zero sequence network from T2) 2. How are Z1, Z2 and Z0 at bus 3 obtained? A. Series sum of element reactances B. From diagonal element Z33 of the bus impedance matrices of each network C. From the generator values only D. By averaging the line reactances Answer: B. Thevenin impedance at a bus is the diagonal element: Z1 = Z2 = j0.2200 and Z0 = j0.3032. (Key point: Thevenin impedances) 3. For an LLG fault through Zf, how are the networks connected? A. All in series B. Positive in series with the parallel of negative and (zero + 3Zf) C. All in parallel D. Positive and negative in parallel, zero open Answer: B. Ia1 = Vf / [Z1 + Z2 (Z0 + 3Zf)/(Z2 + Z0 + 3Zf)]. (Key point: LLG network connection) 4. With Z0 + 3Zf = j0.6032, what is Ia1? A. -j2.624 pu B. -j4.545 pu C. -j1.923 pu D. -j0.701 pu Answer: A. Total = 0.2200 + 0.2200x0.6032/0.8232 = 0.3812, so Ia1 = 1/j0.3812 = -j2.624 pu. (Key point: Positive sequence current) 5. Which check confirms the sequence currents of an LLG fault? A. Ia1 = Ia2 = Ia0 B. Ia1 + Ia2 + Ia0 = Ia = 0 C. Ia0 = 0 D. Ia2 = -Ia1 Answer: B. Phase a is healthy, so the sum -j2.6236 + j1.9225 + j0.7011 = 0. 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