--- title: "SLG fault: sequence networks in series" subject: "Power System Analysis II" chapter: 5 key: "semester_5/power_system_analysis_ii:slg-sequence-networks-series" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/slg-sequence-networks-series site: BCT Engineering --- # SLG fault: sequence networks in series Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems - Asked in: 2081 Bhadra, 8 marks; 2078 Kartik, 8 marks; 2073 Chaitra, 8 marks; 2072 Kartik, 6 marks; 2071 Shrawan, 6 marks (asked 5 times) - Video: [Single Line to Ground Fault Derivation | LG Fault | SLG Fault| Fault Analysis| Power System Analysis](https://www.youtube.com/watch?v=L-GoTzpeojw) — Simple n Easy Learning - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/slg-sequence-networks-series - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5 ## Question Starting from a suitable point, show that the three sequence networks are connected in series during a single line to ground fault in a 3-phase power system (unloaded generator), and develop the mathematical expression for the fault current. ## Short answer Boundary conditions give I_a0 = I_a1 = I_a2 and V_a = 3Z_f I_a1. Substituting V_a1 = E_a – Z_1 I_a1, etc., yields I_a1 = E_a / (Z_1+Z_2+Z_0+3Z_f). The same current flows through all, so the three sequence networks connect in series. Fault current I_f = 3E_a/(Z_1+Z_2+Z_0+3Z_f). ## Answer In a single line to ground (L-G) fault, the three sequence currents of the faulted phase are equal and the sequence voltages add to $3Z_fI_{a1}$, so the positive, negative and zero sequence networks are connected in **series**, and $I_f = 3E_a/(Z_1+Z_2+Z_0+3Z_f)$. **Starting point:** an unloaded star-connected generator with neutral grounded through $Z_n$, generating balanced EMFs $E_a$, $a^2E_a$, $aE_a$; fault on phase a through fault impedance $Z_f$. #### Sequence networks of an unloaded generator The generator EMFs are balanced (positive sequence only), so with $Z_0 = Z_{g0} + 3Z_n$: $$ V_{a0} = -Z_0I_{a0}, \qquad V_{a1} = E_a - Z_1I_{a1}, \qquad V_{a2} = -Z_2I_{a2} $$ ```text Positive Negative Zero F1 o F2 o F0 o |Ia1 ^ |Ia2 ^ |Ia0 ^ Z1 Z2 Z0 | | | (Ea) | | | | | N1 o N2 o N0 o ``` #### Boundary conditions (fault on phase a through $Z_f$) ```text a o----------+ b o--- (Ib=0) c o--- (Ic=0) | Ia = If Zf | ground ``` $$ I_b = 0, \qquad I_c = 0, \qquad V_a = Z_fI_a $$ #### Sequence currents $$ \begin{bmatrix} I_{a0}\\I_{a1}\\I_{a2} \end{bmatrix} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}\begin{bmatrix} I_a\\0\\0 \end{bmatrix} \quad\Rightarrow\quad I_{a0} = I_{a1} = I_{a2} = \frac{I_a}{3} $$ So the **same current flows in all three sequence networks**, which means they are in series. #### Voltage condition $$ V_a = V_{a0} + V_{a1} + V_{a2} = Z_fI_a = 3Z_fI_{a1} $$ Substituting the network equations: $$ -Z_0I_{a1} + E_a - Z_1I_{a1} - Z_2I_{a1} = 3Z_fI_{a1} $$ $$ I_{a1} = I_{a2} = I_{a0} = \frac{E_a}{Z_1 + Z_2 + Z_0 + 3Z_f} $$ #### Fault current $$ I_f = I_a = 3I_{a1} = \frac{3E_a}{Z_1 + Z_2 + Z_0 + 3Z_f} $$ For a bolted fault ($Z_f = 0$): $I_f = \dfrac{3E_a}{Z_1+Z_2+Z_0}$. #### Interconnection of sequence networks ```text F1 --(Ea)--[Z1]-- N1 == F2 --[Z2]-- N2 == F0 --[Z0]-- N0 | | +------------------------[ 3Zf ]------------------------+ common current Ia1 = Ia2 = Ia0 ``` The three networks are connected in **series** with $3Z_f$: $N_1$ is joined to $F_2$ (the "==" links), $N_2$ is joined to $F_0$, and $N_0$ returns to $F_1$ through $3Z_f$. The common current $I_{a1}$ flows through all of them. Both conditions ($I_{a0} = I_{a1} = I_{a2}$ and $V_{a0}+V_{a1}+V_{a2} = 3Z_fI_{a1}$) are satisfied only by this series connection. #### Remarks - The neutral current is $I_n = 3I_{a0} = I_f$; it flows only if the neutral is grounded. For an ungrounded neutral, $Z_0 = \infty$ and $I_f = 0$. - If $Z_0$ is small (solidly grounded), the L-G fault current can exceed the three-phase fault current $E_a/Z_1$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For an L-G fault on phase a with Ib = Ic = 0, what is the relation between the sequence currents? A. Ia0 = Ia1 = Ia2 = Ia/3 B. Ia0 = 0 C. Ia1 = -Ia2 D. Ia0 + Ia1 + Ia2 = 0 Answer: A. Transforming (Ia, 0, 0) gives the same current Ia/3 in all three sequences, which suggests a series connection. (Key point: Equal sequence currents) 2. With fault impedance Zf, the sequence voltages at the fault satisfy: A. Va0 = Va1 = Va2 B. Va0 + Va1 + Va2 = 0 always C. Va1 = Va2 and Va0 = 0 D. Va0 + Va1 + Va2 = 3 Zf Ia1 Answer: D. Va = Zf Ia = 3 Zf Ia1 and Va = Va0 + Va1 + Va2. (Key point: Voltage condition) 3. Which expression is the L-G fault current? A. Ea/(Z1 + Z2 + Z0) B. 3Ea/(Z1 + Z2 + Z0 + 3Zf) C. Ea/(Z1 + Z2 + Z0 + 3Zf) D. 3Ea/(Z1 + Z2 + Z0 + Zf) Answer: B. Ia1 = Ea/(Z1+Z2+Z0+3Zf) and If = Ia = 3Ia1. (Key point: Fault current) 4. What does Z0 contain for a generator grounded through Zn? A. Zg0 only B. Zg0 + Zn C. Zg0 + 3Zn D. 3Zg0 + Zn Answer: C. Z0 = Zg0 + 3Zn since neutral current is 3Ia0. (Key point: Zero sequence impedance) ## Related in Power System Analysis II - [L-G fault of unloaded generator: equal currents](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-unloaded-generator-derivation) — Power System Analysis II, chapter 5 - [SLG fault on bus with three alternators](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-three-alternators-common-bus) — Power System Analysis II, chapter 5 - [L-G fault and 3-phase fault, 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-30mva-voltages-3phase) — Power System Analysis II, chapter 5 - [SLG fault at bus 3, currents and voltages](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-fault-bus3-three-generators) — Power System Analysis II, chapter 5 - [SLG fault on 20 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-fault-20mva-generator) — Power System Analysis II, chapter 5 - [DLG fault sequence network connection](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Adlg-sequence-network-connection) — Power System Analysis II, chapter 5, asked 3× ## Related in other subjects - [Single-phase cycloconverter with mathematical expression](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asingle-phase-cyclo-math) — Power Electronics, chapter 6 - [Delta connected coils line current and power](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Adelta-coils-7ohm-0-03h-400v) — Basic Electrical Engineering, chapter 8 - [Temporary overvoltage for SLG fault at line middle](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Atov-slg-fault-middle) — High Voltage Engineering, chapter 3 - [Fault current: 50 MW and 40 MW generators](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-50mw-40mw-equal-current) — Power Plant Equipment, chapter 4