--- title: "Line-to-line fault current calculation" subject: "Power System Analysis II" chapter: 5 key: "semester_5/power_system_analysis_ii:line-to-line-fault-calculation" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/line-to-line-fault-calculation site: BCT Engineering --- # Line-to-line fault current calculation Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems - Asked in: 2075 Asoj, 6 marks; 2074 Asoj, 8 marks; 2074 Chaitra, 9 marks; 2068 Chaitra, 8 marks (asked 4 times) - Video: [LINE TO LINE FAULT ON AN UNLOADED SYNCHRONOUS GENERATOR](https://www.youtube.com/watch?v=DnX0an2FgWc) — P SUMALATHA - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/line-to-line-fault-calculation - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5 ## Question Explain (derive) with necessary mathematical expressions and diagrams how fault current and bus voltages are calculated when a line-to-line fault occurs in a 3-phase power system network. Draw a diagram showing interconnection of sequence networks for this type of fault. ## Short answer For a line-to-line fault, I_a0=0 and I_a2 = –I_a1, so positive and negative networks connect in parallel opposition through Z_f. I_a1 = V_f/(Z_1+Z_2+Z_f). Fault current I_f = –j√3 V_f/(Z_1+Z_2+Z_f). Sequence voltages at the fault bus are V_a1=V_f–Z_1 I_a1, V_a2=Z_2 I_a1; voltages at other buses follow from the Z-bus matrices. ## Answer In a line-to-line (L-L) fault the zero sequence current is zero, $I_{a1} = -I_{a2}$, and the positive and negative sequence networks are connected in parallel (opposition) through $Z_f$ at the fault point. Fault current is $I_f = -j\sqrt3\,E_a/(Z_1+Z_2+Z_f)$. **Model:** the network is represented at the fault bus k by its Thevenin sequence equivalents: prefault voltage $V_f = V_k(0)$ and impedances $Z_1 = Z_{kk}^{(1)}$, $Z_2 = Z_{kk}^{(2)}$, $Z_0 = Z_{kk}^{(0)}$ (from the sequence Z-bus matrices). For a generator terminal fault, $V_f = E_a$ and $Z$ are the machine's own values. #### Boundary conditions (fault between phases b and c through $Z_f$) ```text a o----------- Ia = 0 b o-----+ Ib = If --> Zf c o-----+ Ic = -Ib ``` $$ I_a = 0, \qquad I_b + I_c = 0, \qquad V_b - V_c = Z_fI_b $$ #### Sequence currents $$ \begin{aligned} I_{a0} &= \tfrac13(I_a + I_b + I_c) = \tfrac13(0 + I_b - I_b) = 0 \\ I_{a1} &= \tfrac13(I_a + aI_b + a^2I_c) = \tfrac13(a - a^2)I_b \\ I_{a2} &= \tfrac13(I_a + a^2I_b + aI_c) = \tfrac13(a^2 - a)I_b \end{aligned} $$ Hence $$ I_{a0} = 0, \qquad I_{a2} = -I_{a1} $$ #### Voltage condition $$ \begin{aligned} V_b - V_c &= (a^2 - a)V_{a1} + (a - a^2)V_{a2} = (a^2 - a)(V_{a1} - V_{a2}) \\ Z_fI_b &= Z_f\,(a^2I_{a1} + aI_{a2}) = Z_f(a^2 - a)I_{a1} \end{aligned} $$ so $$ V_{a1} - V_{a2} = Z_fI_{a1} $$ With $V_{a1} = V_f - Z_1I_{a1}$ and $V_{a2} = -Z_2I_{a2} = Z_2I_{a1}$: $$ I_{a1} = -I_{a2} = \frac{V_f}{Z_1 + Z_2 + Z_f} $$ #### Fault current $$ \begin{aligned} I_b &= I_{a0} + a^2I_{a1} + aI_{a2} = (a^2 - a)I_{a1} = -j\sqrt3\,I_{a1} \\ I_f &= I_b = -I_c = \frac{-j\sqrt3\,V_f}{Z_1 + Z_2 + Z_f} \end{aligned} $$ #### Voltages at the fault bus $$ \begin{aligned} V_{a1} &= V_f - Z_1I_{a1}, \quad V_{a2} = Z_2I_{a1}, \quad V_{a0} = 0 \\ V_a &= V_{a1} + V_{a2}, \quad V_b = a^2V_{a1} + aV_{a2}, \quad V_c = aV_{a1} + a^2V_{a2} \end{aligned} $$ For a bolted fault ($Z_f = 0$), $V_{a1} = V_{a2}$, so $V_a = 2V_{a1}$ and $V_b = V_c = -V_{a1}$. #### Voltages at other buses (i) of a network Using the sequence bus impedance matrices, with the fault at bus k: $$ V_i^{(1)} = V_i(0) - Z_{ik}^{(1)}I_{a1}, \qquad V_i^{(2)} = -Z_{ik}^{(2)}I_{a2}, \qquad V_i^{(0)} = 0 $$ Phase voltages follow from $V_{abc} = AV_{012}$ (with the $\pm30^\circ$ phase shift of Δ-Y transformers applied to the positive and negative sequence values when crossing them). Line currents then follow from $I_{ij} = (V_i - V_j)/z_{ij}$ in each sequence network. #### Interconnection of sequence networks ```text Ia1 --> <-- Ia2 +--(Ea)--Z1---o F1--Zf--o F2---Z2--+ | | N1 ------------------------------- N2 Zero sequence network: not connected (Ia0 = 0) ``` The positive and negative sequence networks are connected **in parallel (in opposition)** through $Z_f$; the zero sequence network is left open. #### Steps in summary 1. Form $Z_1$, $Z_2$ at the fault point (Thevenin), and take $V_f$ = prefault voltage. 2. $I_{a1} = -I_{a2} = V_f/(Z_1+Z_2+Z_f)$, $I_{a0} = 0$. 3. $I_f = -j\sqrt3 I_{a1}$. 4. Find sequence voltages, then phase voltages at the fault and other buses. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which set of conditions describes an L-L fault between phases b and c through Zf? A. Ib = Ic = 0, Va = Zf Ia B. Ia = 0, Ib = -Ic, Vb - Vc = Zf Ib C. Va = Vb = Vc = 0 D. Ia = Ib = Ic Answer: B. Phase a is healthy, fault current flows b to c, and the voltage across the fault is Zf Ib. (Key point: Boundary conditions) 2. What do the boundary conditions give for the sequence currents? A. Ia0 = Ia1 = Ia2 B. Ia1 = Ia2 = 0 C. Ia0 = Ia1 + Ia2 D. Ia0 = 0 and Ia2 = -Ia1 Answer: D. Ia + Ib + Ic = 0 gives Ia0 = 0, and the transform gives Ia2 = -Ia1. (Key point: Sequence relations) 3. How are the sequence networks connected for an L-L fault? A. Positive and negative in parallel (opposition) through Zf, zero network open B. All three in series C. All three in parallel D. Positive and zero in series, negative open Answer: A. Va1 - Va2 = Zf Ia1 and Ia1 = -Ia2 gives positive and negative linked through Zf; no zero network. (Key point: Network interconnection) 4. Which is the L-L fault current magnitude? A. Vf/(Z1 + Z2 + Z0) B. 3Vf/(Z1 + Z2 + Zf) C. sqrt(3) Vf/(Z1 + Z2 + Zf) D. Vf/(Z1 + Z2 + 3Zf) Answer: C. Ib = (a^2 - a)Ia1 = -j sqrt(3) Ia1 with Ia1 = Vf/(Z1+Z2+Zf). 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