--- title: "Neutral reactance and L-G fault severity" subject: "Power System Analysis II" chapter: 5 key: "semester_5/power_system_analysis_ii:generator-neutral-reactance-lg-severity" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/generator-neutral-reactance-lg-severity site: BCT Engineering --- # Neutral reactance and L-G fault severity Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems - Asked in: 2070 Asar, 5+3+3+3 marks; 2070 Chaitra, 5+3+3+3 marks (asked 2 times) - Video: [3- Ph. Fault V/S 1- Ph. to Earth Fault ||Which One is Severe || Fault Current in Transmission System](https://www.youtube.com/watch?v=RbpG77PcbKU) — RELAY PROTECTION ELECTRICAL - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/generator-neutral-reactance-lg-severity - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5 ## Question A three phase synchronous generator whose neutral is grounded through a reactance Xn has balanced emfs and sequence reactance as X1, X2 and X0 such that X1 = X2 >> X0. i) Derive the expression for fault current for solid line to ground fault on phase a. ii) Draw the sequence networks and their interconnection for above fault. iii) Show that if neutral is solidly grounded, L-G fault is more severe than 3-phase fault at the terminal of generator. iv) Find the limiting value of neutral grounding reactance so that fault current in L-G and 3 phase faults are equal. ## Short answer For a solid L-G fault, the fault current is I_f = 3E_a / [j(X1 + X2 + X0 + 3X_n)]. With solid grounding (X_n=0) and X0 << X1, I_LG = 3E_a/(2X1+X0) exceeds the three-phase current E_a/X1, so L-G faults are more severe. The limiting neutral reactance to equalize L-G and three-phase fault currents is X_n = (X1 - X0)/3. ## Answer For an L-G fault on a generator grounded through $X_n$, $I_f = 3E_a/[j(X_1 + X_2 + X_0 + 3X_n)]$; with solid grounding and $X_0 < X_1$ this exceeds the 3-phase current $E_a/X_1$, and the two are equal when $X_n = (X_1 - X_0)/3$. #### i) Fault current for solid L-G fault on phase a Generator sequence equations (neutral reactance appears as $3X_n$ in the zero sequence network): $$ V_{a1} = E_a - jX_1I_{a1}, \quad V_{a2} = -jX_2I_{a2}, \quad V_{a0} = -j(X_0 + 3X_n)I_{a0} $$ Boundary conditions: $I_b = I_c = 0$, $V_a = 0$. From $I_s = A^{-1}I_p$: $I_{a0} = I_{a1} = I_{a2} = I_a/3$. From $V_a = V_{a0} + V_{a1} + V_{a2} = 0$: $$ E_a - jI_{a1}(X_1 + X_2 + X_0 + 3X_n) = 0 \;\Rightarrow\; I_{a1} = \frac{E_a}{j(X_1 + X_2 + X_0 + 3X_n)} $$ $$ I_f = I_a = 3I_{a1} = \frac{3E_a}{j(X_1 + X_2 + X_0 + 3X_n)} $$ #### ii) Sequence networks and interconnection ```text Positive Negative Zero F1 o F2 o F0 o |jX1 |jX2 |jX0 (Ea) | |j3Xn | | | N1 o N2 o N0 o L-G fault: networks in series Ia1--> +-(Ea)-jX1--F1--N2--jX2--F2--N0--j3Xn--jX0--F0-+ | | N1 --------------------------------------------+ ``` (F1 joined to N2, F2 to N0, F0 back to N1: a single series loop carrying $I_{a1}$.) #### iii) Solidly grounded: L-G more severe than 3-phase fault With $X_n = 0$ and $X_1 = X_2$: $$ |I_{LG}| = \frac{3E_a}{2X_1 + X_0}, \qquad |I_{3\phi}| = \frac{E_a}{X_1} = \frac{3E_a}{3X_1} $$ Since $X_0 \ll X_1$, $2X_1 + X_0 < 3X_1$, so $$ |I_{LG}| > |I_{3\phi}| $$ For example, $X_1 = X_2 = 0.2$, $X_0 = 0.05$ pu: $I_{LG} = 3/0.45 = 6.67$ pu while $I_{3\phi} = 1/0.2 = 5$ pu. Hence an L-G fault at the terminals of a solidly grounded generator is more severe, and the winding bracing must be designed for it, or the neutral grounded through a reactor. #### iv) Limiting neutral reactance For equal fault currents: $$ \frac{3E_a}{2X_1 + X_0 + 3X_n} = \frac{E_a}{X_1} \;\Rightarrow\; 3X_1 = 2X_1 + X_0 + 3X_n $$ $$ X_n = \frac{X_1 - X_0}{3} $$ With $X_n$ larger than this value, the L-G fault current is less than the three-phase fault current. In the example above, $X_n = (0.2 - 0.05)/3 = 0.05$ pu. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the L-G fault current on phase a for a generator grounded through Xn? A. 3Ea/j(X1 + X2 + X0 + 3Xn) B. Ea/j(X1 + X2 + X0 + Xn) C. 3Ea/j(X1 + X2 + X0 + Xn) D. Ea/jX1 Answer: A. Ia1 = Ea/j(X1+X2+X0+3Xn) and If = 3Ia1. (Key point: L-G fault current with Xn) 2. How are the sequence networks connected for a solid L-G fault? A. In parallel B. Positive in series with negative and zero in parallel C. In series D. Only the positive network is used Answer: C. Equal sequence currents mean a single series loop. (Key point: Network connection) 3. With Xn = 0 and X1 = X2, why is the L-G fault more severe than the 3-phase fault? A. X0 is larger than X1 B. 3E/(2X1 + X0) > E/X1 since X0 < X1 makes 2X1 + X0 < 3X1 C. The EMF is larger D. Xn is added Answer: B. The denominator 2X1 + X0 is less than 3X1 when X0 < X1, so the L-G current is bigger. (Key point: Severity comparison) 4. For equal L-G and 3-phase currents, the neutral reactance is: A. Xn = X1 + X0 B. Xn = X0/3 C. Xn = X1 - X0 D. Xn = (X1 - X0)/3 Answer: D. Setting 3E/(2X1+X0+3Xn) = E/X1 gives 3Xn = X1 - X0. 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